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== 4.3#h(0.6em)Logarithms


    
      
=== Introduction


      In this section, we introduce a new mathematical tool called a #strong[logarithm], which will help us solve exponential equations.


      Suppose that a colony of bacteria doubles in size every day. If the colony starts with #math.equation(block: false, alt: "50")[$50$] bacteria, how long will it be before there are #math.equation(block: false, alt: "800")[$800$] bacteria? We answered questions of this type in Exponential Functions by writing and solving an exponential equation. The function


      #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 50 times 2 to the power t")[$P ( t ) = 50 ⋅ 2^(t)$]
      gives the number of bacteria present on day #math.equation(block: false, alt: "t")[$t$], so we must solve the equation


      #math.equation(block: true, alt: "800 equals 50 times 2 to the power t")[$800 = 50 ⋅ 2^(t)$]
      Dividing both sides by 50 yields


      #math.equation(block: true, alt: "16 equals 2 to the power t")[$16 = 2^(t)$]
      The solution of this equation is the answer to the following question:


      To what power must we raise #math.equation(block: false, alt: "2")[$2$] in order to get #math.equation(block: false, alt: "16")[$16$]?


      The value of #math.equation(block: false, alt: "t")[$t$] that solves the equation is called the #strong[base] #math.equation(block: false, alt: "2")[$2$] #strong[logarithm] of #math.equation(block: false, alt: "16")[$16$]. Because #math.equation(block: false, alt: "2 to the power 4 equals 16")[$2^(4) = 16$], the base #math.equation(block: false, alt: "2")[$2$] logarithm of #math.equation(block: false, alt: "16")[$16$] is #math.equation(block: false, alt: "4")[$4$]. We write this as


      #math.equation(block: true, alt: "log sub 2 ⁡ open parenthesis 16 close parenthesis equals 4")[$log_(2) ( 16 ) = 4$]
      In other words, we solve an exponential equation by computing a logarithm. You can check that #math.equation(block: false, alt: "t equals 4")[$t = 4$]
	solves the problem stated above:


      #math.equation(block: true, alt: "P open parenthesis 4 close parenthesis equals 50 times 2 to the power 4 equals 800")[$P ( 4 ) = 50 ⋅ 2^(4) = 800$]
      Thus, the unknown exponent is called a logarithm. In general, for positive values of #math.equation(block: false, alt: "b")[$b$] and #math.equation(block: false, alt: "x")[$x$], we make the following definition.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Definition of Logarithm]


        For #math.equation(block: false, alt: "b greater than 0 , b not equal to 1")[$b > 0 , b ≠ 1$], the #strong[base #math.equation(block: false, alt: "b")[$b$] logarithm of #math.equation(block: false, alt: "x")[$x$]],
	
	written #math.equation(block: false, alt: "log sub b ⁡ open parenthesis x close parenthesis")[$log_(b) ( x )$] (or also written #math.equation(block: false, alt: "log sub b ⁡ x")[$log_(b) x$]), is the exponent to which #math.equation(block: false, alt: "b")[$b$] must be raised in order to yield #math.equation(block: false, alt: "x")[$x$].
]


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
It will help to keep in mind that a logarithm is just an exponent.
]


      Some logarithms, like some square roots, are easy to evaluate, while others require a calculator. We will start with the easy ones.


      #examplebox("Example 1")[][
Compute the logarithms.


          + #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 9 close parenthesis")[$log_(3) ( 9 )$]
+ #math.equation(block: false, alt: "log sub 5 ⁡ open parenthesis 125 close parenthesis")[$log_(5) ( 125 )$]
+ #math.equation(block: false, alt: "log sub 4 ⁡ open parenthesis the fraction 1 over 16 close parenthesis")[$log_(4) ( display(frac(1, 16)) )$]
+ #math.equation(block: false, alt: "log sub 5 ⁡ open parenthesis the square root of 5 close parenthesis")[$log_(5) ( sqrt(5) )$]


        
        #solutionbox[
+ To evaluate #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 9 close parenthesis")[$log_(3) ( 9 )$], we ask what exponent on base #math.equation(block: false, alt: "3")[$3$] will produce #math.equation(block: false, alt: "9")[$9$]. Or, 3 to what power equals 9? The exponent we need is #math.equation(block: false, alt: "2")[$2$], so
        #math.equation(block: true, alt: "log sub 3 ⁡ open parenthesis 9 close parenthesis equals 2 because 3 squared equals 9")[$log_(3) ( 9 ) = 2 " " " " " " "because" " " " " " " 3^(2) = 9$]
    We use similar reasoning to compute the other logarithms.
+ #math.equation(block: false, alt: "log sub 5 ⁡ open parenthesis 125 close parenthesis equals 3 because 5 cubed equals 125")[$log_(5) ( 125 ) = 3 " " " " " because " " " " " 5^(3) = 125$]
+ #math.equation(block: false, alt: "log sub 4 ⁡ open parenthesis the fraction 1 over 16 close parenthesis equals minus 2 because 4 to the power minus 2 equals the fraction 1 over 16")[$log_(4) ( display(frac(1, 16)) ) = − 2 " " " " " because " " " " " 4^(− 2) = display(frac(1, 16))$]
+ #math.equation(block: false, alt: "log sub 5 ⁡ open parenthesis the square root of 5 close parenthesis equals the fraction 1 over 2 because 5 to the power 1 / 2 equals the square root of 5")[$log_(5) ( sqrt(5) ) = display(frac(1, 2)) " " " " " because " " " " " 5^(1 / 2) = sqrt(5)$]
]
]


      
        Find each logarithm.


        + #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 81 close parenthesis equals")[$log_(3) ( 81 ) =$]\_\_\_\_\_
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis the fraction 1 over 1000 close parenthesis equals")[$log_(10) ( display(frac(1, 1000)) ) =$]\_\_\_\_\_


        #solutionbox[
+ #math.equation(block: false, alt: "4")[$4$]
+ #math.equation(block: false, alt: "minus 3")[$− 3$]
]


      
      
        Find each logarithm.


        + #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 81 close parenthesis")[$log_(3) ( 81 )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis the fraction 1 over 1000 close parenthesis")[$log_(10) ( display(frac(1, 1000)) )$]


        #solutionbox[
+ #math.equation(block: false, alt: "4")[$4$]
+ #math.equation(block: false, alt: "minus 3")[$− 3$]
]


      
      
        A logarithm is the same as


        \_\_\_\_\_


        #solutionbox[
an exponent
]


      
      
        A logarithm is the same as


        + an exponent.
+ a coefficient.
+ a quotient.
+ a radical.


      
      From the definition of a logarithm and the examples above, we see that the following two statements are equivalent.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Logarithms and Exponents: Conversion Equations]


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


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


      In other words, the logarithm, #math.equation(block: false, alt: "y")[$y$], is the same as the #emph[exponent] in #math.equation(block: false, alt: "x equals b to the power y")[$x = b^(y)$]. We see again that #emph[a logarithm is an exponent]; it is the exponent to which #math.equation(block: false, alt: "b")[$b$] must be raised to yield #math.equation(block: false, alt: "x")[$x$].


      These equations allow us to convert from logarithmic to exponential form, or vice versa. You should memorize the conversion equations, because we will use them frequently.


