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== 5.2#h(0.6em)Logarithmic Functions


    
      
=== Logarithms and Exponents


      Before we look at logarithmic functions, let's quickly review exponents and logs. For a particular base, let's say 5, taking a logarithm is the opposite operation for raising to a power. For example, if we raise base 5 to a power of #math.equation(block: false, alt: "2")[$2$], we get


      #math.equation(block: true, alt: "5 squared equals 25 and thus log sub 5 ⁡ open parenthesis 25 close parenthesis equals 2")[$5^(2) = 25 " " " " " " " " "and thus" " " " " " " " " log_(5) ( 25 ) = 2$]
      We can say that a logarithm is actually an exponent. Asking for the log base 5 of 25 is asking "what power of 5, or what exponent on base 5 will give me 25?"


      #examplebox("Example 1")[][
Write each logarithmic equation as an equivalent exponential equation.


          + #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 81 close parenthesis equals 4")[$log_(3) ( 81 ) = 4$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 32 close parenthesis equals 5")[$log_(b) ( 32 ) = 5$]


        
        #solutionbox[
+ The logarithm asks "To what power must I raise 3 to get 81?" The base is 3 and the logarithm (or exponent) is 4, so #math.equation(block: false, alt: "3 to the power 4 equals 81")[$3^(4) = 81$].
+ The logarithm asks "To what power must I raise #math.equation(block: false, alt: "b")[$b$] to get 32?" The base is #math.equation(block: false, alt: "b")[$b$] and the logarithm (or exponent) is 5, so #math.equation(block: false, alt: "b to the power 5 equals 32")[$b^(5) = 32$].
]
]


      For a more thorough review of logarithms  you can refer to Section 4.3.1.


      
        Which of these could you use to  estimate the value of #math.equation(block: false, alt: "log sub 5 ⁡ 378")[$log_(5) 378$] ?


        \_\_\_\_\_


        #solutionbox[
step1
]


      
      
        Which of these could you use to estimate the value of #math.equation(block: false, alt: "log sub 5 ⁡ open parenthesis 378 close parenthesis")[$log_(5) ( 378 )$] ?


        + Find multiples of 5.
+ Find the fifth root of 378.
+ Find powers of 5.
+ Divide 378 by 5.


      
      Now we'll consider functions defined in terms of logarithms, or #strong[logarithmic functions]. For example,


      #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis")[$f ( x ) = log_(2) ( x )$]
      is a logarithmic function. In order to understand logarithmic functions better, we first investigate how they are related to more familiar functions, the exponential functions.


    
    
      
=== Inverse of the Exponential Function


      Inverse functions are really a generalization of inverse operations. For example, raising to the #math.equation(block: false, alt: "n")[$n$]th power and taking #math.equation(block: false, alt: "n")[$n$]th roots are inverse operations. In fact, we use the following rule to define cube roots:


      #math.equation(block: true, alt: "the cube root of b equals a if and only if a cubed equals b")[$root(3, b) = a " " " " " " " " " if and only if " " " " " " " " " a^(3) = b$]
      Compare this rule to the definition of inverse functions from Inverse Functions.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Inverse Functions]


        Suppose #math.equation(block: false, alt: "g")[$g$] is the #strong[inverse function] for #math.equation(block: false, alt: "f")[$f$]. Then


        #math.equation(block: true, alt: "g open parenthesis b close parenthesis equals a if and only if f open parenthesis a close parenthesis equals b")[$g ( b ) = a " " " " " " " " " " " if and only if " " " " " " " " " " " f ( a ) = b$]
]


      In this case, #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the cube root of x")[$" " g ( x ) = root(3, x) " "$] and #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed")[$" " f ( x ) = x^(3) " "$], and the equations above tell us that the two functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed")[$" " f ( x ) = x^(3) " "$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the cube root of x")[$" " g ( x ) = root(3, x) " "$] are inverse functions.


      In Exponential Functions, we saw that a similar rule relates the operations of raising a base #math.equation(block: false, alt: "b")[$b$] to a power and taking a base #math.equation(block: false, alt: "b")[$b$] logarithm, because they are inverse operations.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Conversion Formulas for Logarithms]


        For any base #math.equation(block: false, alt: "b greater than 0 , b not equal to 1")[$b > 0 , b ≠ 1$],


        #math.equation(block: true, alt: "log sub b ⁡ open parenthesis x close parenthesis equals y if and only if b to the power y equals x")[$bold(log)_(bold(italic(b))) ( bold(italic(x)) ) = bold(italic(y)) " " " " " " " " " if and only if " " " " " " " " " bold(italic(b))^(bold(italic(y))) = bold(italic(x))$]
]


      We can now define the #strong[logarithmic function], #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub b ⁡ open parenthesis x close parenthesis")[$" " g ( x ) = log_(b) ( x )$], that takes the log base #math.equation(block: false, alt: "b")[$b$] of its input values. The conversion formulas tell us that the log function, #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub b ⁡ open parenthesis x close parenthesis")[$" " g ( x ) = log_(b) ( x )$], is the inverse of the exponential function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals b to the power x")[$" " f ( x ) = b^(x)$].


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Logarithmic Function]


        The #strong[logarithmic function] base #math.equation(block: false, alt: "b")[$b$],


        #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals log sub b ⁡ open parenthesis x close parenthesis")[$" " g ( x ) = log_(b) ( x ) " "$]
        , 
    is the inverse of the exponential function of the same base, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals b to the power x")[$f ( x ) = b^(x)$].
]


    
    
      
=== Graphs of Logarithmic Functions


      What does the graph of a log function look like? We can use exponential functions to help us.


      We can obtain a table of values for #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis")[$" " g ( x ) = log_(2) ( x ) " "$] by making a table for #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$" " f ( x ) = 2^(x) " "$] and then interchanging the columns, as shown in the tables below. You can see that the graphs of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$" " f ( x ) = 2^(x) " "$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis")[$" " g ( x ) = log_(2) ( x )$], shown in the figure, are symmetric about the line #math.equation(block: false, alt: "y equals x")[$y = x$].


      #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$f ( x ) = 2^(x)$]]),
  [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "the fraction 1 over 4")[$display(frac(1, 4))$]],
  [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]],
  [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]],
  [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]],
  [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]],
))


      #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis")[$g ( x ) = log_(2) ( x )$]]),
  [#math.equation(block: false, alt: "the fraction 1 over 4")[$display(frac(1, 4))$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]],
  [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]],
  [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]],
  [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "1")[$1$]],
  [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "2")[$2$]],
))


      #figure(figph[2^x and log\_2(x) on the same grid], alt: "2^x and log_2(x) on the same grid", caption: none)


      The same procedure works for graphing log functions with any base: If we want to find values for the function #math.equation(block: false, alt: "y equals log sub b ⁡ open parenthesis x close parenthesis")[$" " y = log_(b) ( x )$], we can find the values for the exponential function #math.equation(block: false, alt: "y equals b to the power x")[$" " y = b^(x)$], and then interchange the #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] values in each ordered pair.


      #examplebox("Example 2")[][
Graph the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 10 to the power x")[$" " f ( x ) = 10^(x) " "$] and its inverse #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis")[$" " g ( x ) = log_(10) ( x ) " "$] on the same axes.


        
        #solutionbox[
We start by making a table of values for the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 10 to the power x")[$f ( x ) = 10^(x)$]. We can make a table of values for the inverse function, #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis")[$" " g ( x ) = log_(10) ( x ) " "$], by interchanging the components of each ordered pair in the table for #math.equation(block: false, alt: "f")[$f$].


          #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]]),
  [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "0.01")[$0.01$]],
  [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0.1")[$0.1$]],
  [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]],
  [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "10")[$10$]],
  [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "100")[$100$]],
))


          #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$]]),
  [#math.equation(block: false, alt: "0.01")[$0.01$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]],
  [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]],
  [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]],
  [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "1")[$1$]],
  [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "2")[$2$]],
))


          #figure(figph[graph of common log and exponential], alt: "graph of common log and exponential", caption: none)


          We plot each set of points and connect them with smooth curves to obtain the graphs shown above.
]
]


      
        Make a table of values and graph the function #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals log sub 4 ⁡ x")[$h ( x ) = log_(4) x$].


        #figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction 1 over 4")[$frac(1, 4)$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "16")[$16$]]),
  [#math.equation(block: false, alt: "log sub 4 ⁡ x")[$log_(4) x$]], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_],
))


        #solutionbox[
#figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction 1 over 4")[$frac(1, 4)$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "16")[$16$]]),
  [#math.equation(block: false, alt: "log sub 4 ⁡ x")[$log_(4) x$]], [#math.equation(block: false, alt: "a")[$"a"$]], [#math.equation(block: false, alt: "b")[$"b"$]], [#math.equation(block: false, alt: "c")[$"c"$]], [#math.equation(block: false, alt: "d")[$"d"$]], [#math.equation(block: false, alt: "e")[$"e"$]],
))


          A graph is below.
]


        #figure(figph[log], alt: "log", caption: none)


      
      
        Complete the table of values and graph the function #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals log sub 4 ⁡ open parenthesis x close parenthesis")[$" " h ( x ) = log_(4) ( x )$].


        #figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction 1 over 4")[$" " display(frac(1, 4)) " "$]], [#math.equation(block: false, alt: "1")[$" " 1 " "$]], [#math.equation(block: false, alt: "2")[$" " 2 " "$]], [#math.equation(block: false, alt: "4")[$" " 4 " "$]], [#math.equation(block: false, alt: "16")[$" " 16 " "$]]),
  [#math.equation(block: false, alt: "log sub 4 ⁡ open parenthesis x close parenthesis")[$log_(4) ( x )$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$],
))


        #solutionbox[
#figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction 1 over 4")[$" " display(frac(1, 4)) " "$]], [#math.equation(block: false, alt: "1")[$" " 1 " "$]], [#math.equation(block: false, alt: "2")[$" " 2 " "$]], [#math.equation(block: false, alt: "4")[$" " 4 " "$]], [#math.equation(block: false, alt: "16")[$" " 16 " "$]]),
  [#math.equation(block: false, alt: "log sub 4 ⁡ open parenthesis x close parenthesis")[$log_(4) ( x )$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]],
))


          #figure(figph[graph of log base 4], alt: "graph of log base 4", caption: none)
]


      
      
        What is the #math.equation(block: false, alt: "y")[$y$]-intercept of the graph of #math.equation(block: false, alt: "y equals log sub 5 ⁡ x")[$y = log_(5) x$] ?


        \_\_\_\_\_


        #solutionbox[
There is none: the graph of #math.equation(block: false, alt: "y equals log sub 5 ⁡ open parenthesis x close parenthesis")[$y = log_(5) ( x )$] has no #math.equation(block: false, alt: "y")[$y$]-intercept.
]


      
      
        What is the #math.equation(block: false, alt: "y")[$y$]-intercept of the graph of #math.equation(block: false, alt: "y equals log sub 5 ⁡ open parenthesis x close parenthesis")[$" " y = log_(5) ( x )$] ?


        + #math.equation(block: false, alt: "open parenthesis 0 , 1 close parenthesis")[$( 0 , 1 )$]
+ #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]
+ #math.equation(block: false, alt: "open parenthesis 0 , 5 close parenthesis")[$( 0 , 5 )$]
+ There is none.


      
      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Did you notice that the graphs of the log functions do not have any points with negative #math.equation(block: false, alt: "x")[$x$]-coordinates? This is because an exponential function has no negative (or zero) output values, so a log function has no negative or zero input values.


        For example, suppose that #math.equation(block: false, alt: "log sub 5 ⁡ open parenthesis minus 25 close parenthesis equals x")[$" " log_(5) ( − 25 ) = x " "$]. Then #math.equation(block: false, alt: "5 to the power x equals minus 25")[$" " 5^(x) = − 25 " "$]. But this is impossible, because #math.equation(block: false, alt: "5 to the power x")[$5^(x)$] cannot be negative. Thus, the log of a negative number (or zero) is undefined. In other words,


        #math.equation(block: true, alt: "We cannot take the log of a negative number or zero.")[$"We cannot take the log of a negative number or zero."$]
]


      You can also see that while an exponential growth function increases very rapidly for positive input values, its inverse, the logarithmic function, grows extremely slowly.


      In addition, the logarithmic function #math.equation(block: false, alt: "y equals log sub b ⁡ open parenthesis x close parenthesis")[$" " y = log_(b) ( x ) " "$] has the following properties.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Logarithmic Functions #math.equation(block: false, alt: "y equals log sub b ⁡ open parenthesis x close parenthesis")[$" " y = log_(b) ( x )$]]


        + Domain: all positive real numbers
+ Range: all real numbers
+ #math.equation(block: false, alt: "x")[$x$]-intercept: #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]
+ #math.equation(block: false, alt: "y")[$y$]-intercept: none
+ Vertical asymptote at #math.equation(block: false, alt: "x equals 0")[$x = 0$]
+ The graphs of #math.equation(block: false, alt: "y equals log sub b ⁡ open parenthesis x close parenthesis")[$" " y = log_(b) ( x ) " "$] and #math.equation(block: false, alt: "y equals b to the power x")[$" " y = b^(x) " "$] are symmetric about the line #math.equation(block: false, alt: "y equals x")[$y = x$].
]


      
        The domain of the function #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub 3 ⁡ open parenthesis x close parenthesis")[$g ( x ) = log_(3) ( x )$] is


        \_\_\_\_\_


        #solutionbox[
The domain of the function #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub 3 ⁡ open parenthesis x close parenthesis")[$g ( x ) = log_(3) ( x )$] is all positive numbers.
]


      
      
        The domain of the function #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub 3 ⁡ open parenthesis x close parenthesis")[$" " g ( x ) = log_(3) ( x )$] is


        + all real numbers.
+ all multiples of 3.
+ all non-negative numbers.
+ all positive numbers.