      As special cases of the equivalence in (1), we can compute the following useful logarithms.
For any base #math.equation(block: false, alt: "b greater than 0 , b not equal to 1")[$b > 0 , b ≠ 1$],


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Some Useful Logarithms]


        #math.equation(block: true, alt: "log sub b ⁡ open parenthesis b close parenthesis, equals 1 because b to the power 1 equals b; log sub b ⁡ open parenthesis 1 close parenthesis, equals 0 because b to the power 0 equals 1; log sub b ⁡ open parenthesis b to the power x close parenthesis, equals x because b to the power x equals b to the power x")[$log_(b) ( b ) & = 1 " " " " " " " because " " " " " " " b^(1) = b \ log_(b) ( 1 ) & = 0 " " " " " " " because " " " " " " " b^(0) = 1 \ log_(b) ( b^(x) ) & = x " " " " " " " because " " " " " " " b^(x) = b^(x)$]
]


      #examplebox("Example 2")[][
+ #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis 2 close parenthesis equals 1")[$log_(2) ( 2 ) = 1$]
+ #math.equation(block: false, alt: "log sub 5 ⁡ open parenthesis 1 close parenthesis equals 0")[$log_(5) ( 1 ) = 0$]
+ #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 3 to the power 4 close parenthesis equals 4")[$log_(3) ( 3^(4) ) = 4$]
]


      
        Find each logarithm.


        + #math.equation(block: false, alt: "log sub n ⁡ open parenthesis 1 close parenthesis equals")[$log_(n) ( 1 ) =$]\_\_\_\_\_
+ #math.equation(block: false, alt: "log sub n ⁡ open parenthesis n cubed close parenthesis equals")[$log_(n) ( n^(3) ) =$]\_\_\_\_\_


        #solutionbox[
+ #math.equation(block: false, alt: "0")[$0$]
+ #math.equation(block: false, alt: "3")[$3$]
]


      
      
        Find each logarithm.


        + #math.equation(block: false, alt: "log sub n ⁡ open parenthesis 1 close parenthesis")[$log_(n) ( 1 )$]
+ #math.equation(block: false, alt: "log sub n ⁡ open parenthesis n cubed close parenthesis")[$log_(n) ( n^(3) )$]


        #solutionbox[
+ #math.equation(block: false, alt: "0")[$0$]
+ #math.equation(block: false, alt: "3")[$3$]
]


      
      
        #math.equation(block: false, alt: "log sub b ⁡ b equals 1")[$log_(b) b = 1$] because


        \_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "b to the power 1 equals b")[$b^(1) = b$]
]


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


        + #math.equation(block: false, alt: "the fraction b over b equals 1")[$display(frac(b, b)) = 1$]
+ #math.equation(block: false, alt: "the bth root of b equals 1")[$root(b, b) = 1$]
+ #math.equation(block: false, alt: "b to the power 1 equals b")[$b^(1) = b$]
+ #math.equation(block: false, alt: "1 to the power b equals 1")[$1^(b) = 1$]


      
    
    
      
=== Using the Conversion Equations


      We use logarithms to solve exponential equations, just as we use square roots to solve quadratic equations. Consider the two equations


      #math.equation(block: true, alt: "x squared equals 25 and 2 to the power x equals 8")[$x^(2) = 25 " " " " " " " " " and " " " " " " " " " 2^(x) = 8$]
      We solve the first equation by taking a square root, and we solve the second equation by computing a logarithm:


      #math.equation(block: true, alt: "x equals plus or minus the square root of 25 equals plus or minus 5 and x equals log sub 2 ⁡ open parenthesis 8 close parenthesis equals 3")[$x = ± sqrt(25) = ± 5 " " " " " " " " " and " " " " " " " " " x = log_(2) ( 8 ) = 3$]
      The operation of taking a base #math.equation(block: false, alt: "b")[$b$] logarithm is the inverse operation for raising the base #math.equation(block: false, alt: "b")[$b$] to a power, just as extracting square roots is the inverse of squaring a number.


      Every exponential equation can be rewritten in logarithmic form by using the conversion equations. Thus,


      #math.equation(block: true, alt: "3 equals log sub 2 ⁡ open parenthesis 8 close parenthesis and 8 equals 2 cubed")[$3 = log_(2) ( 8 ) " " " " " " " " " and " " " " " " " " " 8 = 2^(3)$]
      are equivalent statements, just as


      #math.equation(block: true, alt: "5 equals the square root of 25 and 25 equals 5 squared")[$5 = sqrt(25) " " " " " " " " " and " " " " " " " " " 25 = 5^(2)$]
      are equivalent statements. Rewriting an equation in logarithmic form is a basic  strategy for finding its solution.


      
        To find #math.equation(block: false, alt: "log sub 5 ⁡ 31")[$log_(5) 31$] means to find an exponent #math.equation(block: false, alt: "x")[$x$] that satisfies the equation


        \_\_\_\_\_


        #solutionbox[
#math.equation(block: true, alt: "5 to the power x equals 31")[$5^(x) = 31$]
]


      
      
        To find #math.equation(block: false, alt: "log sub 5 ⁡ open parenthesis 31 close parenthesis")[$log_(5) ( 31 )$] means to find an exponent #math.equation(block: false, alt: "x")[$x$] that satisfies the equation


        + #math.equation(block: false, alt: "x to the power 5 equals 31")[$x^(5) = 31$]
+ #math.equation(block: false, alt: "the 5th root of 31 equals x")[$root(5, 31) = x$]
+ #math.equation(block: false, alt: "the 5th root of x equals 31")[$root(5, x) = 31$]
+ #math.equation(block: false, alt: "5 to the power x equals 31")[$5^(x) = 31$]


      
      #examplebox("Example 3")[][
Rewrite each equation in logarithmic form.


          + #math.equation(block: false, alt: "2 to the power minus 1 equals the fraction 1 over 2")[$2^(− 1) = display(frac(1, 2))$]
+ #math.equation(block: false, alt: "a to the power 1 / 5 equals 2.8")[$a^(1 / 5) = 2.8$]
+ #math.equation(block: false, alt: "6 to the power 1.5 equals T")[$6^(1.5) = T$]
+ #math.equation(block: false, alt: "M to the power v equals 3 K")[$M^(v) = 3 K$]


        
        #solutionbox[
First identify the base #math.equation(block: false, alt: "b")[$b$], and then the exponent or logarithm #math.equation(block: false, alt: "y")[$y$]. Use the conversion equations to rewrite #math.equation(block: false, alt: "b to the power y equals x")[$b^(y) = x$] in the form #math.equation(block: false, alt: "log sub b ⁡ open parenthesis x close parenthesis equals y")[$log_(b) ( x ) = y$].


          + The base is #math.equation(block: false, alt: "2")[$2$] and the exponent is #math.equation(block: false, alt: "minus 1")[$− 1$]. Thus, #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis the fraction 1 over 2 close parenthesis equals minus 1")[$log_(2) ( display(frac(1, 2)) ) = − 1$].
+ The base is #math.equation(block: false, alt: "a")[$a$] and the exponent is #math.equation(block: false, alt: "the fraction 1 over 5")[$display(frac(1, 5))$]. Thus, #math.equation(block: false, alt: "log sub a ⁡ open parenthesis 2.8 close parenthesis equals the fraction 1 over 5")[$log_(a) ( 2.8 ) = display(frac(1, 5))$].
+ The base is #math.equation(block: false, alt: "6")[$6$] and the exponent is #math.equation(block: false, alt: "1.5")[$1.5$]. Thus, #math.equation(block: false, alt: "log sub 6 ⁡ open parenthesis T close parenthesis equals 1.5")[$log_(6) ( T ) = 1.5$].
+ The base is #math.equation(block: false, alt: "M")[$M$] and the exponent is #math.equation(block: false, alt: "v")[$v$]. Thus, #math.equation(block: false, alt: "log sub M ⁡ open parenthesis 3 K close parenthesis equals v")[$log_(M) ( 3 K ) = v$].
]
]


      
        Rewrite each equation in logarithmic form.


        + #math.equation(block: false, alt: "8 to the power minus 1 / 3 equals the fraction 1 over 2")[$8^(− 1 / 3) = display(frac(1, 2))$] #linebreak() The given equation is equivalent to one of the form #math.equation(block: false, alt: "log sub b ⁡ open parenthesis m close parenthesis equals n")[$log_(b) ( m ) = n$], where #linebreak() #math.equation(block: false, alt: "b equals")[$b =$]\_\_\_\_\_ #linebreak() #math.equation(block: false, alt: "m equals")[$m =$]\_\_\_\_\_, and #linebreak() #math.equation(block: false, alt: "n equals")[$n =$]\_\_\_\_\_
+ #math.equation(block: false, alt: "5 to the power x equals 46")[$5^(x) = 46$] #linebreak() The given equation is equivalent to one of the form #math.equation(block: false, alt: "log sub b ⁡ m equals n")[$log_(b) m = n$], where #linebreak() #math.equation(block: false, alt: "b equals")[$b =$]\_\_\_\_\_ #linebreak() #math.equation(block: false, alt: "m equals")[$m =$]\_\_\_\_\_, and #linebreak() #math.equation(block: false, alt: "n equals")[$n =$]\_\_\_\_\_


        #solutionbox[
+ #math.equation(block: false, alt: "log sub 8 ⁡ open parenthesis the fraction 1 over 2 close parenthesis equals the fraction minus 1 over 3")[$log_(8) ( display(frac(1, 2)) ) = display(frac(− 1, 3))$]
+ #math.equation(block: false, alt: "log sub 5 ⁡ open parenthesis 46 close parenthesis equals x")[$log_(5) ( 46 ) = x$]
]


      
      
        Rewrite each equation in logarithmic form.


        + #math.equation(block: false, alt: "8 to the power minus 1 / 3 equals the fraction 1 over 2")[$8^(− 1 / 3) = display(frac(1, 2))$]
+ #math.equation(block: false, alt: "5 to the power x equals 46")[$5^(x) = 46$]


        #solutionbox[
+ #math.equation(block: false, alt: "log sub 8 ⁡ open parenthesis the fraction 1 over 2 close parenthesis equals the fraction minus 1 over 3")[$log_(8) ( display(frac(1, 2)) ) = display(frac(− 1, 3))$]
+ #math.equation(block: false, alt: "log sub 5 ⁡ open parenthesis 46 close parenthesis equals x")[$log_(5) ( 46 ) = x$]
]


      
      
        If #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis x close parenthesis")[$log_(2) ( x )$] is negative, what can you say about #math.equation(block: false, alt: "x")[$x$]?


        \_\_\_\_\_


      
      
        If #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis x close parenthesis")[$log_(2) ( x )$] is negative, what can you say about #math.equation(block: false, alt: "x")[$x$]?


      
    
    
      
=== Approximating Logarithms


      Suppose we would like to solve the equation


      #math.equation(block: true, alt: "2 to the power x equals 26")[$2^(x) = 26$]
      The solution of this equation is #math.equation(block: false, alt: "x equals log sub 2 ⁡ open parenthesis 26 close parenthesis")[$x = log_(2) ( 26 )$], but can we find a decimal approximation for this value? There is no integer power of #math.equation(block: false, alt: "2")[$2$] that equals #math.equation(block: false, alt: "26")[$26$], because


      #math.equation(block: true, alt: "2 to the power 4, equals 16; and 2 to the power 5, equals 32")[$2^(4) & = 16 \ "and " " " " " " " " " 2^(5) & = 32$]
      Thus, #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis 26 close parenthesis")[$log_(2) ( 26 )$] must be between #math.equation(block: false, alt: "4")[$4$] and #math.equation(block: false, alt: "5")[$5$]. We can use trial and error to find the value of #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis 26 close parenthesis")[$log_(2) ( 26 )$] to the nearest tenth. Use your calculator to make a table of values for #math.equation(block: false, alt: "y equals 2 to the power x")[$y = 2^(x)$], starting with #math.equation(block: false, alt: "x equals 4")[$x = 4$] and using increments of #math.equation(block: false, alt: "0.1")[$0.1$].


      #figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "2 to the power x")[$2^(x)$]], [], [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "2 to the power x")[$2^(x)$]]),
  [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "2 to the power 4 equals 16")[$2^(4) = 16$]], [], [#math.equation(block: false, alt: "4.5")[$4.5$]], [#math.equation(block: false, alt: "2 to the power 4.5 equals 22.627")[$2^(4.5) = 22.627$]],
  [#math.equation(block: false, alt: "4.1")[$4.1$]], [#math.equation(block: false, alt: "2 to the power 4.1 equals 17.148")[$2^(4.1) = 17.148$]], [], [#math.equation(block: false, alt: "4.6")[$4.6$]], [#math.equation(block: false, alt: "2 to the power 4.6 equals 24.251")[$2^(4.6) = 24.251$]],
  [#math.equation(block: false, alt: "4.2")[$4.2$]], [#math.equation(block: false, alt: "2 to the power 4.2 equals 18.379")[$2^(4.2) = 18.379$]], [], [#math.equation(block: false, alt: "4.7")[$4.7$]], [#math.equation(block: false, alt: "2 to the power 4.7 equals 25.992")[$2^(4.7) = 25.992$]],
  [#math.equation(block: false, alt: "4.3")[$4.3$]], [#math.equation(block: false, alt: "2 to the power 4.3 equals 19.698")[$2^(4.3) = 19.698$]], [], [#math.equation(block: false, alt: "4.8")[$4.8$]], [#math.equation(block: false, alt: "2 to the power 4.8 equals 27.858")[$2^(4.8) = 27.858$]],
  [#math.equation(block: false, alt: "4.4")[$4.4$]], [#math.equation(block: false, alt: "2 to the power 4.4 equals 21.112")[$2^(4.4) = 21.112$]], [], [#math.equation(block: false, alt: "4.9")[$4.9$]], [#math.equation(block: false, alt: "2 to the power 4.9 equals 29.857")[$2^(4.9) = 29.857$]],
))


      From the table we see that #math.equation(block: false, alt: "26")[$26$] is between #math.equation(block: false, alt: "2 to the power 4.7")[$2^(4.7)$] and #math.equation(block: false, alt: "2 to the power 4.8")[$2^(4.8)$], and is closer to #math.equation(block: false, alt: "2 to the power 4.7")[$2^(4.7)$]. To the nearest tenth, #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis 26 close parenthesis approximately equals 4.7")[$log_(2) ( 26 ) ≈ 4.7$].


      Trial and error can be a time-consuming process. In Example 4, we illustrate a graphical method for estimating the value of a logarithm.


      #examplebox("Example 4")[][
Approximate #math.equation(block: false, alt: "log sub 3 ⁡ 7")[$log_(3) 7$] to the nearest hundredth.


        
        #solutionbox[
If #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 7 close parenthesis equals x")[$log_(3) ( 7 ) = x$], then #math.equation(block: false, alt: "3 to the power x equals 7")[$3^(x) = 7$]. We will use the graph of #math.equation(block: false, alt: "y equals 3 to the power x")[$y = 3^(x)$] to approximate a solution to #math.equation(block: false, alt: "3 to the power x equals 7")[$3^(x) = 7$].