      
      
        Why does the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log ⁡ open parenthesis x close parenthesis")[$f ( x ) = log ( x )$] grow so slowly?


        \_\_\_\_\_


      
      
        Why does the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log ⁡ open parenthesis x close parenthesis")[$" " f ( x ) = log ( x ) " "$] grow so slowly?


      
    
    
      
=== Modeling with Logarithmic Functions


      We can use the LOG key on a calculator to evaluate the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis")[$" " f ( x ) = log_(10) ( x )$].


      #examplebox("Example 3")[][
Let #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis")[$" " f ( x ) = log_(10) ( x )$]. Evaluate the following expressions.


          + #math.equation(block: false, alt: "f open parenthesis 35 close parenthesis")[$f ( 35 )$]
+ #math.equation(block: false, alt: "f open parenthesis minus 8 close parenthesis")[$f ( − 8 )$]
+ #math.equation(block: false, alt: "2 f open parenthesis 16 close parenthesis plus 1")[$2 f ( 16 ) + 1$]


        
        #solutionbox[
+ #math.equation(block: false, alt: "f open parenthesis 35 close parenthesis equals log sub 10 ⁡ open parenthesis 35 close parenthesis approximately equals 1.544")[$f ( 35 ) = log_(10) ( 35 ) ≈ 1.544$]
+ Because #math.equation(block: false, alt: "minus 8")[$− 8$] is not in the domain of #math.equation(block: false, alt: "f")[$f$], #math.equation(block: false, alt: "f open parenthesis minus 8 close parenthesis")[$f ( − 8 )$], or #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis minus 8 close parenthesis")[$log_(10) ( − 8 )$], is undefined.
+ #math.equation(block: false, alt: "2 f open parenthesis 16 close parenthesis plus 1 equals 2 open parenthesis log sub 10 ⁡ open parenthesis 16 close parenthesis close parenthesis plus 1 approximately equals 2 open parenthesis 1.204 close parenthesis plus 1 equals 3.408")[$2 f ( 16 ) + 1 = 2 ( log_(10) ( 16 ) ) + 1 ≈ 2 ( 1.204 ) + 1 = 3.408$]
]
]


      
        Which statement is true?


        \_\_\_\_\_


        #solutionbox[
We cannot take the log of a negative number.
]


      
      
        Which statement is true?


        + The log of a number is never negative.
+ We cannot take the log of a negative number.
+ The log of a fraction is called a common log.
+ The log of 0 is 1.


      
      
        The formula #math.equation(block: false, alt: "T equals the fraction log ⁡ open parenthesis 2 close parenthesis times t sub i over 3 log ⁡ open parenthesis D sub f / D sub 0 close parenthesis")[$T = display(frac(log ( 2 ) ⋅ t_(i), 3 log ( D_(f) / D_(0) )))$] is used by X-ray technicians to calculate the doubling time of a malignant tumor. #math.equation(block: false, alt: "D sub 0")[$D_(0)$] is the diameter of the tumor when first detected, #math.equation(block: false, alt: "D sub f")[$D_(f)$] is its diameter at the next reading, and #math.equation(block: false, alt: "t sub i")[$t_(i)$] is the time interval between readings, in days. Calculate the doubling time of the following tumor: its diameter when first detected was 1 cm, and 7 days later its diameter was 1.05 cm.


        \_\_\_\_\_ days


        #solutionbox[
33 days
]


      
      
        The formula


        #math.equation(block: true, alt: "T equals the fraction log ⁡ open parenthesis 2 close parenthesis times t sub i over 3 log ⁡ open parenthesis D sub f / D sub 0 close parenthesis")[$T = display(frac(log ( 2 ) ⋅ t_(i), 3 log ( D_(f) / D_(0) )))$]
        is used by X-ray technicians to calculate the doubling time of a malignant tumor. #math.equation(block: false, alt: "D sub 0")[$D_(0)$] is the diameter of the tumor when first detected, #math.equation(block: false, alt: "D sub f")[$D_(f)$] is its diameter at the next reading, and #math.equation(block: false, alt: "t sub i")[$t_(i)$] is the time interval between readings, in days.


        Calculate the doubling time of the following tumor: its diameter when first detected was 1 cm, and 7 days later its diameter was 1.05 cm.


        #solutionbox[
#math.equation(block: false, alt: "T equals the fraction log ⁡ open parenthesis 2 close parenthesis times 7 over 3 log ⁡ open parenthesis 1.05 / 1 close parenthesis equals")[$T = display(frac(log ( 2 ) ⋅ 7, 3 log ( 1.05 / 1 ))) =$] 33 days
]


      
      Logarithmic functions are useful for modeling increasing functions  that slow down as the input increases.


      #examplebox("Example 4")[][
Life expectancy at birth is the average number of years a newborn child is expected to live. In 1900, the average life expectancy at birth in the U.S. was 47.3 years, and in 1910 it had risen to 50.0 years. During rest of the twentieth century, life expectancy was modeled by the formula


          #math.equation(block: true, alt: "L open parenthesis x close parenthesis equals 19.13 plus 28.34 log ⁡ open parenthesis x close parenthesis")[$L ( x ) = 19.13 + 28.34 log ( x )$]
          where #math.equation(block: false, alt: "x")[$x$] is the number of years after 1900.


          + Graph the life expectancy function for the years 1910 to 2000.
+ The life expectancy in 1950 was 68.2 years. What does the function #math.equation(block: false, alt: "L open parenthesis x close parenthesis")[$L ( x )$] predict for life expectancy in 1950?
+ According to the model, how much did life expectancy increase between 1920 and 1930? How much did it increase between 1990 and 2000?


        
        #solutionbox[
+ We can make a table of values for #math.equation(block: false, alt: "L open parenthesis x close parenthesis")[$L ( x )$] and plot points to obtain the graph below, which also shows the actual data points for life expectancy for the decades from 1910 to 2000.#figure(table(
  columns: 11, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "70")[$70$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "90")[$90$]], [#math.equation(block: false, alt: "100")[$100$]]),
  [#math.equation(block: false, alt: "L open parenthesis x close parenthesis")[$L ( x )$]], [#math.equation(block: false, alt: "47.5")[$47.5$]], [#math.equation(block: false, alt: "56.1")[$56.1$]], [#math.equation(block: false, alt: "61.1")[$61.1$]], [#math.equation(block: false, alt: "64.6")[$64.6$]], [#math.equation(block: false, alt: "67.4")[$67.4$]], [#math.equation(block: false, alt: "69.6")[$69.6$]], [#math.equation(block: false, alt: "71.5")[$71.5$]], [#math.equation(block: false, alt: "73.1")[$73.1$]], [#math.equation(block: false, alt: "74.6")[$74.6$]], [#math.equation(block: false, alt: "75.9")[$75.9$]],
))

#figure(figph[average life expectancy], alt: "average life expectancy", caption: none)
+ We substitute #math.equation(block: false, alt: "x equals 50")[$x = 50$] into the function to find
        #math.equation(block: true, alt: "L open parenthesis 50 close parenthesis equals 19.13 plus 28.34 log ⁡ open parenthesis 50 close parenthesis approximately equals 67.4")[$L ( 50 ) = 19.13 + 28.34 log ( 50 ) ≈ 67.4$]
    The function predicts a life expectancy of 67.4 years in 1950.
+ Between 1920 and 1930, life expectancy increased from 56.1 to 61.1, or 5 years. Between 1990 and 2000 it increased from 74.6 to 75.9, or 1.3 years.
]
]


      In the previous Example, we see that although life expectancy has been increasing over time, it has been slowing down or leveling off. In fact, life expectancy in the US actually declined slightly from 78.94 in 2013 to 78.81 in 2018. (What factors may have contributed to this decline?) By 2024 it had rebounded to 79.25. It remains to be seen how well the model predicts life expectancy in the 21st century.


      
        The CDC (Centers for Disease Control and Prevention) provides Growth Charts for the average height and weight of children from age 2 to 20. The average height of girl children is given in centimeters by


        #math.equation(block: true, alt: "H open parenthesis t close parenthesis equals 49.29 plus 91.3 log ⁡ t")[$H ( t ) = 49.29 + 91.3 log t$]
        where #math.equation(block: false, alt: "t")[$t$] is age in years.


        + Graph the height function for #math.equation(block: false, alt: "2 less than or equal to t less than or equal to 20")[$2 ≤ t ≤ 20$].
+ Use the height function to complete the table.#figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]]),
  [#math.equation(block: false, alt: "H open parenthesis t close parenthesis")[$H ( t )$]], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_],
))
+ How much is a girl's height expected to increase between the ages of 5 and 10? \_\_\_\_\_ cm.  #linebreak() Between the ages of 15 and 20?  \_\_\_\_\_ cm.


        #solutionbox[
+ A graph is below.
+ #figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]]),
  [#math.equation(block: false, alt: "H open parenthesis t close parenthesis")[$H ( t )$]], [H0], [H1], [H2], [H3], [H4],
))
+ c1 cm, c2 cm
]


        Graph for part (a):


        #figure(figph[curve], alt: "curve", caption: none)


      
      
        The CDC (Centers for Disease Control and Prevention) provides Growth Charts for the average height and weight of children from age 2 to 20. The average height of girl children is given in centimeters by


        #math.equation(block: true, alt: "H open parenthesis t close parenthesis equals 49.29 plus 91.3 log ⁡ open parenthesis t close parenthesis")[$H ( t ) = 49.29 + 91.3 log ( t )$]
        where #math.equation(block: false, alt: "t")[$t$] is age in years.


        + Graph the height function for #math.equation(block: false, alt: "2 less than or equal to t less than or equal to 20")[$2 ≤ t ≤ 20$].
+ Use the height function to complete the table.#figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]]),
  [#math.equation(block: false, alt: "H open parenthesis t close parenthesis")[$H ( t )$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$],
))
+ How much is a girl's height expected to increase between the ages of 5 and 10? Between the ages of 15 and 20?


        
          + #figure(figph[logrithmic curve], alt: "logrithmic curve", caption: none)
+ #figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]]),
  [#math.equation(block: false, alt: "H open parenthesis t close parenthesis")[$H ( t )$]], [#math.equation(block: false, alt: "77")[$77$]], [#math.equation(block: false, alt: "113")[$113$]], [#math.equation(block: false, alt: "141")[$141$]], [#math.equation(block: false, alt: "157")[$157$]], [#math.equation(block: false, alt: "168")[$168$]],
))
+ 28 cm, 11 cm


        
      
    
    
      
=== Logarithmic Equations


      A #strong[logarithmic equation] is one in which the variable appears inside of a logarithm. For example,


      #math.equation(block: true, alt: "log sub 4 ⁡ open parenthesis x close parenthesis equals 3")[$log_(4) ( x ) = 3$]
      is a log equation. To solve a log equation, we can use the conversion equations to rewrite the equation in exponential form.


      #examplebox("Example 5")[][
Solve for #math.equation(block: false, alt: "x")[$x$].


          + #math.equation(block: false, alt: "2 open parenthesis log sub 3 ⁡ open parenthesis x close parenthesis close parenthesis minus 1 equals 4")[$2 ( log_(3) ( x ) ) − 1 = 4$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 2 x plus 100 close parenthesis equals 3")[$log_(10) ( 2 x + 100 ) = 3$]


        
        #solutionbox[
+ We isolate the logarithm, then rewrite the equation in exponential form:
            #math.equation(block: true, alt: "2 open parenthesis log sub 3 ⁡ open parenthesis x close parenthesis close parenthesis, equals 5, Divide both sides by 5.; log sub 3 ⁡ open parenthesis x close parenthesis, equals the fraction 5 over 2, Convert to exponential form.; x, equals 3 to the power 5 / 2")[$2 ( log_(3) ( x ) ) & = 5 &  & "Divide both sides by 5." \ log_(3) ( x ) & = frac(5, 2) &  & "Convert to exponential form." \ x & = 3^(5 / 2)$]
+ First, we convert the equation to exponential form.
            #math.equation(block: true, alt: "2 x plus 100 equals 10 cubed equals 1000")[$2 x + 100 = 10^(3) = 1000$]
    Now we can solve for #math.equation(block: false, alt: "x")[$x$] to find #math.equation(block: false, alt: "2 x equals 900")[$2 x = 900$], or #math.equation(block: false, alt: "x equals 450")[$x = 450$].
]
]


      
        Solve for the unknown value in each equation.


        + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 2 close parenthesis equals the fraction 1 over 2")[$log_(b) ( 2 ) = display(frac(1, 2))$] #linebreak() #math.equation(block: false, alt: "b equals")[$b =$]\_\_\_\_\_
+ #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 2 x minus 1 close parenthesis equals 4")[$log_(3) ( 2 x − 1 ) = 4$] #linebreak() #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_


        #solutionbox[
+ #math.equation(block: false, alt: "b equals 4")[$b = 4$]
+ #math.equation(block: false, alt: "x equals 41")[$x = 41$]
]


      
      
        Solve for the unknown value in each equation.


        + #math.equation(block: false, alt: "log sub b ⁡ open parenthesis 2 close parenthesis equals the fraction 1 over 2")[$log_(b) ( 2 ) = display(frac(1, 2))$]
+ #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 2 x minus 1 close parenthesis equals 4")[$log_(3) ( 2 x − 1 ) = 4$]


        #solutionbox[
+ #math.equation(block: false, alt: "b equals 4")[$b = 4$]
+ #math.equation(block: false, alt: "x equals 41")[$x = 41$]
]


      
      #examplebox("Example 6")[][
If #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis")[$" " f ( x ) = log_(10) ( x )$], find #math.equation(block: false, alt: "x")[$x$] so that #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals minus 3.2")[$f ( x ) = − 3.2$].


        
        #solutionbox[
We must solve the equation #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x close parenthesis equals minus 3.2")[$" " log_(10) ( x ) = − 3.2$]. Rewriting the equation in exponential form yields


          #math.equation(block: true, alt: "x equals 10 to the power minus 3.2 approximately equals 0.00063")[$x = 10^(− 3.2) ≈ 0.00063$]
]
]


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Evaluating #math.equation(block: false, alt: "10 to the power x")[$10^(x)$]]


        In Example, the expression #math.equation(block: false, alt: "10 to the power minus 3.2")[$10^(− 3.2)$] can be evaluated in two different ways with a calculator. We can use the ^ key and press


        #math.equation(block: false, alt: "10")[$10$] ^ (-) #math.equation(block: false, alt: "3.2")[$3.2$] ENTER


        which gives 6.30957344 E -4, or approximately #math.equation(block: false, alt: "0.00063")[$0.00063$]. Alternatively, because #math.equation(block: false, alt: "10 to the power x")[$10^(x)$] is the inverse function for #math.equation(block: false, alt: "log ⁡ x")[$log x$], we can press


        2nd LOG (-) #math.equation(block: false, alt: "3.2")[$3.2$] ENTER


        which gives the same answer as before.
]


      
        Imagine the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis")[$f ( x ) = log_(10) ( x )$]. How far must you travel along the #math.equation(block: false, alt: "x")[$x$]-axis until the #math.equation(block: false, alt: "y")[$y$]-coordinate reaches a height of 5.25?


        Answer: Until #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_


        Do not enter commas, that is, enter "10000" rather than "10,000".


        #solutionbox[
#math.equation(block: true, alt: "x equals 177 , 827.941")[$x = 177 , 827.941$]
]


      
      
        Imagine the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis")[$" " f ( x ) = log_(10) ( x )$]. How far must you travel along the #math.equation(block: false, alt: "x")[$x$]-axis until the #math.equation(block: false, alt: "y")[$y$]-coordinate reaches a height of 5.25?


        #solutionbox[
#math.equation(block: true, alt: "x equals 177 , 827.941")[$x = 177 , 827.941$]
]


      
      If an equation contains more than one log, we must first combine any expressions involving logs into a single logarithm.


      #examplebox("Example 7")[][
Solve #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x plus 1 close parenthesis plus log sub 10 ⁡ open parenthesis x minus 2 close parenthesis equals 1")[$" " " " log_(10) ( x + 1 ) + log_(10) ( x − 2 ) = 1$].


        
        #solutionbox[
We use Property (1) of logarithms (see Properties of Logarithms) to rewrite the left-hand side as a single logarithm:


          #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis x plus 1 close parenthesis open parenthesis x minus 2 close parenthesis equals 1")[$log_(10) ( x + 1 ) ( x − 2 ) = 1$]
          Once the left-hand side is expressed as a single logarithm, we can rewrite the equation in exponential form as


          #math.equation(block: true, alt: "open parenthesis x plus 1 close parenthesis open parenthesis x minus 2 close parenthesis equals 10 to the power 1")[$( x + 1 ) ( x − 2 ) = 10^(1)$]
          Simplifying the right side gives us a quadratic equation to solve.


          #math.equation(block: true, alt: "x squared minus x minus 2, equals 10, Subtract 10 from both sides.; x squared minus x minus 12, equals 0, Factor the left side.; open parenthesis x minus 4 close parenthesis open parenthesis x plus 3 close parenthesis, equals 0, Apply the zero-factor principle.")[$x^(2) − x − 2 & = 10 &  & "Subtract 10 from both sides." \ x^(2) − x − 12 & = 0 &  & "Factor the left side." \ ( x − 4 ) ( x + 3 ) & = 0 &  & "Apply the zero-factor principle."$]
          We find #math.equation(block: false, alt: "x equals 4")[$x = 4$] or #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$].  But let us check both of these values in the original equation.


          For #math.equation(block: false, alt: "x equals 4")[$x = 4$] we have


          #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis 4 plus 1 close parenthesis open parenthesis 4 minus 2 close parenthesis equals log sub 10 ⁡ open parenthesis 5 times 2 close parenthesis equals log sub 10 ⁡ open parenthesis 10 close parenthesis equals 1")[$log_(10) ( 4 + 1 ) ( 4 − 2 ) = log_(10) ( 5 ⋅ 2 ) = log_(10) ( 10 ) = 1$]
          so #math.equation(block: false, alt: "x equals 4")[$x = 4$] is a solution.


          But #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$] is not a solution of the original equation, because neither #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x plus 1 close parenthesis")[$log_(10) ( x + 1 )$] nor #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x minus 2 close parenthesis")[$log_(10) ( x − 2 )$] is defined for #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$]. (Remember that we cannot take a logarithm of a negative number or zero.) We say that the apparent solution #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$] is #strong[extraneous], and the only solution of the original equation is #math.equation(block: false, alt: "4")[$4$].
]
]


      #strong[Extraneous solutions] can arise whenever we solve a logarithmic equation, especially if there is more than one apparent solution. Therefore, we should always check that a possible solution does not cause one of the logarithms to be undefined. Here are guidelines for solving a logarithmic equation.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Steps for Solving Logarithmic Equations]


        + Use the properties of logarithms to combine all logs into one log.
+ Isolate the log on one side of the equation.
+ Convert the equation to exponential form.
+ Solve for the variable.
+ Check for extraneous solutions.
]


      
        Which of these is the first step in solving the equation #math.equation(block: false, alt: "log ⁡ open parenthesis x close parenthesis plus log ⁡ open parenthesis x minus 1 close parenthesis equals 2")[$log ( x ) + log ( x − 1 ) = 2$] ?


        \_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "log ⁡ open parenthesis x open parenthesis x minus 1 close parenthesis close parenthesis equals 2")[$log ( x ( x − 1 ) ) = 2$]
]


      
      
        Which of these is the first step in solving the equation #math.equation(block: false, alt: "log ⁡ x plus log ⁡ open parenthesis x minus 1 close parenthesis equals 2")[$" " log x + log ( x − 1 ) = 2$] ?


        + #math.equation(block: false, alt: "log ⁡ open parenthesis 2 x minus 1 close parenthesis equals 2")[$log ( 2 x − 1 ) = 2$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis x close parenthesis equals 2 minus log ⁡ open parenthesis x minus 1 close parenthesis")[$log ( x ) = 2 − log ( x − 1 )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis x open parenthesis x minus 1 close parenthesis close parenthesis equals 2")[$log ( x ( x − 1 ) ) = 2$]
+ #math.equation(block: false, alt: "x plus x minus 1 equals 2 to the power x")[$x + x − 1 = 2^(x)$]


      
      
        Solve #math.equation(block: false, alt: "log sub 2 ⁡ x plus log sub 2 ⁡ open parenthesis x minus 2 close parenthesis equals 3")[$log_(2) x + log_(2) ( x − 2 ) = 3$].


        #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_  \[Separate multiple solutions with commas when appropriate.\]


        
          Rewrite the left side as a single logarithm.


          Rewrite the equation in exponential form.


          Solve for #math.equation(block: false, alt: "x")[$x$].


          Check for extraneous solutions.


        
        #solutionbox[
#math.equation(block: true, alt: "x equals 4")[$x = 4$]
]


      
      
        Solve #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis x close parenthesis plus log sub 2 ⁡ open parenthesis x minus 2 close parenthesis equals 3")[$" " log_(2) ( x ) + log_(2) ( x − 2 ) = 3$].


        #solutionbox[
Follow the steps:


          Rewrite the left side as a single logarithm.


          Rewrite the equation in exponential form.


          Solve for #math.equation(block: false, alt: "x")[$x$].


          Check for extraneous solutions.


          The solution is #math.equation(block: false, alt: "x equals 4")[$x = 4$].
]


      
      
        + We cannot take a logarithm of \_\_\_\_\_.
+ After solving a logarithmic equation, we must check for  \_\_\_\_\_.
+ If an equation contains more than one log, we must first combine them into  \_\_\_\_\_.
+ If there is only one log involved, we write the equation in  \_\_\_\_\_ form.


        #solutionbox[
+ a negative number or zero
+ extraneous solutions
+ a single logarithm
+ exponential form
]


      
      
        Fill in the blanks to complete each statement.


        + We cannot take a logarithm of \_\_\_\_\_\_.
+ After solving a logarithmic equation, we must check for \_\_\_\_\_\_.
+ If an equation contains more than one log, we must first combine them into \_\_\_\_\_\_.
+ If there is only one log involved, we rewrite the equation in \_\_\_\_\_\_ form.


      
      
        What is an extraneous solution?


        \_\_\_\_\_


      
      
        What is an extraneous solution?


      
    
    
      
=== More About Inverse Functions


      Let's take a closer look at the relationship between functions and their inverse functions. In Section 5.1 we saw that an inverse function "undoes" the effects of the function, and vice versa. In other words, if we apply a function and then its inverse to an input, we return to that input. For example, consider the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x cubed")[$" " f ( x ) = x^(3) " "$] and its inverse function #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the cube root of x")[$" " g ( x ) = root(3, x)$]. We'll start with an input of #math.equation(block: false, alt: "x equals 5")[$x = 5$], first apply the function #math.equation(block: false, alt: "f")[$f$], and then apply the function #math.equation(block: false, alt: "g")[$g$] to the output. In function notation, that operation looks like this.


      #math.equation(block: true, alt: "g open parenthesis f open parenthesis 5 close parenthesis close parenthesis equals g open parenthesis 5 cubed close parenthesis equals g open parenthesis 125 close parenthesis equals the cube root of 125 equals 5")[$g ( bold(italic(f)) ( 5 ) ) = g ( 5^(3) ) = g ( 125 ) = root(3, 125) = 5$]
      We start by applying the innermost function, namely #math.equation(block: false, alt: "f")[$f$], to get #math.equation(block: false, alt: "5 cubed equals 125")[$5^(3) = 125$], and then apply #math.equation(block: false, alt: "g")[$g$], to get #math.equation(block: false, alt: "the cube root of 125 equals 5")[$root(3, 125) = 5$]. We have returned to out original input.


      
        In Example of Section 5.1 we found that the inverse of the function #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals 6 plus 2 t")[$f ( t ) = 6 + 2 t$] is #math.equation(block: false, alt: "g open parenthesis t close parenthesis equals the fraction t minus 6 over 2.")[$g ( t ) = display(frac(t − 6, 2)) .$]


        + Show that #math.equation(block: false, alt: "g open parenthesis f open parenthesis 7 close parenthesis close parenthesis equals 7.")[$g ( f ( 7 ) ) = 7.$]
+ Show that #math.equation(block: false, alt: "f open parenthesis g open parenthesis 30 close parenthesis close parenthesis equals 30.")[$f ( g ( 30 ) ) = 30.$]


        #solutionbox[
In each case, start by evaluating the innermost function.
]


      
      These examples illustrate a general rule about inverse functions.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Functions and Inverse Functions]


        Suppose #math.equation(block: false, alt: "g")[$g$] is the inverse function for #math.equation(block: false, alt: "f")[$f$]. Then


        #math.equation(block: true, alt: "g open parenthesis f open parenthesis x close parenthesis close parenthesis equals x and f open parenthesis g open parenthesis y close parenthesis close parenthesis equals y")[$g ( f ( x ) ) = x " " " " " " " and " " " " " " " f ( g ( y ) ) = y$]
        as long as #math.equation(block: false, alt: "x")[$x$] is in the domain of #math.equation(block: false, alt: "f")[$f$], and #math.equation(block: false, alt: "y")[$y$] is in the domain of #math.equation(block: false, alt: "g")[$g$].
]


      Because a logarithmic function is the inverse of the exponential function with the same base, each undoes the effect of the other. For example, the function #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis")[$" " g ( x ) = log_(2) ( x ) " "$] is the inverse of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$" " f ( x ) = 2^(x)$]. So, if we start with #math.equation(block: false, alt: "x equals 3")[$x = 3$], apply #math.equation(block: false, alt: "f")[$f$], and then apply #math.equation(block: false, alt: "g")[$g$] to the result, we return to the original number, 3.


      #math.equation(block: true, alt: "x equals 3 → exponential function Apply the f open parenthesis 3 close parenthesis equals 2 cubed equals 8 → log function Apply the g open parenthesis 8 close parenthesis equals log sub 2 ⁡ 8 equals 3 number Original")[$x = 3 limits(→)^("Apply the")_("exponential function") f ( 3 ) = 2^(3) = 8 limits(→)^("Apply the")_("log function") g ( 8 ) = log_(2) 8 = limits(3)^("Original")_("number")$]
      And because #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals b to the power x")[$" " f ( x ) = b^(x) " "$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub b ⁡ open parenthesis x close parenthesis")[$" " g ( x ) = log_(b) ( x ) " "$] are inverse functions, we can write these operations in one expression as


      #math.equation(block: true, alt: "g open parenthesis f open parenthesis 3 close parenthesis close parenthesis equals log sub 2 ⁡ open parenthesis 2 cubed close parenthesis equals 3")[$g ( f ( 3 ) ) = log_(2) ( 2^(3) ) = 3$]
      We evaluate the expression starting with the inside function, #math.equation(block: false, alt: "f open parenthesis 3 close parenthesis equals 2 cubed equals 8")[$" " f ( 3 ) = 2^(3) = 8$], and then compute the log base 2 of the result. Applying the exponential function and then the log function with the same base returne us to the original input.


      .


      Because the log and the exponential are inverse functions, similar calculations hold for any value of #math.equation(block: false, alt: "x")[$x$] and any base #math.equation(block: false, alt: "b greater than 0")[$b > 0$], so that #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b to the power x close parenthesis equals x")[$" " log_(b) ( b^(x) ) = x$].


      #examplebox("Example 8")[][
Simplify each expression.