          We graph #math.equation(block: false, alt: "Y sub 1 equals 3")[$Y_(1) = 3$]^ X and #math.equation(block: false, alt: "Y sub 2 equals 7")[$Y_(2) = 7$] in the standard window (ZOOM 6) to obtain the graph shown below. Next we activate the intersect feature to find that the two graphs intersect at the point #math.equation(block: false, alt: "open parenthesis 1.7712437 , 7 close parenthesis")[$( 1.7712437 , 7 )$]. Because this point lies on the graph of #math.equation(block: false, alt: "y equals 3 to the power x")[$y = 3^(x)$] , we know that


          #math.equation(block: true, alt: "3 to the power 1.7712437 approximately equals 7 , or log sub 3 ⁡ open parenthesis 7 close parenthesis approximately equals 1.7712437")[$3^(1.7712437) ≈ 7 " " " " " " ", or " " " " " " " log_(3) ( 7 ) ≈ 1.7712437$]
          To the nearest hundredth, #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 7 close parenthesis approximately equals 1.77")[$log_(3) ( 7 ) ≈ 1.77$].


          #figure(figph[GC intersection of exponential curve and horizontal line], alt: "GC intersection of exponential curve and horizontal line", caption: none)
]
]


      
        + Rewrite the equation #math.equation(block: false, alt: "3 to the power x equals 90")[$3^(x) = 90$] in logarithmic form. #linebreak() The given equation is equivalent to one of the form #math.equation(block: false, alt: "log sub b ⁡ m equals n")[$log_(b) m = n$], where #linebreak() #math.equation(block: false, alt: "b equals")[$b =$]\_\_\_\_\_ #linebreak() #math.equation(block: false, alt: "m equals")[$m =$]\_\_\_\_\_, and #linebreak() #math.equation(block: false, alt: "n equals")[$n =$]\_\_\_\_\_
+ Use a graph to approximate the solution to the equation in part (a). Round your answer to three decimal places. #linebreak() #math.equation(block: false, alt: "x approximately equals")[$x ≈$]\_\_\_\_\_


        #solutionbox[
+ #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 90 close parenthesis equals x")[$log_(3) ( 90 ) = x$]
+ #math.equation(block: false, alt: "x approximately equals 4.096")[$x ≈ 4.096$]
]


      
      
        + Rewrite the equation #math.equation(block: false, alt: "3 to the power x equals 90")[$3^(x) = 90$] in logarithmic form.
+ Use a graph to approximate the solution to the equation in part (a). Round your answer to three decimal places.


        #solutionbox[
+ #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 90 close parenthesis equals x")[$log_(3) ( 90 ) = x$]
+ #math.equation(block: false, alt: "x approximately equals 4.096")[$x ≈ 4.096$]
]


      
    
    
      
=== Base 10 Logarithms


      Some logarithms are used so frequently in applications that their values are programmed into scientific and graphing calculators. These are the base #math.equation(block: false, alt: "10")[$10$] logarithms, such as


      #math.equation(block: true, alt: "log sub 10 ⁡ 1000 equals 3 and log sub 10 ⁡ 0.01 equals minus 2")[$log_(10) 1000 = 3 " " " " " " " and " " " " " " " log_(10) 0.01 = − 2$]
      Base #math.equation(block: false, alt: "10")[$10$] logarithms are called #strong[common logarithms], and the subscript #math.equation(block: false, alt: "10")[$10$] is often omitted, so that #math.equation(block: false, alt: "log ⁡ open parenthesis x close parenthesis")[$log ( x )$] (or simply #math.equation(block: false, alt: "log ⁡ x")[$log x$]) is understood to mean #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x close parenthesis")[$log_(10) ( x )$].


      
        If #math.equation(block: false, alt: "log ⁡ open parenthesis x close parenthesis equals 2.5")[$log ( x ) = 2.5$], what can you say about #math.equation(block: false, alt: "x")[$x$]?


        \_\_\_\_\_


        #solutionbox[
#math.equation(block: false, alt: "x")[$x$] is between 100 and 1000
]


      
      
        If #math.equation(block: false, alt: "log ⁡ open parenthesis x close parenthesis equals 2.5")[$log ( x ) = 2.5$], what can you say about #math.equation(block: false, alt: "x")[$x$]?


        + #math.equation(block: false, alt: "x equals 25")[$x = 25$]
+ #math.equation(block: false, alt: "x")[$x$] is between 20 and 30.
+ #math.equation(block: false, alt: "x")[$x$] is between 100 and 1000.
+ #math.equation(block: false, alt: "x equals the fraction 1 over 2.5 equals 0.4")[$x = display(frac(1, 2.5)) = 0.4$]


      
      To evaluate a base #math.equation(block: false, alt: "10")[$10$] logarithm, we use the LOG key on a calculator. Many logarithms are irrational numbers, and the calculator gives as many digits as its display allows. We can then round off to the desired accuracy.


      #examplebox("Example 5")[][
Approximate the following logarithms to #math.equation(block: false, alt: "2")[$2$] decimal places.


          + #math.equation(block: false, alt: "log ⁡ open parenthesis 6.5 close parenthesis")[$log ( 6.5 )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis 256 close parenthesis")[$log ( 256 )$]


        
        #solutionbox[
+ The keying sequence LOG #math.equation(block: false, alt: "6.5")[$6.5$] )ENTER produces the display#figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "log ⁡ open parenthesis 6.5 close parenthesis")[$log ( 6.5 )$]], [], []),
  [], [], [#math.equation(block: false, alt: ".812913566")[$.812913566$]],
))

 #linebreak() so #math.equation(block: false, alt: "log ⁡ open parenthesis 6.5 close parenthesis approximately equals 0.81")[$log ( 6.5 ) ≈ 0.81$].
+ The keying sequence LOG #math.equation(block: false, alt: "256")[$256$] ) ENTER yields #math.equation(block: false, alt: "2.408239965")[$2.408239965$], so #math.equation(block: false, alt: "log ⁡ open parenthesis 256 close parenthesis approximately equals 2.41")[$log ( 256 ) ≈ 2.41$].
]
]


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
We can check the approximations found in Example with our conversion equations. Remember that a logarithm is an exponent, and in this example the base is #math.equation(block: false, alt: "10")[$10$]. We find that


        #math.equation(block: true, alt: "10 to the power 0.81, approximately equals 6.45654229; and, 10 to the power 2.41, approximately equals 257.0395783")[$&  & 10^(0.81) & ≈ 6.45654229 \ "and" &  & 10^(2.41) & ≈ 257.0395783$]
        so our approximations are reasonable, although you can see that rounding a logarithm to #math.equation(block: false, alt: "2")[$2$] decimal places does lose some accuracy.


        For this reason, #emph[rounding logarithms to #math.equation(block: false, alt: "4")[$4$] decimal places is customary.]
]


      
        + Evaluate #math.equation(block: false, alt: "log ⁡ open parenthesis 250 close parenthesis")[$log ( 250 )$], and round your answer to two decimal places. Check your answer using the conversion equations. #linebreak() #math.equation(block: false, alt: "log ⁡ open parenthesis 250 close parenthesis approximately equals")[$log ( 250 ) ≈$]\_\_\_\_\_ to two decimal places
+ Evaluate #math.equation(block: false, alt: "log ⁡ open parenthesis 250 close parenthesis")[$log ( 250 )$], and round your answer to four decimal places. Check your answer using the conversion equations. #linebreak() #math.equation(block: false, alt: "log ⁡ open parenthesis 250 close parenthesis approximately equals")[$log ( 250 ) ≈$]\_\_\_\_\_ to four decimal places


        #solutionbox[
+ #math.equation(block: false, alt: "2.40")[$2.40$]
+ #math.equation(block: false, alt: "2.3979")[$2.3979$]
]


      
      
        + Evaluate #math.equation(block: false, alt: "log ⁡ open parenthesis 250 close parenthesis")[$log ( 250 )$], and round your answer to two decimal places. Check your answer using the conversion equations.
+ Evaluate #math.equation(block: false, alt: "log ⁡ open parenthesis 250 close parenthesis")[$log ( 250 )$], and round your answer to four decimal places. Check your answer using the conversion equations.