          + #math.equation(block: false, alt: "log sub 4 ⁡ open parenthesis 4 to the power 6 close parenthesis")[$log_(4) ( 4^(6) )$]
+ #math.equation(block: false, alt: "log sub 8 ⁡ open parenthesis 8 to the power 2 a plus 3 close parenthesis")[$log_(8) ( 8^(2 a + 3) )$]


        
        #solutionbox[
+ In this expression, we start with #math.equation(block: false, alt: "6")[$6$], apply the exponential function with base #math.equation(block: false, alt: "4")[$4$], and then take a logarithm base #math.equation(block: false, alt: "4")[$4$]. Because the logarithm is the inverse of the exponential function, we return to the original number, #math.equation(block: false, alt: "6")[$6$].
		#math.equation(block: true, alt: "x equals 6 → exponential function Apply the 4 to the power 6 → log function Apply the log sub 4 ⁡ open parenthesis 4 to the power 6 close parenthesis equals 6 number Original")[$x = 6 limits(→)^("Apply the")_("exponential function") 4^(6) limits(→)^("Apply the")_("log function") log_(4) ( 4^(6) ) = limits(6)^("Original")_("number")$]
+ The input of the exponential function is the expression #math.equation(block: false, alt: "2 a plus 3")[$2 a + 3$]. Because the bases of the log and the exponential function are both #math.equation(block: false, alt: "8")[$8$], they are inverse functions, and applying them in succession returns us to the original input. Thus, #math.equation(block: false, alt: "log sub 8 ⁡ open parenthesis 8 to the power 2 a plus 3 close parenthesis equals 2 a plus 3")[$log_(8) ( 8^(2 a + 3) ) = 2 a + 3$].
]
]


      
        Simplify each expression.


        + #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 10 to the power 6 close parenthesis equals")[$log_(10) ( 10^(6) ) =$]\_\_\_\_\_
+ #math.equation(block: false, alt: "log sub w ⁡ open parenthesis w to the power x plus 1 close parenthesis equals")[$log_(w) ( w^(x + 1) ) =$]\_\_\_\_\_ , for #math.equation(block: false, alt: "w greater than 0 , w not equal to 1")[$w > 0 , " " w ≠ 1$]


        #solutionbox[
+ #math.equation(block: false, alt: "6")[$6$]
+ #math.equation(block: false, alt: "w plus 1")[$w + 1$]
]


      
      
        Simplify each expression.


        + #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 10 to the power 6 close parenthesis")[$log_(10) ( 10^(6) )$]
+ #math.equation(block: false, alt: "log sub w ⁡ open parenthesis w to the power x plus 1 close parenthesis ,")[$log_(w) ( w^(x + 1) ) ,$] for #math.equation(block: false, alt: "w greater than 0 , w not equal to 1")[$w > 0 , " " w ≠ 1$]


        #solutionbox[
+ #math.equation(block: false, alt: "6")[$6$]
+ #math.equation(block: false, alt: "w plus 1")[$w + 1$]
]


      
      We can also apply the two functions in the opposite order. For example,


      #math.equation(block: true, alt: "2 to the power log sub 2 ⁡ open parenthesis 8 close parenthesis equals 8")[$2^(log_(2) ( 8 )) = 8$]
      To see that this equation is true, we simplify the exponent first. We start with #math.equation(block: false, alt: "8")[$8$], and apply the log base #math.equation(block: false, alt: "2")[$2$] function. Because #math.equation(block: false, alt: "log sub 2 ⁡ 8 equals 3")[$log_(2) 8 = 3$], we have


      #math.equation(block: true, alt: "8 → log function Apply the log sub 2 ⁡ 8 equals 3 → exponential function Apply the equals 2 to the power log sub 2 ⁡ 8 equals 2 cubed equals 8 number Original")[$8 limits(→)^("Apply the")_("log function") log_(2) 8 = 3 limits(→)^("Apply the")_("exponential function") = 2^(log_(2) 8) = 2^(3) = limits(8)^("Original")_("number")$]
      Using function notation, the caluclation above looks ike this.


      #math.equation(block: true, alt: "f open parenthesis g open parenthesis 8 close parenthesis close parenthesis equals 2 to the power log sub 2 ⁡ open parenthesis 8 close parenthesis equals 8")[$f ( g ( 8 ) ) = 2^(log_(2) ( 8 )) = 8$]
      (Remember the order of operations: do what's inside of parentheses first, to get #math.equation(block: false, alt: "g open parenthesis 8 close parenthesis equals log sub 2 ⁡ open parenthesis 8 close parenthesis")[$g ( 8 ) = log_(2) ( 8 )$].) In other words, applying first the log function and then the exponential function returns the original input value.


      Of course, a similar equation holds for any positive value of #math.equation(block: false, alt: "x")[$x$] and any base #math.equation(block: false, alt: "b greater than 0 , b not equal to 1")[$b > 0 , b ≠ 1$]:


      #math.equation(block: true, alt: "b to the power log sub b ⁡ x equals x")[$bold(italic(b))^(bold(log)_(bold(italic(b))) bold(italic(x))) = bold(italic(x))$]
      #examplebox("Example 9")[][
Simplify each expression.


          + #math.equation(block: false, alt: "10 to the power log sub 10 ⁡ open parenthesis 1000 close parenthesis")[$10^(log_(10) ( 1000 ))$]
+ #math.equation(block: false, alt: "Q to the power log sub Q ⁡ open parenthesis 25 close parenthesis , for Q greater than 0 , Q not equal to 1")[$Q^(log_(Q) ( 25 )) ", for " Q > 0 , Q ≠ 1$]


        
        #solutionbox[
+ Think of simplifying the exponent first:  #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 1000 close parenthesis equals 3")[$log_(10) ( 1000 ) = 3$]. Then
                #math.equation(block: true, alt: "10 to the power log sub 10 ⁡ open parenthesis 1000 close parenthesis equals 10 cubed equals 1000")[$10^(log_(10) ( 1000 )) = 10^(3) = 1000$]
    We first found the exponent on 10 that gives 1000, then we raised 10 to that exponent (namely 3), so we get 1000.
+ The expression #math.equation(block: false, alt: "log sub Q ⁡ 25")[$log_(Q) 25$] asks for the exponent on #math.equation(block: false, alt: "Q")[$Q$] that gives #math.equation(block: false, alt: "25")[$25$]. Let's call that exponent #math.equation(block: false, alt: "k")[$k$], so that #math.equation(block: false, alt: "log sub Q ⁡ open parenthesis 25 close parenthesis equals k")[$log_(Q) ( 25 ) = k$]. Then #math.equation(block: false, alt: "Q to the power log sub Q ⁡ open parenthesis 25 close parenthesis equals Q to the power k")[$Q^(log_(Q) ( 25 )) = Q^(k)$], which gives us #math.equation(block: false, alt: "25")[$25$].
]
]


      
        Simplify each expression.


        + #math.equation(block: false, alt: "4 to the power log sub 4 ⁡ open parenthesis 64 close parenthesis equals")[$4^(log_(4) ( 64 )) =$]\_\_\_\_\_
+ #math.equation(block: false, alt: "2 to the power log sub 2 ⁡ open parenthesis x squared plus 1 close parenthesis equals")[$2^(log_(2) ( x^(2) + 1 )) =$]\_\_\_\_\_


        #solutionbox[
+ #math.equation(block: false, alt: "64")[$64$]
+ #math.equation(block: false, alt: "x squared plus 1")[$x^(2) + 1$]
]


      
      
        Simplify each expression.


        + #math.equation(block: false, alt: "4 to the power log sub 4 ⁡ open parenthesis 64 close parenthesis")[$4^(log_(4) ( 64 ))$]
+ #math.equation(block: false, alt: "2 to the power log sub 2 ⁡ open parenthesis x squared plus 1 close parenthesis")[$2^(log_(2) ( x^(2) + 1 ))$]


        #solutionbox[
+ #math.equation(block: false, alt: "64")[$64$]
+ #math.equation(block: false, alt: "x squared plus 1")[$x^(2) + 1$]
]


      
      We summarize these relationships as follows.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Exponential and Logarithmic Functions]


        Because #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals b to the power x")[$" " f ( x ) = b^(x) " "$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub b ⁡ open parenthesis x close parenthesis")[$" " g ( x ) = log_(b) ( x ) " "$] are inverse functions for #math.equation(block: false, alt: "b greater than 0 , b not equal to 1")[$b > 0 , b ≠ 1$],


        #math.equation(block: true, alt: "log sub b ⁡ open parenthesis b to the power x close parenthesis equals x , for all x and b to the power log sub b ⁡ open parenthesis x close parenthesis equals x , for x greater than 0")[$bold(log)_(bold(italic(b))) ( bold(italic(b))^(bold(italic(x))) ) = bold(italic(x)) ", for all " x " " " " " " " " " " " and " " " " " " " " " " " bold(italic(b))^(bold(log)_(bold(italic(b))) ( bold(italic(x)) )) = bold(italic(x)) ", for " x > 0$]
]


      In Chapter 4 we solved exponential equations by using the conversion equations to rewrite them in logarithmic form. The fact that #math.equation(block: false, alt: "log sub b ⁡ b to the power x equals x")[$" " log_(b) b^(x) = x " "$] gives us another way to think of the solution; we can take the log of both sides of the equation. We'll use this method in the next Example. Recall from Inverse Function Notation in Section 5.1 that the inverse function for a function #math.equation(block: false, alt: "f")[$f$] is often denoted by #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$].


      #examplebox("Example 10")[][
+ Find the inverse of the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x minus 3 minus 4")[$f ( x ) = 2^(x − 3) − 4$].
+ Graph #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] on the same grid.
+ State the domain and range of #math.equation(block: false, alt: "f")[$f$] and of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$].


        
        #solutionbox[
+ We write the function as #math.equation(block: false, alt: "y equals 2 to the power x minus 3 minus 4")[$" " y = 2^(x − 3) − 4$], and solve for #math.equation(block: false, alt: "x")[$x$] in terms of #math.equation(block: false, alt: "y")[$y$]. First, we isolate the power:
			#math.equation(block: true, alt: "y plus 4, equals 2 to the power x minus 3, Take the log base 2 of each side.; log sub 2 ⁡ open parenthesis y plus 4 close parenthesis, equals log sub 2 ⁡ open parenthesis 2 close parenthesis to the power x minus 3, Simplify the right side.; log sub 2 ⁡ open parenthesis y plus 4 close parenthesis, equals x minus 3; 3 plus log sub 2 ⁡ open parenthesis y plus 4 close parenthesis, equals x")[$y + 4 & = 2^(x − 3) &  & "Take the log base 2 of each side." \ log_(2) ( y + 4 ) & = log_(2) attach(( 2 ), t: x − 3) &  & "Simplify the right side." \  \ log_(2) ( y + 4 ) & = x − 3 \ 3 + log_(2) ( y + 4 ) & = x$]
	The inverse function is #math.equation(block: false, alt: "f to the power minus 1 open parenthesis y close parenthesis equals 3 plus log sub 2 ⁡ open parenthesis y plus 4 close parenthesis")[$" " f^(− 1) ( y ) = 3 + log_(2) ( y + 4 )$]. However, to graph both #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] on the same grid, we write the inverse function as #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals 3 plus log sub 2 ⁡ open parenthesis x plus 4 close parenthesis")[$" " f^(− 1) ( x ) = 3 + log_(2) ( x + 4 )$].
+ To graph #math.equation(block: false, alt: "f")[$f$], we translate the graph of #math.equation(block: false, alt: "y equals 2 to the power x")[$" " y = 2^(x) " "$] by #math.equation(block: false, alt: "3")[$3$] units to the right and #math.equation(block: false, alt: "4")[$4$] units down. The graph of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] looks like the graph of #math.equation(block: false, alt: "y equals log sub 2 ⁡ open parenthesis x close parenthesis")[$y = log_(2) ( x )$], but shifted #math.equation(block: false, alt: "4")[$4$] units to the left and #math.equation(block: false, alt: "3")[$3$] units up. The graphs are shown below, along with the line #math.equation(block: false, alt: "y equals x")[$y = x$].#figure(figph[graph of translated exponential and inverse], alt: "graph of translated exponential and inverse", caption: none)
+ The function #math.equation(block: false, alt: "f")[$f$] is a translation of an exponential function, and its domain consists of all real numbers. Because the graph is shifted #math.equation(block: false, alt: "4")[$4$] units down, the range of #math.equation(block: false, alt: "f")[$f$] is #math.equation(block: false, alt: "open parenthesis minus 4 , ∞ close parenthesis")[$( − 4 , upright(∞) )$]. #linebreak() Because the log of a negative number or zero is undefined, for #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals 3 plus log sub 2 ⁡ open parenthesis x plus 4 close parenthesis")[$" " f^(− 1) ( x ) = 3 + log_(2) ( x + 4 )$], we must have #math.equation(block: false, alt: "x plus 4 greater than 0")[$x + 4 > 0$], so the domain of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is #math.equation(block: false, alt: "x greater than minus 4")[$x > − 4$]. We can verify on the graph that the range of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] includes all real numbers. Thus,
			#math.equation(block: true, alt: "Domain open parenthesis f close parenthesis equals all real numbers equals Range open parenthesis f to the power minus 1 close parenthesis; Range open parenthesis f close parenthesis equals open parenthesis minus 4 , ∞ close parenthesis equals Domain open parenthesis f to the power minus 1 close parenthesis")[$& "Domain" ( f ) = "all real numbers" = "Range" ( f^(− 1) ) \  & "Range" ( f ) = ( − 4 , upright(∞) ) = "Domain" ( f^(− 1) )$]
]
]


      
        + Find the inverse function for #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 log ⁡ open parenthesis x plus 1 close parenthesis")[$f ( x ) = 2 log ( x + 1 )$]. #linebreak() #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals")[$f^(− 1) ( x ) =$]\_\_\_\_\_
+ Graph #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] in the window
                    #math.equation(block: true, alt: "Xmin, equals minus 6, Xmax equals 6; Ymin, equals minus 4, Ymax equals 4")[$"Xmin" & = − 6 &  & "Xmax" = 6 \ "Ymin" & = − 4 &  & "Ymax" = 4$]
+ State the domain and range of #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]. Use "inf" for #math.equation(block: false, alt: "∞")[$upright(∞)$]. #linebreak() Domain of #math.equation(block: false, alt: "f")[$f$]: \_\_\_\_\_; Range of #math.equation(block: false, alt: "f")[$f$]: \_\_\_\_\_  #linebreak() Domain of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]: \_\_\_\_\_; Range of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]: \_\_\_\_\_


        #solutionbox[
+ #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals 10 to the power x / 2 minus 1")[$f^(− 1) ( x ) = 10^(x / 2) − 1$]
+ A graph is below.
+ Domain of #math.equation(block: false, alt: "f")[$f$]:  #math.equation(block: false, alt: "open parenthesis minus 1 , ∞ close parenthesis")[$( − 1 , upright(∞) )$]; Range of #math.equation(block: false, alt: "f")[$f$]: all real numbers; Domain of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]: all real numbers; Range of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]:  #math.equation(block: false, alt: "open parenthesis minus 1 , ∞ close parenthesis")[$( − 1 , upright(∞) )$]
]


        #figure(figph[log and inverse], alt: "log and inverse", caption: none)


      
      
        + Find the inverse function for #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 log ⁡ open parenthesis x plus 1 close parenthesis")[$f ( x ) = 2 log ( x + 1 )$].
+ Graph #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] in the window
                    #math.equation(block: true, alt: "Xmin, equals minus 6, Xmax equals 6; Ymin, equals minus 4, Ymax equals 4")[$"Xmin" & = − 6 &  & "Xmax" = 6 \ "Ymin" & = − 4 &  & "Ymax" = 4$]
+ State the domain and range of #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$].