        #solutionbox[
+ #math.equation(block: false, alt: "2.40")[$2.40$]
+ #math.equation(block: false, alt: "2.3979")[$2.3979$]
]


      
      
        Explain how to estimate #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x close parenthesis")[$log_(10) ( x )$] between two integers.


        \_\_\_\_\_


      
      
        Explain how to estimate #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x close parenthesis")[$log_(10) ( x )$] between two integers.


      
    
    
      
=== Solving Exponential Equations


      We can now solve any exponential equation with base #math.equation(block: false, alt: "10")[$10$]. For instance, to solve the equation


      #math.equation(block: true, alt: "16 times 10 to the power t equals 360")[$16 ⋅ 10^(t) = 360$]
      , 
we first divide both sides by #math.equation(block: false, alt: "16")[$16$] to obtain


      #math.equation(block: true, alt: "10 to the power t equals 22.5")[$10^(t) = 22.5$]
      Then we convert the equation to logarithmic form and evaluate:


      #math.equation(block: true, alt: "t equals log sub 10 ⁡ open parenthesis 22.5 close parenthesis approximately equals 1.352182518")[$t = log_(10) ( 22.5 ) ≈ 1.352182518$]
      To #math.equation(block: false, alt: "4")[$4$] decimal places, the solution is #math.equation(block: false, alt: "1.3522")[$1.3522$].


      To solve exponential equations involving powers of 10, we can use the following steps.


      #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Steps for Solving Base 10 Exponential Equations]


        + Isolate the power on one side of the equation.
+ Rewrite the equation in logarithmic form.
+ Use a calculator, if necessary, to evaluate the logarithm.
+ Solve for the variable.
]


      #examplebox("Example 6")[][
Solve the equation #math.equation(block: false, alt: "38 equals 95 minus 15 times 10 to the power 0.4 x")[$" " " " 38 = 95 − 15 ⋅ 10^(0.4 x)$]


        
        #solutionbox[
First, we isolate the power of #math.equation(block: false, alt: "10")[$10$]: We subtract #math.equation(block: false, alt: "95")[$95$] from both sides of the equation and divide by #math.equation(block: false, alt: "minus 15")[$− 15$] to obtain


          #math.equation(block: true, alt: "minus 57, equals minus 15 times 10 to the power 0.4 x, Divide by minus 15.; 3.8, equals 10 to the power 0.4 x")[$− 57 & = − 15 ⋅ 10^(0.4 x) &  & "Divide by " − 15 . \ 3.8 & = 10^(0.4 x)$]
          Next, we convert the equation to logarithmic form as


          #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis 3.8 close parenthesis equals 0.4 x")[$log_(10) ( 3.8 ) = 0.4 x$]
          Solving for #math.equation(block: false, alt: "x")[$x$] yields


          #math.equation(block: true, alt: "the fraction log sub 10 ⁡ open parenthesis 3.8 close parenthesis over 0.4 equals x")[$frac(log_(10) ( 3.8 ), 0.4) = x$]
          We can evaluate this expression on the calculator by entering


          LOG #math.equation(block: false, alt: "3.8")[$3.8$] ) ÷ #math.equation(block: false, alt: "0.4")[$0.4$] ENTER


          which yields #math.equation(block: false, alt: "1.449458992")[$1.449458992$]. Thus, to four decimal places, #math.equation(block: false, alt: "x approximately equals 1.4495")[$x ≈ 1.4495$].
]
]


      #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Be careful when using a calculator to evaluate expressions involving logs. We can evaluate a single logarithm like #math.equation(block: false, alt: "log ⁡ open parenthesis 3.8 close parenthesis")[$log ( 3.8 )$] by entering LOG #math.equation(block: false, alt: "3.8")[$3.8$]ENTER without an ending parenthesis, so that the calculator shows


        #figure(table(
  columns: 3, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "log ⁡ open parenthesis 3.8")[$log \( 3.8$]], [], []),
  [], [], [#math.equation(block: false, alt: ".5795835966")[$.5795835966$]],
))


        But if we want to evaluate #math.equation(block: false, alt: "the fraction log ⁡ 3.8 over 0.4")[$display(frac(log 3.8, 0.4))$], we must enclose #math.equation(block: false, alt: "3.8")[$3.8$] in parentheses, as shown in Example. If we omit the parenthesis after #math.equation(block: false, alt: "3.8")[$3.8$], the calculator will interpret the expression as #math.equation(block: false, alt: "log ⁡ open parenthesis the fraction 3.8 over 0.4 close parenthesis")[$log ( display(frac(3.8, 0.4)) )$], which is not the expression we wanted.
]


      
        What is the first step in solving the equation #math.equation(block: false, alt: "5 open parenthesis 10 close parenthesis to the power x equals 12")[$5 ( 10 )^(x) = 12$]?


        \_\_\_\_\_


        #solutionbox[
Divide both sides by 5.
]


      
      
        What is the first step in solving the equation #math.equation(block: false, alt: "5 open parenthesis 10 close parenthesis to the power x equals 12")[$5 ( 10 )^(x) = 12$]?


        + Multiply 5 times 10.
+ Take the log of both sides.
+ Get zero on one side of the equation.
+ Divide both sides by 5.


      
      
        Solve #math.equation(block: false, alt: "12 minus 30 open parenthesis 10 to the power minus 0.2 x close parenthesis equals 11.25")[$12 − 30 ( 10^(− 0.2 x) ) = 11.25$]


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


        #solutionbox[
#math.equation(block: true, alt: "x equals minus 5 log ⁡ open parenthesis 0.025 close parenthesis approximately equals 8.01")[$x = − 5 log ( 0.025 ) ≈ 8.01$]
]


      
      
        Solve #math.equation(block: false, alt: "12 minus 30 open parenthesis 10 to the power minus 0.2 x close parenthesis equals 11.25")[$12 − 30 ( 10^(− 0.2 x) ) = 11.25$]


        #solutionbox[
#math.equation(block: true, alt: "x equals minus 5 log ⁡ open parenthesis 0.025 close parenthesis approximately equals 8.01")[$x = − 5 log ( 0.025 ) ≈ 8.01$]
]


      
    
    
      
=== Application to Exponential Models


      We have seen that exponential functions are used to describe some applications of growth and decay, #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 b to the power t")[$P ( t ) = P_(0) b^(t)$]. There are two common questions that arise in connection with exponential models:


      + Given a value of #math.equation(block: false, alt: "t")[$t$], what is the corresponding value of #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$]?
+ Given a value of #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$], what is find the corresponding value of #math.equation(block: false, alt: "t")[$t$]?


      To answer the first question, we evaluate the function #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$] at the appropriate value. To answer the second question, we must solve an exponential equation, and this usually involves logarithms.


      #examplebox("Example 7")[][
The value of a large tractor originally worth \$#math.equation(block: false, alt: "30 , 000")[$30 , 000$] depreciates exponentially according to the formula


          #math.equation(block: true, alt: "V open parenthesis t close parenthesis equals 30 , 000 open parenthesis 10 close parenthesis to the power minus 0.04 t")[$V ( t ) = 30 , 000 ( 10 )^(− 0.04 t)$]
          where #math.equation(block: false, alt: "t")[$t$] is in years. When will the tractor be worth half its original value?