        #solutionbox[
+ #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals 10 to the power x / 2 minus 1")[$f^(− 1) ( x ) = 10^(x / 2) − 1$]
+ #figure(figph[log and inverse], alt: "log and inverse", caption: none)
+ Domain of #math.equation(block: false, alt: "f")[$f$]:  #math.equation(block: false, alt: "open parenthesis minus 1 , ∞ close parenthesis")[$( − 1 , upright(∞) )$]; Range of #math.equation(block: false, alt: "f")[$f$]: all real numbers; Domain of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]: all real numbers; Range of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]:  #math.equation(block: false, alt: "open parenthesis minus 1 , ∞ close parenthesis")[$( − 1 , upright(∞) )$]
]


      
      
        Compare the graphs of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log ⁡ the fraction 1 over x")[$f ( x ) = log display(frac(1, x))$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals minus log ⁡ open parenthesis x close parenthesis")[$g ( x ) = − log ( x )$], and explain.


        \_\_\_\_\_


      
      
        Compare the graphs of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis log ⁡ the fraction 1 over x close parenthesis")[$f ( x ) = ( log display(frac(1, x)) )$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals minus log ⁡ open parenthesis x close parenthesis")[$g ( x ) = − log ( x )$], and explain.


      
    
    
      
=== Section Summary


      
        
==== Vocabulary


        Look up the definitions of new terms in the Glossary.


        - Logarithmic function
- Logarithmic equation
- Extraneous solution


      
      
        
==== CONCEPTS


        + We define the #strong[logarithmic function], #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub b ⁡ open parenthesis x close parenthesis")[$g ( x ) = log_(b) ( x )$], which takes the log base #math.equation(block: false, alt: "b")[$b$] of its input values. The log function #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub b ⁡ x")[$g ( x ) = log_(b) x$] is the inverse of the exponential function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals b to the power x")[$f ( x ) = b^(x)$].
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Exponential and Logarithmic Functions]

Because #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals b to the power x")[$f ( x ) = b^(x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub b ⁡ open parenthesis x close parenthesis")[$g ( x ) = log_(b) ( x )$] are inverse functions for #math.equation(block: false, alt: "b greater than 0 , b not equal to 1")[$b > 0 , b ≠ 1$],

#math.equation(block: true, alt: "log sub b ⁡ b to the power x equals x , for all x and b to the power log sub b ⁡ open parenthesis x close parenthesis equals x , for x greater than 0")[$bold(log)_(bold(italic(b))) bold(italic(b))^(bold(italic(x))) = bold(italic(x)) ", for all " x " " " " " " " " " " " and " " " " " " " " " " " bold(italic(b))^(bold(log)_(bold(italic(b))) ( bold(italic(x)) )) = bold(italic(x)) ", for " x > 0$]
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Logarithmic Functions #math.equation(block: false, alt: "y equals log sub b ⁡ open parenthesis x close parenthesis")[$y = log_(b) ( x )$]]

+ Domain: all positive real numbers
+ Range: all real numbers
+ #math.equation(block: false, alt: "x")[$x$]-intercept: #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]
+ #math.equation(block: false, alt: "y")[$y$]-intercept: none
+ Vertical asymptote at #math.equation(block: false, alt: "x equals 0")[$x = 0$]
+ The graphs of #math.equation(block: false, alt: "y equals log sub b ⁡ open parenthesis x close parenthesis")[$y = log_(b) ( x )$] and #math.equation(block: false, alt: "y equals b to the power x")[$y = b^(x)$] are symmetric about the line #math.equation(block: false, alt: "y equals x")[$y = x$].
]
+ A #strong[logarithmic equation] is one in which the variable appears inside of a logarithm. We can solve logarithmic equations by converting to exponential form.
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Steps for Solving Logarithmic Equations]

+ Use the properties of logarithms to combine all logs into one log.
+ Isolate the log on one side of the equation.
+ Convert the equation to exponential form.
+ Solve for the variable.
+ Check for extraneous solutions.
]


      
      
        
==== STUDY QUESTIONS


        + Can the output of the function #math.equation(block: false, alt: "y equals log sub b ⁡ open parenthesis x close parenthesis")[$y = log_(b) ( x )$] be negative?
+ Francine says that #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis the fraction 1 over x close parenthesis equals minus log sub 2 ⁡ open parenthesis x close parenthesis")[$log_(2) ( display(frac(1, x)) ) = − log_(2) ( x )$]. Is she correct? Why or why not?
+ Sketch a typical logarithmic function.
+ Simplify: 
			+ #math.equation(block: false, alt: "10 to the power log ⁡ open parenthesis 13 close parenthesis")[$10^(log ( 13 ))$]
+ #math.equation(block: false, alt: "7 to the power log sub 7 ⁡ open parenthesis 13 close parenthesis")[$7^(log_(7) ( 13 ))$]
+ Why is the following attempt to solve the equation incorrect?
			#math.equation(block: true, alt: "Solve: log ⁡ open parenthesis x close parenthesis plus log ⁡ open parenthesis x plus 1 close parenthesis, equals 2; x plus x plus 1, equals 10 squared")[$"Solve:" #hide($b l a n k$) log ( x ) + log ( x + 1 ) & = 2 \ x + x + 1 & = 10^(2)$]


      
      
        
==== SKILLS


        Practice each skill in the Homework problems listed.


        + Evaluate log functions: \#1–16, 27 and 28
+ Simplify expressions involving logs: \#15 and 16, 19, and 20
+ Graph logarithmic functions and transformations of log functions: \#1–4, 25–28
+ Find formulas for inverse functions: \#17–24
+ Solve logarithmic equations: \#29–54
+ Solve formulas involving logs: \#55–60


      
    
    
      
=== Homework 5.2


      In Problems 1–4,


      + Make tables of values for each exponential function and its inverse logarithmic function.
+ Graph both functions on the same set of axes.


      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$f ( x ) = 2^(x)$]


        
          + #figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]]),
  [#math.equation(block: false, alt: "2 to the power x")[$2^(x)$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]],
))

#figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]]),
  [#math.equation(block: false, alt: "log sub 2 ⁡ x")[$log_(2) x$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]],
))
+ #figure(figph[exponential and log functions], alt: "exponential and log functions", caption: none)


        
      
      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 3 to the power x")[$f ( x ) = 3^(x)$]


      
      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 3 close parenthesis to the power x")[$f ( x ) = attach(( display(frac(1, 3)) ), t: x)$]


        
          + #figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]]),
  [#math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power x")[$attach(( frac(1, 3) ), t: x)$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction 1 over 3")[$frac(1, 3)$]],
))

#figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction 1 over 3")[$frac(1, 3)$]]),
  [#math.equation(block: false, alt: "log sub 1 / 3 ⁡ x")[$log_(1 / 3) x$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]],
))
+ #figure(figph[exponential and log functions], alt: "exponential and log functions", caption: none)


        
      
      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 2 close parenthesis to the power x")[$f ( x ) = attach(( display(frac(1, 2)) ), t: x)$]


      
      
        + How large must #math.equation(block: false, alt: "x")[$x$] be before the graph of #math.equation(block: false, alt: "y equals log sub 10 ⁡ open parenthesis x close parenthesis")[$y = log_(10) ( x )$] reaches a height of #math.equation(block: false, alt: "4")[$4$]?
+ How large must #math.equation(block: false, alt: "x")[$x$] be before the graph of #math.equation(block: false, alt: "y equals log sub 10 ⁡ open parenthesis x close parenthesis")[$y = log_(10) ( x )$] reaches a height of #math.equation(block: false, alt: "8")[$8$]?


        
          + #math.equation(block: false, alt: "x equals 10 , 000")[$x = 10 , 000$]
+ #math.equation(block: false, alt: "x equals 10 to the power 8")[$x = 10^(8)$]


        
      
      
        + How large must #math.equation(block: false, alt: "x")[$x$] be before the graph of #math.equation(block: false, alt: "y equals log sub 2 ⁡ open parenthesis x close parenthesis")[$y = log_(2) ( x )$] reaches a height of #math.equation(block: false, alt: "5")[$5$]?
+ How large must #math.equation(block: false, alt: "x")[$x$] be before the graph of #math.equation(block: false, alt: "y equals log sub 2 ⁡ open parenthesis x close parenthesis")[$y = log_(2) ( x )$] reaches a height of #math.equation(block: false, alt: "10")[$10$]?


      
      
        For what values of #math.equation(block: false, alt: "x")[$x$] is #math.equation(block: false, alt: "y equals log sub 10 ⁡ open parenthesis x close parenthesis less than minus 2")[$y = log_(10) ( x ) < − 2$]?


        
          #math.equation(block: true, alt: "0 less than open parenthesis x close parenthesis less than 0.01")[$0 < ( x ) < 0.01$]


        
      
      
        For what values of #math.equation(block: false, alt: "x")[$x$] is #math.equation(block: false, alt: "y equals log sub 2 ⁡ open parenthesis x close parenthesis less than minus 3")[$y = log_(2) ( x ) < − 3$]?


      
      In Problems 9–14, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis")[$f ( x ) = log_(10) ( x )$]. Evaluate.


      
        + #math.equation(block: false, alt: "f open parenthesis 487 close parenthesis plus f open parenthesis 206 close parenthesis")[$f ( 487 ) + f ( 206 )$]
+ #math.equation(block: false, alt: "f open parenthesis 487 plus 206 close parenthesis")[$f ( 487 + 206 )$]


        
          + #math.equation(block: false, alt: "log ⁡ 100 , 322 approximately equals 5.001")[$log 100 , 322 ≈ 5.001$]
+ #math.equation(block: false, alt: "log ⁡ 693 approximately equals 2.841")[$log 693 ≈ 2.841$]


        
      
      
        + #math.equation(block: false, alt: "f open parenthesis 93 close parenthesis plus f open parenthesis 1500 close parenthesis")[$f ( 93 ) + f ( 1500 )$]
+ #math.equation(block: false, alt: "f open parenthesis 93 plus 1500 close parenthesis")[$f ( 93 + 1500 )$]


      
      
        + #math.equation(block: false, alt: "f open parenthesis minus 7 close parenthesis")[$f ( − 7 )$]
+ #math.equation(block: false, alt: "6 f open parenthesis 28 close parenthesis")[$6 f ( 28 )$]


        
          + #math.equation(block: false, alt: "log ⁡ open parenthesis minus 7 close parenthesis")[$log ( − 7 )$] is undefined.
+ #math.equation(block: false, alt: "6 log ⁡ 28 approximately equals 8.683")[$6 log 28 ≈ 8.683$]


        
      
      
        + #math.equation(block: false, alt: "f open parenthesis 0 close parenthesis")[$f ( 0 )$]
+ #math.equation(block: false, alt: "3 f open parenthesis 41 close parenthesis")[$3 f ( 41 )$]


      
      
        + #math.equation(block: false, alt: "18 minus 5 f open parenthesis 3 close parenthesis")[$18 − 5 f ( 3 )$]
+ #math.equation(block: false, alt: "the fraction 2 over 5 plus f open parenthesis 0.6 close parenthesis")[$display(frac(2, 5 + f ( 0.6 )))$]


        
          + #math.equation(block: false, alt: "15.614")[$15.614$]
+ #math.equation(block: false, alt: "0.419")[$0.419$]


        
      
      
        + #math.equation(block: false, alt: "15 minus 4 f open parenthesis 7 close parenthesis")[$15 − 4 f ( 7 )$]
+ #math.equation(block: false, alt: "the fraction 3 over 2 plus f open parenthesis 0.2 close parenthesis")[$display(frac(3, 2 + f ( 0.2 )))$]


      
      
        Let #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 to the power x")[$f ( x ) = 3^(x)$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals log sub 3 ⁡ open parenthesis x close parenthesis")[$g ( x ) = log_(3) ( x )$].


        + Compute #math.equation(block: false, alt: "f open parenthesis 4 close parenthesis")[$f ( 4 )$].
+ Compute #math.equation(block: false, alt: "g open bracket f open parenthesis 4 close parenthesis close bracket")[$g [ f ( 4 ) ]$].
+ Explain why #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 3 to the power x close parenthesis equals x")[$log_(3) ( 3^(x) ) = x$] for any #math.equation(block: false, alt: "x")[$x$].
+ Compute #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 3 to the power 1.8 close parenthesis")[$log_(3) ( 3^(1.8) )$].
+ Simplify #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 3 to the power a close parenthesis")[$log_(3) ( 3^(a) )$].


        
          + #math.equation(block: false, alt: "81")[$81$]
+ #math.equation(block: false, alt: "4")[$4$]
+ Definition of logarithm base #math.equation(block: false, alt: "3")[$3$]
+ #math.equation(block: false, alt: "1.8")[$1.8$]
+ #math.equation(block: false, alt: "a")[$a$]


        
      
      
        Let #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis")[$f ( x ) = log_(2) ( x )$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 to the power x")[$g ( x ) = 2^(x)$].


        + Compute #math.equation(block: false, alt: "f open parenthesis 32 close parenthesis")[$f ( 32 )$].
+ Compute #math.equation(block: false, alt: "g open bracket f open parenthesis 32 close parenthesis close bracket")[$g [ f ( 32 ) ]$].
+ Explain why #math.equation(block: false, alt: "2 to the power log sub 2 ⁡ open parenthesis x close parenthesis equals x")[$2^(log_(2) ( x )) = x$] for any #math.equation(block: false, alt: "x greater than 0")[$x > 0$].
+ Compute #math.equation(block: false, alt: "2 to the power log sub 2 ⁡ open parenthesis 6 close parenthesis")[$2^(log_(2) ( 6 ))$].
+ Simplify #math.equation(block: false, alt: "2 to the power log sub 2 ⁡ open parenthesis Q close parenthesis")[$2^(log_(2) ( Q ))$].


      
      
        + If #math.equation(block: false, alt: "h open parenthesis r close parenthesis equals log sub 2 ⁡ open parenthesis r close parenthesis")[$h ( r ) = log_(2) ( r )$], find #math.equation(block: false, alt: "h to the power minus 1 open parenthesis 8 close parenthesis")[$h^(− 1) ( 8 )$].
+ If #math.equation(block: false, alt: "H open parenthesis w close parenthesis equals 3 to the power w")[$H ( w ) = 3^(w)$], find #math.equation(block: false, alt: "H to the power minus 1 open parenthesis the fraction 1 over 9 close parenthesis")[$H^(− 1) ( display(frac(1, 9)) )$].


        
          + #math.equation(block: false, alt: "2 to the power 8")[$2^(8)$]
+ #math.equation(block: false, alt: "minus 2")[$− 2$]


        
      
      
        + If #math.equation(block: false, alt: "g open parenthesis z close parenthesis equals log sub 3 ⁡ open parenthesis z close parenthesis")[$g ( z ) = log_(3) ( z )$], find #math.equation(block: false, alt: "g to the power minus 1 open parenthesis minus 3 close parenthesis")[$g^(− 1) ( − 3 )$].
+ If #math.equation(block: false, alt: "G open parenthesis q close parenthesis equals 2 to the power q")[$G ( q ) = 2^(q)$], find #math.equation(block: false, alt: "G to the power minus 1 open parenthesis 1 close parenthesis")[$G^(− 1) ( 1 )$].


      
      For Problems 19–20, simplify.


      
        + #math.equation(block: false, alt: "10 to the power log ⁡ open parenthesis 2 k close parenthesis")[$10^(log ( 2 k ))$]
+ #math.equation(block: false, alt: "10 to the power 3 log ⁡ open parenthesis x close parenthesis")[$10^(3 log ( x ))$]
+ #math.equation(block: false, alt: "open parenthesis the square root of 10 close parenthesis to the power log ⁡ open parenthesis x close parenthesis")[$( sqrt(10) )^(log ( x ))$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis 100 to the power m close parenthesis")[$log ( 100^(m) )$]


        
          + #math.equation(block: false, alt: "2 k")[$2 k$]
+ #math.equation(block: false, alt: "x cubed")[$x^(3)$]
+ #math.equation(block: false, alt: "the square root of x")[$sqrt(x)$]
+ #math.equation(block: false, alt: "2 m")[$2 m$]


        
      
      
        + #math.equation(block: false, alt: "log ⁡ open parenthesis 10 to the power open parenthesis 1 minus x close parenthesis close parenthesis")[$log ( 10^(( 1 − x )) )$]
+ #math.equation(block: false, alt: "100 to the power log ⁡ open parenthesis 2 x close parenthesis")[$100^(log ( 2 x ))$]
+ #math.equation(block: false, alt: "open parenthesis 0.1 close parenthesis to the power log ⁡ open parenthesis x minus 1 close parenthesis")[$( 0.1 )^(log ( x − 1 ))$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis 10 to the power log ⁡ open parenthesis 10 close parenthesis close parenthesis")[$log ( 10^(log ( 10 )) )$]


      
      
        + What is the domain of the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4 plus log sub 3 ⁡ open parenthesis x minus 9 close parenthesis")[$f ( x ) = 4 + log_(3) ( x − 9 )$]?
+ Find a formula for #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis")[$f^(− 1) ( x )$].


        
          + #math.equation(block: false, alt: "open parenthesis 9 , ∞ close parenthesis")[$( 9 , upright(∞) )$]
+ #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals 3 to the power x minus 4 plus 9")[$f^(− 1) ( x ) = 3^(x − 4) + 9$]


        
      
      
        + What is the domain of the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 1 minus log sub 2 ⁡ open parenthesis 16 minus 4 x close parenthesis")[$f ( x ) = 1 − log_(2) ( 16 − 4 x )$]?
+ Find a formula for #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis")[$f^(− 1) ( x )$].


      
      
        + Find the inverse of the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 100 minus 4 to the power x plus 2")[$f ( x ) = 100 − 4^(x + 2)$].
+ Show that #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] undoes the effect of #math.equation(block: false, alt: "f")[$f$] on #math.equation(block: false, alt: "x equals 1")[$x = 1$].
+ Show that #math.equation(block: false, alt: "f")[$f$] undoes the effect of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] on #math.equation(block: false, alt: "x equals 84")[$x = 84$].


        
          + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals log sub 4 ⁡ open parenthesis 100 minus x close parenthesis minus 2")[$f^(− 1) ( x ) = log_(4) ( 100 − x ) − 2$]
+ #math.equation(block: false, alt: "f to the power minus 1 open parenthesis f open parenthesis 1 close parenthesis close parenthesis equals f to the power minus 1 open parenthesis 36 close parenthesis equals log sub 4 ⁡ open parenthesis 64 close parenthesis minus 2 equals 1")[$f^(− 1) ( f ( 1 ) ) = f^(− 1) ( 36 ) = log_(4) ( 64 ) − 2 = 1$]
+ #math.equation(block: false, alt: "f open parenthesis f to the power minus 1 open parenthesis 84 close parenthesis close parenthesis equals f open parenthesis 0 close parenthesis equals 100 minus 4 squared equals 84")[$f ( f^(− 1) ( 84 ) ) = f ( 0 ) = 100 − 4^(2) = 84$]


        
      
      
        + Find the inverse of the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 5 plus 2 to the power minus x")[$f ( x ) = 5 + 2^(− x)$].
+ Show that #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] undoes the effect of #math.equation(block: false, alt: "f")[$f$] on #math.equation(block: false, alt: "x equals minus 2")[$x = − 2$].
+ Show that #math.equation(block: false, alt: "f")[$f$] undoes the effect of #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] on #math.equation(block: false, alt: "x equals 6")[$x = 6$].


      
      For Problems 25–26, match each graph to its equation.


      
        + #math.equation(block: false, alt: "y equals log sub 2 ⁡ open parenthesis x minus 3 close parenthesis")[$y = log_(2) ( x − 3 )$]
+ #math.equation(block: false, alt: "y equals 3 plus log sub 2 ⁡ open parenthesis x close parenthesis")[$y = 3 + log_(2) ( x )$]
+ #math.equation(block: false, alt: "y equals 2 minus log sub 2 ⁡ x")[$y = 2 − log_(2) x$]
+ #math.equation(block: false, alt: "y equals log sub 2 ⁡ open parenthesis x plus 4 close parenthesis minus 1")[$y = log_(2) ( x + 4 ) − 1$]


        + #figure(figph[transformed log], alt: "transformed log", caption: none)
+ #figure(figph[transformed log], alt: "transformed log", caption: none)
+ #figure(figph[transformed log], alt: "transformed log", caption: none)
+ #figure(figph[transformed log], alt: "transformed log", caption: none)


        
          + IV
+ I
+ II
+ III


        
      
      
        + #math.equation(block: false, alt: "y equals 5 log ⁡ open parenthesis x close parenthesis")[$y = 5 log ( x )$]
+ #math.equation(block: false, alt: "y equals log ⁡ open parenthesis the fraction x over 2 close parenthesis")[$y = log ( display(frac(x, 2)) )$]
+ #math.equation(block: false, alt: "y equals log ⁡ open parenthesis the fraction 1 over x close parenthesis")[$y = log ( display(frac(1, x)) )$]
+ #math.equation(block: false, alt: "y equals log ⁡ open parenthesis minus x close parenthesis")[$y = log ( − x )$]


        + #figure(figph[transformed log], alt: "transformed log", caption: none)
+ #figure(figph[transformed log], alt: "transformed log", caption: none)
+ #figure(figph[transformed log], alt: "transformed log", caption: none)
+ #figure(figph[transformed log], alt: "transformed log", caption: none)


      
      
        In a psychology experiment, volunteers were asked to memorize a list of nonsense words, then 24 hours later were tested to see how many of the words they recalled. On average, the subjects had forgotten #math.equation(block: false, alt: "20 %")[$20 upright(%)$] of the words. The researchers found that the more lists their volunteers memorized, the larger the fraction of words they were unable to recall. (Source: Underwood, #emph[Scientific American], vol. 210, no. 3)


        #figure(table(
  columns: 7, align: left, inset: 6pt,
  table.header([Number of lists, #math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "20")[$20$]]),
  [Percent forgotten, #math.equation(block: false, alt: "F")[$F$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "55")[$55$]], [#math.equation(block: false, alt: "66")[$66$]], [#math.equation(block: false, alt: "74")[$74$]], [#math.equation(block: false, alt: "80")[$80$]],
))


        + Plot the data. What sort of function seems to fit the data points?
+ Psychologists often describe rates of forgetting by logarithmic functions. Graph the function 
        #math.equation(block: true, alt: "f open parenthesis n close parenthesis equals 16.6 plus 46.3 log ⁡ open parenthesis n close parenthesis")[$f ( n ) = 16.6 + 46.3 log ( n )$]
        on the same graph with your data. Comment on the fit.
+ What happens to the function #math.equation(block: false, alt: "f open parenthesis n close parenthesis")[$f ( n )$] as #math.equation(block: false, alt: "n")[$n$] grows increasingly large? Does this behavior accurately reflect the situation being modeled?


        
          + #figure(figph[data points and log curve], alt: "data points and log curve", caption: none)
+ The graph resembles a logarithmic function. The (translated) log function is close to the points but appears too steep at first and not steep enough after #math.equation(block: false, alt: "n equals 15")[$n = 15$]. Overall, it is a good fit.
+ #math.equation(block: false, alt: "f")[$f$] grows (more and more slowly) without bound. #math.equation(block: false, alt: "f")[$f$] will eventually exceed #math.equation(block: false, alt: "100")[$100$] per cent, but no one can forget more than #math.equation(block: false, alt: "100 %")[$100 upright(%)$] of what is learned.


        
      
      
        The water velocity at any point in a stream or river is related to the logarithm of the depth at that point. For the Hoback River near Bondurant, Wyoming,


        #math.equation(block: true, alt: "v equals 2.63 plus 1.03 log ⁡ open parenthesis d close parenthesis")[$v = 2.63 + 1.03 log ( d )$]
        where #math.equation(block: false, alt: "v")[$v$] is the velocity of the water, in feet per second, and #math.equation(block: false, alt: "d")[$d$] is the vertical distance from the stream bed, in feet, at that point. For Pole Creek near Pinedale, Wyoming,


        #math.equation(block: true, alt: "v equals 1.96 plus 0.65 log ⁡ open parenthesis d close parenthesis")[$v = 1.96 + 0.65 log ( d )$]
        Both streams are #math.equation(block: false, alt: "1.2")[$1.2$] feet deep at the locations mentioned. (Source: Leopold, Luna, Wolman, and Gordon, 1992)


        + Complete the table of values for each stream.#figure(table(
  columns: 7, align: left, inset: 6pt,
  table.header([Distance from bed (feet)], [#math.equation(block: false, alt: "0.2")[$0.2$]], [#math.equation(block: false, alt: "0.4")[$0.4$]], [#math.equation(block: false, alt: "0.6")[$0.6$]], [#math.equation(block: false, alt: "0.8")[$0.8$]], [#math.equation(block: false, alt: "1.0")[$1.0$]], [#math.equation(block: false, alt: "1.2")[$1.2$]]),
  [Velocity, Hoback #linebreak() River, (ft/sec) #linebreak()], [], [], [], [], [], [],
  [Velocity, Pole Creek (ft/sec)], [], [], [], [], [], [],
))
+ If you double the distance from the bed, by how much does the velocity increase in each stream?
+ Plot both functions on the same graph.
+ The average velocity of the entire stream can be closely approximated as follows: Measure the velocity at #math.equation(block: false, alt: "20 %")[$20 upright(%)$] of the total depth of the stream from the surface and at #math.equation(block: false, alt: "80 %")[$80 upright(%)$] of the total depth, then average these two values. Find the average velocity for the Hoback River and for Pole Creek.


      
      In Problems 29–30, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log sub 10 ⁡ open parenthesis x close parenthesis")[$f ( x ) = log_(10) ( x )$]. Solve for #math.equation(block: false, alt: "x")[$x$].