        
        #solutionbox[
We want to find the value of #math.equation(block: false, alt: "t")[$t$] for which #math.equation(block: false, alt: "V open parenthesis t close parenthesis equals 15 , 000")[$V ( t ) = 15 , 000$]. That is, we want to solve the equation


          #math.equation(block: true, alt: "15 , 000 equals 30 , 000 open parenthesis 10 close parenthesis to the power minus 0.04 t")[$15 , 000 = 30 , 000 ( 10 )^(− 0.04 t)$]
          We divide both sides by 30,000 to obtain


          #math.equation(block: true, alt: "0.5 equals 10 to the power minus 0.04 t")[$0.5 = 10^(− 0.04 t)$]
          We convert the equation to logarithmic form as


          #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis 0.5 close parenthesis equals minus 0.04 t")[$log_(10) ( 0.5 ) = − 0.04 t$]
          and divide by #math.equation(block: false, alt: "minus 0.04")[$− 0.04$] to obtain


          #math.equation(block: true, alt: "the fraction log sub 10 ⁡ open parenthesis 0.5 close parenthesis over minus 0.04 equals t")[$frac(log_(10) ( 0.5 ), − 0.04) = t$]
          To evaluate this expression, we key in


          LOG #math.equation(block: false, alt: "0.5")[$0.5$] ) ÷ (-) #math.equation(block: false, alt: "0.04")[$0.04$] ENTER


          to find #math.equation(block: false, alt: "t approximately equals 7.525749892")[$t ≈ 7.525749892$]. The tractor will be worth \$15,000 in approximately #math.equation(block: false, alt: "7 the fraction 1 over 2")[$7 frac(1, 2)$] years.
]
]


      
        The percentage of American homes with computers grew exponentially from 1994 to 1999. For #math.equation(block: false, alt: "t equals 0")[$t = 0$] in 1994, the growth law was


        #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals f")[$P ( t ) = "f"$]
        \[Source: Los Angeles Times, August 20, 1999\]


        + What percent of American homes had computers in 1994? #linebreak() Answer: \_\_\_\_\_%
+ If the percentage of homes with computers continued to grow at the same rate, when did 90% of American homes have a computer? #linebreak() Answer: #math.equation(block: false, alt: "t approximately equals")[$t ≈$]\_\_\_\_\_, which was the year \_\_\_\_\_
+ Do you think that the function #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$] will continue to model the percentage of American homes with computers? Why or why not? #linebreak() \_\_\_\_\_


        #solutionbox[
+ 25.85%
+ #math.equation(block: false, alt: "t approximately equals 10.4")[$t ≈ 10.4$] (year 2004)
+ No, the percent of homes with computers cannot exceed #math.equation(block: false, alt: "100 %")[$100 upright(%)$].
]


      
      
        The percentage of American homes with computers grew exponentially from 1994 to 1999. For #math.equation(block: false, alt: "t equals 0")[$t = 0$] in 1994, the growth law was #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 25.85 open parenthesis 10 close parenthesis to the power 0.052 t .")[$P ( t ) = 25.85 ( 10 )^(0.052 t) .$] \[Source: Los Angeles Times, August 20, 1999\]


        + What percent of American homes had computers in 1994?
+ If the percentage of homes with computers continued to grow at the same rate, when did 90% of American homes have a computer?
+ Do you think that the function #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$] will continue to model the percentage of American homes with computers? Why or why not?


        #solutionbox[
+ 25.85%
+ #math.equation(block: false, alt: "t approximately equals 10.4")[$t ≈ 10.4$] (year 2004)
+ No, the percent of homes with computers cannot exceed #math.equation(block: false, alt: "100 %")[$100 upright(%)$].
]


      
      At this stage, it seems we will only be able to solve exponential equations in which the base is #math.equation(block: false, alt: "10")[$10$]. However, we will see in Properties of Logarithms how the properties of logarithms enable us to solve exponential equations with any base.


      
        The population of rabbits on an island grows according to #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 6 open parenthesis 10 close parenthesis to the power 0.3 t")[$P ( t ) = 6 ( 10 )^(0.3 t)$]. Explain the difference between the two problems: (1) given #math.equation(block: false, alt: "P")[$P$] find #math.equation(block: false, alt: "t")[$t$], and (2) given #math.equation(block: false, alt: "t")[$t$] find #math.equation(block: false, alt: "P")[$P$].


        \_\_\_\_\_


      
      
        The population of rabbits on an island grows according to #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 6 open parenthesis 10 close parenthesis to the power 0.3 t")[$P ( t ) = 6 ( 10 )^(0.3 t)$]. Explain the difference between the two problems: (1) given #math.equation(block: false, alt: "P")[$P$] find #math.equation(block: false, alt: "t")[$t$], and (2) given #math.equation(block: false, alt: "t")[$t$] find #math.equation(block: false, alt: "P")[$P$].


      
    
    
      
=== Section Summary


      
        
==== Vocabulary


        Look up the definitions of new terms in the Glossary.


        - Logarithm
- Common logarithm


      
      
        
==== CONCEPTS


        + We use logarithms to help us solve exponential equations.
+ The #strong[base #math.equation(block: false, alt: "b")[$b$] logarithm of #math.equation(block: false, alt: "x")[$x$]], written #math.equation(block: false, alt: "log sub b ⁡ open parenthesis x close parenthesis")[$log_(b) ( x )$] (or written as #math.equation(block: false, alt: "log sub b ⁡ x")[$log_(b) x$]), is the exponent to which #math.equation(block: false, alt: "b")[$b$] must be raised in order to yield #math.equation(block: false, alt: "x")[$x$].
+ If #math.equation(block: false, alt: "b greater than 0")[$b > 0$] and #math.equation(block: false, alt: "x greater than 0")[$x > 0$],
    		#math.equation(block: true, alt: "y equals log sub b ⁡ open parenthesis x close parenthesis if and only if x equals b to the power y")[$y = log_(b) ( x ) " " " " " " " if and only if " " " " " " " x = b^(y)$]
+ The operation of taking a base #math.equation(block: false, alt: "b")[$b$] logarithm is the inverse operation for raising the base #math.equation(block: false, alt: "b")[$b$] to a power.
+ Base #math.equation(block: false, alt: "10")[$10$] logarithms are called #strong[common logarithms], and #math.equation(block: false, alt: "log ⁡ open parenthesis x close parenthesis")[$log ( x )$] (or #math.equation(block: false, alt: "log ⁡ x")[$log x$]) means #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis x close parenthesis")[$log_(10) ( x )$].
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Steps for Solving Base 10 Exponential Equations]

+ Isolate the power on one side of the equation.
+ Rewrite the equation in logarithmic form.
+ Use a calculator, if necessary, to evaluate the logarithm.
+ Solve for the variable.
]


      
      
        
==== STUDY QUESTIONS


        + To find #math.equation(block: false, alt: "log sub 6 ⁡ 27")[$log_(6) 27$] means to find an exponent  that satisfies the equation \_\_\_\_\_\_\_\_\_\_\_\_.
+ Can a logarithm be a negative number?
+ Evaluate the following logarithms: + #math.equation(block: false, alt: "log sub 8 ⁡ open parenthesis 8 to the power 15 close parenthesis")[$log_(8) ( 8^(15) )$]
+ #math.equation(block: false, alt: "log sub 5 ⁡ open parenthesis 5 to the power the square root of 13 close parenthesis")[$log_(5) ( 5^(sqrt(13)) )$]
+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis b to the power 2.63 close parenthesis")[$log_(b) ( b^(2.63) )$]
+ Guess the solution of #math.equation(block: false, alt: "10 to the power x equals 750")[$10^(x) = 750$]. Now find an approximation correct to four decimal places. Was your guess too big or too small?


      
      
        
==== SKILLS


        Practice each skill in the Homework problems listed.