      
        + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 1.41")[$f ( x ) = 1.41$]
+ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals minus 1.69")[$f ( x ) = − 1.69$]
+ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0.52")[$f ( x ) = 0.52$]


        
          + #math.equation(block: false, alt: "10 to the power 1.41 approximately equals 25.704")[$10^(1.41) ≈ 25.704$]
+ #math.equation(block: false, alt: "10 to the power minus 1.69 approximately equals 0.020417")[$10^(− 1.69) ≈ 0.020417$]
+ #math.equation(block: false, alt: "10 to the power 0.52 approximately equals 3.3113")[$10^(0.52) ≈ 3.3113$]


        
      
      
        + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2.3")[$f ( x ) = 2.3$]
+ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals minus 1.3")[$f ( x ) = − 1.3$]
+ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0.8")[$f ( x ) = 0.8$]


      
      For Problems 31–38, convert the logarithmic equation to exponential form.


      
        #math.equation(block: true, alt: "log sub 16 ⁡ open parenthesis 256 close parenthesis equals w")[$log_(16) ( 256 ) = w$]


        
          #math.equation(block: true, alt: "16 to the power w equals 256")[$16^(w) = 256$]


        
      
      
        #math.equation(block: true, alt: "log sub 9 ⁡ open parenthesis 729 close parenthesis equals y")[$log_(9) ( 729 ) = y$]


      
      
        #math.equation(block: true, alt: "log sub b ⁡ open parenthesis 9 close parenthesis equals minus 2")[$log_(b) ( 9 ) = − 2$]


        
          #math.equation(block: true, alt: "b to the power minus 2 equals 9")[$b^(− 2) = 9$]


        
      
      
        #math.equation(block: true, alt: "log sub b ⁡ open parenthesis 8 close parenthesis equals minus 3")[$log_(b) ( 8 ) = − 3$]


      
      
        #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis A close parenthesis equals minus 2.3")[$log_(10) ( A ) = − 2.3$]


        
          #math.equation(block: true, alt: "10 to the power minus 2.3 equals A")[$10^(− 2.3) = A$]


        
      
      
        #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis C close parenthesis equals minus 4.5")[$log_(10) ( C ) = − 4.5$]


      
      
        #math.equation(block: true, alt: "log sub u ⁡ open parenthesis v close parenthesis equals w")[$log_(u) ( v ) = w$]


        
          #math.equation(block: true, alt: "u to the power w equals v")[$u^(w) = v$]


        
      
      
        #math.equation(block: true, alt: "log sub m ⁡ open parenthesis n close parenthesis equals p")[$log_(m) ( n ) = p$]


      
      For Problems 39–46, solve for the unknown value.


      
        #math.equation(block: true, alt: "log sub b ⁡ open parenthesis 8 close parenthesis equals 3")[$log_(b) ( 8 ) = 3$]


        
          #math.equation(block: true, alt: "b equals 2")[$b = 2$]


        
      
      
        #math.equation(block: true, alt: "log sub b ⁡ open parenthesis 625 close parenthesis equals 4")[$log_(b) ( 625 ) = 4$]


      
      
        #math.equation(block: true, alt: "log sub b ⁡ open parenthesis 10 close parenthesis equals the fraction 1 over 2")[$log_(b) ( 10 ) = display(frac(1, 2))$]


        
          #math.equation(block: true, alt: "b equals 100")[$b = 100$]


        
      
      
        #math.equation(block: true, alt: "log sub b ⁡ open parenthesis 0.1 close parenthesis equals minus 1")[$log_(b) ( 0.1 ) = − 1$]


      
      
        #math.equation(block: true, alt: "log sub 2 ⁡ open parenthesis 3 x minus 1 close parenthesis equals 5")[$log_(2) ( 3 x − 1 ) = 5$]


        
          #math.equation(block: true, alt: "x equals 11")[$x = 11$]


        
      
      
        #math.equation(block: true, alt: "log sub 5 ⁡ open parenthesis 9 minus 4 x close parenthesis equals 3")[$log_(5) ( 9 − 4 x ) = 3$]


      
      
        #math.equation(block: true, alt: "3 log sub 7 ⁡ open parenthesis x close parenthesis plus 5 equals 7")[$3 log_(7) ( x ) + 5 = 7$]


        
          #math.equation(block: true, alt: "x equals 7 to the power 2 / 3")[$x = 7^(2 / 3)$]


        
      
      
        #math.equation(block: true, alt: "5 log sub 2 ⁡ open parenthesis x close parenthesis plus 6 equals minus 14")[$5 log_(2) ( x ) + 6 = − 14$]


      
      For Problems 47–54, solve the logarithmic equation.


      
        #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis x close parenthesis plus log sub 10 ⁡ open parenthesis x plus 21 close parenthesis equals 2")[$log_(10) ( x ) + log_(10) ( x + 21 ) = 2$]


        
          #math.equation(block: true, alt: "x equals 4")[$x = 4$]


        
      
      
        #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis x plus 3 close parenthesis plus log sub 10 ⁡ open parenthesis x close parenthesis equals 1")[$log_(10) ( x + 3 ) + log_(10) ( x ) = 1$]


      
      
        #math.equation(block: true, alt: "log sub 8 ⁡ open parenthesis x plus 5 close parenthesis minus log sub 8 ⁡ open parenthesis 2 close parenthesis equals 1")[$log_(8) ( x + 5 ) − log_(8) ( 2 ) = 1$]


        
          #math.equation(block: true, alt: "x equals 11")[$x = 11$]


        
      
      
        #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis x minus 1 close parenthesis minus log sub 10 ⁡ open parenthesis 4 close parenthesis equals 2")[$log_(10) ( x − 1 ) − log_(10) ( 4 ) = 2$]


      
      
        #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis x plus 2 close parenthesis plus log sub 10 ⁡ open parenthesis x minus 1 close parenthesis equals 1")[$log_(10) ( x + 2 ) + log_(10) ( x − 1 ) = 1$]


        
          #math.equation(block: true, alt: "x equals 3")[$x = 3$]


        
      
      
        #math.equation(block: true, alt: "log sub 4 ⁡ open parenthesis x plus 8 close parenthesis minus log sub 4 ⁡ open parenthesis x plus 2 close parenthesis equals 2")[$log_(4) ( x + 8 ) − log_(4) ( x + 2 ) = 2$]


      
      
        #math.equation(block: true, alt: "log sub 3 ⁡ open parenthesis x minus 2 close parenthesis minus log sub 3 ⁡ open parenthesis x plus 1 close parenthesis equals 3")[$log_(3) ( x − 2 ) − log_(3) ( x + 1 ) = 3$]


        
          No solution


        
      
      
        #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis x plus 3 close parenthesis minus log sub 10 ⁡ open parenthesis x minus 1 close parenthesis equals 1")[$log_(10) ( x + 3 ) − log_(10) ( x − 1 ) = 1$]


      
      For Problems 55–60, solve for the indicated variable.


      
        #math.equation(block: false, alt: "t equals T log sub 10 ⁡ open parenthesis 1 plus the fraction A over k close parenthesis")[$t = T log_(10) ( 1 + display(frac(A, k)) )$], for #math.equation(block: false, alt: "A")[$A$]


        
          #math.equation(block: true, alt: "A equals k open parenthesis 10 to the power t / T minus 1 close parenthesis")[$A = k ( 10^(t / T) − 1 )$]


        
      
      
        #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis R close parenthesis equals log sub 10 ⁡ open parenthesis R sub 0 close parenthesis plus k t")[$log_(10) ( R ) = log_(10) ( R_(0) ) + k t$], for #math.equation(block: false, alt: "R")[$R$]


      
      
        #math.equation(block: false, alt: "N equals N sub 0 log sub b ⁡ open parenthesis k s close parenthesis")[$N = N_(0) log_(b) ( k s )$], for #math.equation(block: false, alt: "s")[$s$]


        
          #math.equation(block: true, alt: "s equals the fraction b to the power N / N sub 0 over k")[$s = display(frac(b^(N / N_(0)), k))$]


        
      
      
        #math.equation(block: false, alt: "T equals the fraction H log sub 10 ⁡ open parenthesis the fraction N over N sub 0 close parenthesis over log sub 10 ⁡ open parenthesis the fraction 1 over 2 close parenthesis")[$T = display(frac(H log_(10) ( display(frac(N, N_(0))) ), log_(10) ( display(frac(1, 2)) )))$], for #math.equation(block: false, alt: "N")[$N$]


      
      
        #math.equation(block: false, alt: "M equals the square root of the fraction log sub 10 ⁡ open parenthesis H close parenthesis over k log sub 10 ⁡ open parenthesis H sub 0 close parenthesis")[$M = sqrt(display(frac(log_(10) ( H ), k log_(10) ( H_(0) ))))$], for #math.equation(block: false, alt: "H")[$H$]


        
          #math.equation(block: true, alt: "H equals open parenthesis H sub 0 close parenthesis to the power k M squared")[$H = ( H_(0) )^(k M^(2))$]


        
      
      
        #math.equation(block: false, alt: "h equals a minus the square root of the fraction log sub 10 ⁡ open parenthesis B close parenthesis over t")[$h = a − sqrt(display(frac(log_(10) ( B ), t)))$], for #math.equation(block: false, alt: "B")[$B$]


      
      
        Choose the graph for each function described below.


        + The area, #math.equation(block: false, alt: "A")[$A$], of a pentagon is a quadratic function of the length, #math.equation(block: false, alt: "l")[$l$], of its side.
+ The strength, #math.equation(block: false, alt: "F")[$F$], of a hurricane varies inversely with its speed, #math.equation(block: false, alt: "s")[$s$].
+ The price of food has increased by #math.equation(block: false, alt: "3 %")[$3 upright(%)$] every year for a decade.
+ The magnitude, #math.equation(block: false, alt: "M")[$M$], of a star is a logarithmic function of its brightness, #math.equation(block: false, alt: "I")[$I$].
+ The speed of the train increased at a constant rate.
+ If you do not practice a foreign language, you lose #math.equation(block: false, alt: "the fraction 1 over 8")[$frac(1, 8)$] of the words in your working vocabulary, #math.equation(block: false, alt: "V")[$V$], each year.


        #figure(figph[six curves], alt: "six curves", caption: none)


        
          + II
+ VI
+ III
+ V
+ I
+ IV


        
      
      
        For each of the functions listed below, select the graph of its inverse function, if possible, from the figures labeled I–VI. (The inverse of one of the functions is not shown.)


        + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$f ( x ) = 2^(x)$]
+ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared , x greater than or equal to 0")[$f ( x ) = x^(2) , " " " " x ≥ 0$]
+ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 2 over x")[$f ( x ) = display(frac(2, x))$]
+ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the square root of x")[$f ( x ) = sqrt(x)$]
+ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals log sub 2 ⁡ open parenthesis x close parenthesis")[$f ( x ) = log_(2) ( x )$]
+ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 2 close parenthesis to the power x")[$f ( x ) = attach(( display(frac(1, 2)) ), t: x)$]


        #figure(figph[two curves], alt: "two curves", caption: none)


        #figure(figph[two curves], alt: "two curves", caption: none)


        #figure(figph[two curves], alt: "two curves", caption: none)


      
      For Problems 63–64, graph the function on the domain #math.equation(block: false, alt: "open bracket minus 4 , 4 close bracket")[$[ − 4 , 4 ]$] and a suitable range. Which have inverses that are also functions?


      
        + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 5 open parenthesis 2 to the power minus x squared close parenthesis")[$f ( x ) = 5 ( 2^(− x^(2)) )$]
+ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x plus 2 to the power minus x")[$f ( x ) = 2^(x) + 2^(− x)$]


        
          + #figure(figph[bell], alt: "bell", caption: none)

 #linebreak() No inverse function
+ #figure(figph[catenary], alt: "catenary", caption: none)

 #linebreak() No inverse function


        
      
      
        + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 5 open parenthesis log ⁡ open parenthesis x close parenthesis close parenthesis squared plus 1")[$f ( x ) = 5 ( log ( x ) )^(2) + 1$]
+ #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 5 log ⁡ open parenthesis x squared plus 1 close parenthesis")[$f ( x ) = 5 log ( x^(2) + 1 )$]


      
      For Problems 65–68, graph the pair of functions on your calculator. Explain the result.


      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals log ⁡ open parenthesis 2 x close parenthesis , g open parenthesis x close parenthesis equals log ⁡ open parenthesis 2 close parenthesis plus log ⁡ open parenthesis x close parenthesis")[$f ( x ) = log ( 2 x ) , " " " " " " g ( x ) = log ( 2 ) + log ( x )$]


        
          #figure(figph[translated log], alt: "translated log", caption: none)


          The functions are equal.


        
      
      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals log ⁡ open parenthesis the fraction x over 3 close parenthesis , g open parenthesis x close parenthesis equals log ⁡ open parenthesis x close parenthesis minus log ⁡ open parenthesis 3 close parenthesis")[$f ( x ) = log ( display(frac(x, 3)) ) , " " " " " " g ( x ) = log ( x ) − log ( 3 )$]


      
      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals log ⁡ open parenthesis the fraction 1 over x close parenthesis , g open parenthesis x close parenthesis equals minus log ⁡ open parenthesis x close parenthesis")[$f ( x ) = log ( display(frac(1, x)) ) , " " " " " " g ( x ) = − log ( x )$]


        
          #figure(figph[translated log], alt: "translated log", caption: none)


          The functions are equal.