        + Compute logs base #math.equation(block: false, alt: "b")[$b$] using the definition: \#1–10, 59–66
+ Convert from exponential to logarithmic form: \#11–22
+ Approximate logarithms: \#23–34
+ Solve exponential equations base 10: \#35–48
+ Solve application problems: \#49–58


      
    
    
      
=== Homework 4.3


      For Problems 1–10, find each logarithm without using a calculator.


      
        + #math.equation(block: false, alt: "log sub 7 ⁡ open parenthesis 49 close parenthesis")[$log_(7) ( 49 )$]
+ #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis 32 close parenthesis")[$log_(2) ( 32 )$]


        
          + #math.equation(block: false, alt: "2")[$2$]
+ #math.equation(block: false, alt: "5")[$5$]


        
      
      
        + #math.equation(block: false, alt: "log sub 4 ⁡ open parenthesis 64 close parenthesis")[$log_(4) ( 64 )$]
+ #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis 27 close parenthesis")[$log_(3) ( 27 )$]


      
      
        + #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis the square root of 3 close parenthesis")[$log_(3) ( sqrt(3) )$]
+ #math.equation(block: false, alt: "log sub 3 ⁡ open parenthesis the fraction 1 over 3 close parenthesis")[$log_(3) ( display(frac(1, 3)) )$]


        
          + #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: "log sub 5 ⁡ open parenthesis the fraction 1 over 5 close parenthesis")[$log_(5) ( display(frac(1, 5)) )$]
+ #math.equation(block: false, alt: "log sub 5 ⁡ open parenthesis the square root of 5 close parenthesis")[$log_(5) ( sqrt(5) )$]


      
      
        + #math.equation(block: false, alt: "log sub 4 ⁡ open parenthesis 4 close parenthesis")[$log_(4) ( 4 )$]
+ #math.equation(block: false, alt: "log sub 6 ⁡ open parenthesis 1 close parenthesis")[$log_(6) ( 1 )$]


        
          + #math.equation(block: false, alt: "1")[$1$]
+ #math.equation(block: false, alt: "0")[$0$]


        
      
      
        + #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 1 close parenthesis")[$log_(10) ( 1 )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 10 to the power minus 6 close parenthesis")[$log_(10) ( 10^(− 6) )$]


      
      
        + #math.equation(block: false, alt: "log sub 8 ⁡ open parenthesis 8 to the power 5 close parenthesis")[$log_(8) ( 8^(5) )$]
+ #math.equation(block: false, alt: "log sub 7 ⁡ open parenthesis 7 to the power 6 close parenthesis")[$log_(7) ( 7^(6) )$]


        
          + #math.equation(block: false, alt: "5")[$5$]
+ #math.equation(block: false, alt: "6")[$6$]


        
      
      
        + #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 10 to the power minus 4 close parenthesis")[$log_(10) ( 10^(− 4) )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 10 to the power minus 6 close parenthesis")[$log_(10) ( 10^(− 6) )$]


      
      
        + #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 0.1 close parenthesis")[$log_(10) ( 0.1 )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 0.001 close parenthesis")[$log_(10) ( 0.001 )$]


        
          + #math.equation(block: false, alt: "minus 1")[$− 1$]
+ #math.equation(block: false, alt: "minus 3")[$− 3$]


        
      
      
        + #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 10 , 000 close parenthesis")[$log_(10) ( 10 , 000 )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 1000 close parenthesis")[$log_(10) ( 1000 )$]


      
      For Problems 11–22, rewrite the equation in logarithmic form.


      
        #math.equation(block: true, alt: "2 to the power 10 equals 1024")[$2^(10) = 1024$]


        
          #math.equation(block: true, alt: "log sub 2 ⁡ open parenthesis 1024 close parenthesis equals 10")[$log_(2) ( 1024 ) = 10$]


        
      
      
        #math.equation(block: true, alt: "11 to the power 4 equals 14 , 641")[$11^(4) = 14 , 641$]


      
      
        #math.equation(block: true, alt: "10 to the power 0.699 approximately equals 5")[$10^(0.699) ≈ 5$]


        
          #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis 5 close parenthesis approximately equals 0.699")[$log_(10) ( 5 ) ≈ 0.699$]


        
      
      
        #math.equation(block: true, alt: "10 to the power minus 0.602 approximately equals 0.25")[$10^(− 0.602) ≈ 0.25$]


      
      
        #math.equation(block: true, alt: "t to the power 3 / 2 equals 16")[$t^(3 / 2) = 16$]


        
          #math.equation(block: true, alt: "log sub t ⁡ open parenthesis 16 close parenthesis equals the fraction 3 over 2")[$log_(t) ( 16 ) = display(frac(3, 2))$]


        
      
      
        #math.equation(block: true, alt: "v to the power 5 / 3 equals 12")[$v^(5 / 3) = 12$]


      
      
        #math.equation(block: true, alt: "0.8 to the power 1.2 equals M")[$0.8^(1.2) = M$]


        
          #math.equation(block: true, alt: "log sub open parenthesis ⁡ 0.8 close parenthesis M equals 1.2")[$log_( 0.8 ) M = 1.2$]


        
      
      
        #math.equation(block: true, alt: "3.7 to the power 2.5 equals Q")[$3.7^(2.5) = Q$]


      
      
        #math.equation(block: true, alt: "x to the power 5 t equals W minus 3")[$x^(5 t) = W − 3$]


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


        
      
      
        #math.equation(block: true, alt: "z to the power minus 3 t equals 2 P plus 5")[$z^(− 3 t) = 2 P + 5$]


      
      
        #math.equation(block: true, alt: "3 to the power minus 0.2 t equals 2 N sub 0")[$3^(− 0.2 t) = 2 N_(0)$]


        
          #math.equation(block: true, alt: "log sub 3 ⁡ open parenthesis 2 N sub 0 close parenthesis equals minus 0.2 t")[$log_(3) ( 2 N_(0) ) = − 0.2 t$]


        
      
      
        #math.equation(block: true, alt: "10 to the power 1.3 t equals 3 M sub 0")[$10^(1.3 t) = 3 M_(0)$]


      
      For Problems 23–26,


      + Solve each equation, writing your answer as a logarithm.
+ Use trial and error to approximate the logarithm to one decimal place.


      
        #math.equation(block: true, alt: "4 to the power x equals 2.5")[$4^(x) = 2.5$]


        
          + #math.equation(block: false, alt: "log sub 4 ⁡ open parenthesis 2.5 close parenthesis")[$log_(4) ( 2.5 )$]
+ #math.equation(block: false, alt: "0.7")[$0.7$]


        
      
      
        #math.equation(block: true, alt: "2 to the power x equals 0.2")[$2^(x) = 0.2$]


      
      
        #math.equation(block: true, alt: "10 to the power x equals 0.003")[$10^(x) = 0.003$]


        
          + #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 0.003 close parenthesis")[$log_(10) ( 0.003 )$]
+ #math.equation(block: false, alt: "minus 2.5")[$− 2.5$]


        
      
      
        #math.equation(block: true, alt: "10 to the power x equals 4500")[$10^(x) = 4500$]


      
      For Problems 27–30,


      + By computing successive powers of the base, trap each log between two integers.
+ Use a graph to approximate each logarithm to the nearest hundredth. (#emph[Hint]: Use the conversion equations to rewrite #math.equation(block: false, alt: "x equals log sub b ⁡ open parenthesis y close parenthesis")[$x = log_(b) ( y )$] as an appropriate exponential equation.)