        
      
      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals log ⁡ open parenthesis x cubed close parenthesis , g open parenthesis x close parenthesis equals 3 log ⁡ x")[$f ( x ) = log ( x^(3) ) , " " " " " " g ( x ) = 3 log x$]


      
      
        + Complete the following table.#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "x squared")[$x^(2)$]], [#math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x close parenthesis")[$log_(10) ( x )$]], [#math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x squared close parenthesis")[$log_(10) ( x^(2) )$]]),
  [#math.equation(block: false, alt: "1")[$1$]], [$#hide($00000$)$], [$#hide($0000000$)$], [$#hide($0000000$)$],
  [#math.equation(block: false, alt: "2")[$2$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$],
  [#math.equation(block: false, alt: "3")[$3$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$],
  [#math.equation(block: false, alt: "4")[$4$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$],
  [#math.equation(block: false, alt: "5")[$5$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$],
  [#math.equation(block: false, alt: "6")[$6$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$],
))
+ Do you notice a relationship between #math.equation(block: false, alt: "log sub 10 ⁡ x")[$log_(10) x$] and #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x squared close parenthesis")[$log_(10) ( x^(2) )$]? State the relationship as an equation.


        
          + #figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "x squared")[$x^(2)$]], [#math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x close parenthesis")[$log_(10) ( x )$]], [#math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x squared close parenthesis")[$log_(10) ( x^(2) )$]]),
  [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]],
  [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "0.301")[$0.301$]], [#math.equation(block: false, alt: "0.602")[$0.602$]],
  [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "0.477")[$0.477$]], [#math.equation(block: false, alt: "0.954")[$0.954$]],
  [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "0.602")[$0.602$]], [#math.equation(block: false, alt: "1.204")[$1.204$]],
  [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "0.699")[$0.699$]], [#math.equation(block: false, alt: "1.398")[$1.398$]],
  [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "36")[$36$]], [#math.equation(block: false, alt: "0.778")[$0.778$]], [#math.equation(block: false, alt: "1.556")[$1.556$]],
))
+ #math.equation(block: false, alt: "log sub 10 ⁡ x squared equals 2 log sub 10 ⁡ open parenthesis x close parenthesis")[$log_(10) x^(2) = 2 log_(10) ( x )$]


        
      
      
        + Complete the following table.#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction 1 over x")[$display(frac(1, x))$]], [#math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x close parenthesis")[$log_(10) ( x )$]], [#math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis the fraction 1 over x close parenthesis")[$log_(10) ( display(frac(1, x)) )$]]),
  [#math.equation(block: false, alt: "1")[$1$]], [$#hide($00000$)$], [$#hide($0000000$)$], [$#hide($0000000$)$],
  [#math.equation(block: false, alt: "2")[$2$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$],
  [#math.equation(block: false, alt: "3")[$3$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$],
  [#math.equation(block: false, alt: "4")[$4$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$],
  [#math.equation(block: false, alt: "5")[$5$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$],
  [#math.equation(block: false, alt: "6")[$6$]], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$],
))
+ Do you notice a relationship between #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x close parenthesis")[$log_(10) ( x )$] and #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis the fraction 1 over x close parenthesis")[$log_(10) ( display(frac(1, x)) )$]? State the relationship as an equation.


      
      In Problems 71 and 72, you found relationships between #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x close parenthesis")[$log_(10) ( x )$] and #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x squared close parenthesis")[$log_(10) ( x^(2) )$], and between #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x close parenthesis")[$log_(10) ( x )$] and #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis the fraction 1 over x close parenthesis")[$log_(10) ( display(frac(1, x)) )$]. Assuming that those relationships hold for any base, complete the following tables and use them to graph the given functions.


      
        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y equals log sub e ⁡ open parenthesis x close parenthesis")[$y = log_(e) ( x )$]]),
  [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]],
  [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "0.693")[$0.693$]],
  [#math.equation(block: false, alt: "4")[$4$]], [],
  [#math.equation(block: false, alt: "16")[$16$]], [],
  [#math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$]], [],
  [#math.equation(block: false, alt: "the fraction 1 over 4")[$frac(1, 4)$]], [],
  [#math.equation(block: false, alt: "the fraction 1 over 16")[$frac(1, 16)$]], [],
))


        
          #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y equals log sub e ⁡ open parenthesis x close parenthesis")[$y = log_(e) ( x )$]]),
  [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]],
  [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "0.693")[$0.693$]],
  [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "1.386")[$1.386$]],
  [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "2.772")[$2.772$]],
  [#math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$]], [#math.equation(block: false, alt: "minus 0.693")[$− 0.693$]],
  [#math.equation(block: false, alt: "the fraction 1 over 4")[$frac(1, 4)$]], [#math.equation(block: false, alt: "minus 1.386")[$− 1.386$]],
  [#math.equation(block: false, alt: "the fraction 1 over 16")[$frac(1, 16)$]], [#math.equation(block: false, alt: "minus 2.772")[$− 2.772$]],
))


          #figure(figph[log base e], alt: "log base e", caption: none)


        
      
      
        #figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y equals log sub f ⁡ open parenthesis x close parenthesis")[$y = log_(f) ( x )$]]),
  [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]],
  [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "0.431")[$0.431$]],
  [#math.equation(block: false, alt: "4")[$4$]], [],
  [#math.equation(block: false, alt: "16")[$16$]], [],
  [#math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$]], [],
  [#math.equation(block: false, alt: "the fraction 1 over 4")[$frac(1, 4)$]], [],
  [#math.equation(block: false, alt: "the fraction 1 over 16")[$frac(1, 16)$]], [],
))


      
    
    
      
=== Investigation


      #notebox("Investigation", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Interest Compounded Continuously]


        We learned in Properties of Logarithms that the amount, #math.equation(block: false, alt: "A")[$A$] (principal plus interest), accumulated in an account with interest compounded #math.equation(block: false, alt: "n")[$n$] times annually is


        #math.equation(block: true, alt: "A equals P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t")[$A = P attach(( 1 + display(frac(r, n)) ), t: n t)$]
        where #math.equation(block: false, alt: "P")[$P$] is the principal invested, #math.equation(block: false, alt: "r")[$r$] is the interest rate, and #math.equation(block: false, alt: "t")[$t$] is the time period, in years.


        + Suppose you keep #math.equation(block: false, alt: "$ 1000")[$upright(\$) 1000$] in an account that pays #math.equation(block: false, alt: "8 %")[$8 upright(%)$] interest. How much is the amount #math.equation(block: false, alt: "A")[$A$] after #math.equation(block: false, alt: "1")[$1$] year if the interest is compounded twice a year? Four times a year?
			#math.equation(block: true, alt: "n equals 2 : A equals 1000 open parenthesis 1 plus the fraction 0.08 over 2 close parenthesis to the power 2 open parenthesis 1 close parenthesis equals; n equals 4 : A equals 1000 open parenthesis 1 plus the fraction 0.08 over 4 close parenthesis to the power 4 open parenthesis 1 close parenthesis equals")[$n = 2 : A = 1000 attach(( 1 + frac(0.08, 2) ), t: 2 ( 1 )) = \ n = 4 : A = 1000 attach(( 1 + frac(0.08, 4) ), t: 4 ( 1 )) =$]
+ What happens to #math.equation(block: false, alt: "A")[$A$] as we increase #math.equation(block: false, alt: "n")[$n$], the number of compounding periods per year? Fill in the table showing the amount in the account for different values of #math.equation(block: false, alt: "n")[$n$].#figure(table(
  columns: 2, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "A")[$A$]]),
  [#math.equation(block: false, alt: "1")[$1$] (annually)], [#math.equation(block: false, alt: "1080")[$1080$]],
  [#math.equation(block: false, alt: "2")[$2$] (semiannually)], [],
  [#math.equation(block: false, alt: "4")[$4$] (quarterly)], [],
  [#math.equation(block: false, alt: "6")[$6$] (bimonthly)], [],
  [#math.equation(block: false, alt: "12")[$12$] (monthly)], [],
  [#math.equation(block: false, alt: "365")[$365$] (daily)], [],
  [#math.equation(block: false, alt: "1000")[$1000$]], [],
  [#math.equation(block: false, alt: "10 , 000")[$10 , 000$]], [],
))

#figure(figph[grid], alt: "grid", caption: none)
+ Plot the values in the table from #math.equation(block: false, alt: "n equals 1")[$n = 1$] to #math.equation(block: false, alt: "n equals 12")[$n = 12$], and connect them with a smooth curve. Describe the curve: What is happening to the value of #math.equation(block: false, alt: "A")[$A$]?
+ In part (2), as you increased the value of #math.equation(block: false, alt: "n")[$n$], the other parameters in the formula stayed the same. In other words, #math.equation(block: false, alt: "A")[$A$] is a function of #math.equation(block: false, alt: "n")[$n$], given by #math.equation(block: false, alt: "A equals 1000 open parenthesis 1 plus the fraction 0.08 over n close parenthesis to the power n")[$A = 1000 attach(( 1 + display(frac(0.08, n)) ), t: n)$]. Use your calculator to graph A on successively larger domains:+ #math.equation(block: false, alt: "Xmin equals 0 , Xmax equals 12 ; Ymin equals 1080 , Ymax equals 1084")[$"Xmin" = 0 , "Xmax" = 12 ; "Ymin" = 1080 , "Ymax" = 1084$]
+ #math.equation(block: false, alt: "Xmin equals 0 , Xmax equals 50 ; Ymin equals 1080 , Ymax equals 1084")[$"Xmin" = 0 , "Xmax" = 50 ; "Ymin" = 1080 , "Ymax" = 1084$]
+ #math.equation(block: false, alt: "Xmin equals 0 , Xmax equals 365 ; Ymin equals 1080 , Ymax equals 1084")[$"Xmin" = 0 , "Xmax" = 365 ; "Ymin" = 1080 , "Ymax" = 1084$]
+ Use the #strong[Trace] feature or the #strong[Table] feature to evaluate #math.equation(block: false, alt: "A")[$A$] for very large values of #math.equation(block: false, alt: "n")[$n$]. Rounded to the nearest penny, what is the largest value of #math.equation(block: false, alt: "A")[$A$] that you can find?
+ As #math.equation(block: false, alt: "n")[$n$] increases, the values of #math.equation(block: false, alt: "A")[$A$] approach a limiting value. Although #math.equation(block: false, alt: "A")[$A$] continues to increase, it does so by smaller and smaller increments and will never exceed #math.equation(block: false, alt: "$ 1083.29")[$upright(\$) 1083.29$]. When the number of compounding periods increases without bound, we call the limiting result #strong[continuous compounding].
+ Is there an easier way to compute #math.equation(block: false, alt: "A")[$A$] under continues compounding? Yes! Compute #math.equation(block: false, alt: "1000 e to the power 0.08")[$1000 e^(0.08)$] on your calculator. (Press 2nd  LN to enter #math.equation(block: false, alt: "e to the power x")[$e^(x)$].) Compare the value to your answer in part (5) for the limiting value. The number #math.equation(block: false, alt: "e")[$e$] is called the #strong[natural base]. We'll compute its value shortly.
+ Repeat your calculations for two other interest rates, #math.equation(block: false, alt: "15 %")[$15 upright(%)$] and (an extremely unrealistic) #math.equation(block: false, alt: "100 %")[$100 upright(%)$], again for an investment of #math.equation(block: false, alt: "$ 1000")[$upright(\$) 1000$] for #math.equation(block: false, alt: "1")[$1$] year. In each case, compare the limiting value of #math.equation(block: false, alt: "A")[$A$], and compare to the value of #math.equation(block: false, alt: "1000 e to the power r")[$1000 e^(r)$].
			+ #figure(table(
  columns: 2, align: left, inset: 6pt,
  [#math.equation(block: false, alt: "r equals 0.15")[$r = 0.15$]], [],
  [#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "A")[$A$]],
  [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "115")[$115$]],
  [#math.equation(block: false, alt: "2")[$2$]], [$#hide($00000$)$],
  [#math.equation(block: false, alt: "4")[$4$]], [],
  [#math.equation(block: false, alt: "6")[$6$]], [],
  [#math.equation(block: false, alt: "12")[$12$]], [],
  [#math.equation(block: false, alt: "3652")[$3652$]], [],
  [#math.equation(block: false, alt: "1000")[$1000$]], [],
  [#math.equation(block: false, alt: "10 , 000")[$10 , 000$]], [],
  [], [],
  [#math.equation(block: false, alt: "1000 e to the power 0.15 equals")[$" " " " 1000 " " e^(0.15) =$]], [],
))
+ #figure(table(
  columns: 2, align: left, inset: 6pt,
  [#math.equation(block: false, alt: "r equals 1")[$r = 1$]], [],
  [#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "A")[$A$]],
  [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "200")[$200$]],
  [#math.equation(block: false, alt: "2")[$2$]], [$#hide($00000$)$],
  [#math.equation(block: false, alt: "4")[$4$]], [],
  [#math.equation(block: false, alt: "6")[$6$]], [],
  [#math.equation(block: false, alt: "12")[$12$]], [],
  [#math.equation(block: false, alt: "3652")[$3652$]], [],
  [#math.equation(block: false, alt: "1000")[$1000$]], [],
  [#math.equation(block: false, alt: "10 , 000")[$10 , 000$]], [],
  [], [],
  [#math.equation(block: false, alt: "1000 e to the power 1 equals")[$" " " " 1000 " " e^(1) =$]], [],
))
+ In part (8b), you have computed an approximation for #math.equation(block: false, alt: "1000 e")[$1000 e$]. What is the value of #math.equation(block: false, alt: "e")[$e$], rounded to #math.equation(block: false, alt: "5")[$5$] decimal places?
+ Complete the table of values. What does #math.equation(block: false, alt: "open parenthesis 1 plus the fraction 1 over n close parenthesis to the power n")[$attach(( 1 + display(frac(1, n)) ), t: n)$] appear to approach as #math.equation(block: false, alt: "n")[$n$] increases?#figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "1000")[$1000$]], [#math.equation(block: false, alt: "10 , 000")[$10 , 000$]], [#math.equation(block: false, alt: "100 , 000")[$100 , 000$]]),
  [#math.equation(block: false, alt: "open parenthesis 1 plus the fraction 1 over n close parenthesis to the power n")[$attach(( 1 + display(frac(1, n)) ), t: n)$]], [], [], [], [],
))
]