      
        #math.equation(block: true, alt: "log sub 10 ⁡ 7")[$log_(10) 7$]


        
          + #math.equation(block: false, alt: "0 less than log sub 10 ⁡ open parenthesis 7 close parenthesis less than 1")[$0 < log_(10) ( 7 ) < 1$]
+ #math.equation(block: false, alt: "0.85")[$0.85$]


        
      
      
        #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis 50 close parenthesis")[$log_(10) ( 50 )$]


      
      
        #math.equation(block: true, alt: "log sub 3 ⁡ open parenthesis 67.9 close parenthesis")[$log_(3) ( 67.9 )$]


        
          + #math.equation(block: false, alt: "3 less than log sub 3 ⁡ open parenthesis 67.9 close parenthesis less than 4")[$3 < log_(3) ( 67.9 ) < 4$]
+ #math.equation(block: false, alt: "3.84")[$3.84$]


        
      
      
        #math.equation(block: true, alt: "log sub 5 ⁡ open parenthesis 86.3 close parenthesis")[$log_(5) ( 86.3 )$]


      
      For Problems 31–34, use a calculator to approximate each logarithm to four decimal places. Make a conjecture about logarithms based on the results of each problem.


      
        + #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 5.43 close parenthesis")[$log_(10) ( 5.43 )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 54.3 close parenthesis")[$log_(10) ( 54.3 )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 543 close parenthesis")[$log_(10) ( 543 )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 5430 close parenthesis")[$log_(10) ( 5430 )$]


        
          + #math.equation(block: false, alt: "0.7348")[$0.7348$]
+ #math.equation(block: false, alt: "1.7348")[$1.7348$]
+ #math.equation(block: false, alt: "2.7348")[$2.7348$]
+ #math.equation(block: false, alt: "3.7348")[$3.7348$]


          When the input to the common logarithm is multiplied by #math.equation(block: false, alt: "10")[$10$], the output is increased by #math.equation(block: false, alt: "1")[$1$].


        
      
      
        + #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 0.625 close parenthesis")[$log_(10) ( 0.625 )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 0.0625 close parenthesis")[$log_(10) ( 0.0625 )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 0.00625 close parenthesis")[$log_(10) ( 0.00625 )$]
+ #math.equation(block: false, alt: "log sub 10 ⁡ open parenthesis 0.000625 close parenthesis")[$log_(10) ( 0.000625 )$]


      
      
        + #math.equation(block: false, alt: "log ⁡ open parenthesis 2 close parenthesis")[$log ( 2 )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis 4 close parenthesis")[$log ( 4 )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis 8 close parenthesis")[$log ( 8 )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis 16 close parenthesis")[$log ( 16 )$]


        
          + #math.equation(block: false, alt: "0.3010")[$0.3010$]
+ #math.equation(block: false, alt: "0.6021")[$0.6021$]
+ #math.equation(block: false, alt: "0.9031")[$0.9031$]
+ #math.equation(block: false, alt: "1.2041")[$1.2041$]


          When the input to the common logarithm is doubled, the output is increased by about #math.equation(block: false, alt: "0.3010")[$0.3010$].


        
      
      
        + #math.equation(block: false, alt: "log ⁡ open parenthesis 4 close parenthesis")[$log ( 4 )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis 0.25 close parenthesis")[$log ( 0.25 )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis 5 close parenthesis")[$log ( 5 )$]
+ #math.equation(block: false, alt: "log ⁡ open parenthesis 0.2 close parenthesis")[$log ( 0.2 )$]


      
      For Problems 35–44, solve for #math.equation(block: false, alt: "x")[$x$]. Round your answers to hundredths.


      
        #math.equation(block: true, alt: "10 to the power minus 3 x equals 5")[$10^(− 3 x) = 5$]


        
          #math.equation(block: true, alt: "minus 0.23")[$− 0.23$]


        
      
      
        #math.equation(block: true, alt: "10 to the power minus 5 x equals 76")[$10^(− 5 x) = 76$]


      
      
        #math.equation(block: true, alt: "25 times 10 to the power 0.2 x equals 80")[$25 ⋅ 10^(0.2 x) = 80$]


        
          #math.equation(block: true, alt: "2.53")[$2.53$]


        
      
      
        #math.equation(block: true, alt: "8 times 10 to the power 1.6 x equals 312")[$8 ⋅ 10^(1.6 x) = 312$]


      
      
        #math.equation(block: true, alt: "12.2 equals 2 open parenthesis 10 to the power 1.4 x close parenthesis minus 11.6")[$12.2 = 2 ( 10^(1.4 x) ) − 11.6$]


        
          #math.equation(block: true, alt: "0.77")[$0.77$]


        
      
      
        #math.equation(block: true, alt: "163 equals 3 open parenthesis 10 to the power 0.7 x close parenthesis minus 49.3")[$163 = 3 ( 10^(0.7 x) ) − 49.3$]


      
      
        #math.equation(block: true, alt: "3 open parenthesis 10 to the power minus 1.5 x close parenthesis minus 14.7 equals 17.1")[$3 ( 10^(− 1.5 x) ) − 14.7 = 17.1$]


        
          #math.equation(block: true, alt: "minus 0.68")[$− 0.68$]


        
      
      
        #math.equation(block: true, alt: "4 open parenthesis 10 to the power minus 0.6 x close parenthesis plus 16.1 equals 28.2")[$4 ( 10^(− 0.6 x) ) + 16.1 = 28.2$]


      
      
        #math.equation(block: true, alt: "80 open parenthesis 1 minus 10 to the power minus 0.2 x close parenthesis equals 65")[$80 ( 1 − 10^(− 0.2 x) ) = 65$]


        
          #math.equation(block: true, alt: "3.63")[$3.63$]


        
      
      
        #math.equation(block: true, alt: "250 open parenthesis 1 minus 10 to the power minus 0.3 x close parenthesis equals 100")[$250 ( 1 − 10^(− 0.3 x) ) = 100$]


      
      In Problems 45–48, each calculation contains an error. Find and correct it.


      
        #math.equation(block: true, alt: "2 times 5 to the power x, equals 848; 10 to the power x, equals 848; x, equals log ⁡ open parenthesis 848 close parenthesis, open parenthesis I n c o r r e c t ! close parenthesis")[$2 ⋅ 5^(x) & = 848 &  &  \ 10^(x) & = 848 \ x & = log ( 848 ) &  & ( I n c o r r e c t ! )$]
        
          #math.equation(block: false, alt: "2 times 5 to the power x not equal to 10 to the power x")[$2 ⋅ 5^(x) ≠ 10^(x)$]; the first step should be to divide both sides of the equation by #math.equation(block: false, alt: "2")[$2$]; #math.equation(block: false, alt: "x equals log sub 5 ⁡ open parenthesis 424 close parenthesis")[$x = log_(5) ( 424 )$].


        
      
      
        #math.equation(block: true, alt: "15 times 10 to the power x, equals 20; 10 to the power x, equals 5; x, equals log ⁡ 5, open parenthesis I n c o r r e c t ! close parenthesis")[$15 ⋅ 10^(x) & = 20 &  &  \ 10^(x) & = 5 \ x & = log 5 &  & ( I n c o r r e c t ! )$]
      
      
        #math.equation(block: true, alt: "10 to the power 4 x, equals 20; 10 to the power x, equals 5; x, equals log ⁡ 5, open parenthesis I n c o r r e c t ! close parenthesis")[$10^(4 x) & = 20 &  &  \ 10^(x) & = 5 \ x & = log 5 &  & ( I n c o r r e c t ! )$]
        
          #math.equation(block: false, alt: "the fraction 10 to the power 4 x over 4 not equal to 10 to the power x")[$frac(10^(4 x), 4) ≠ 10^(x)$]; the first step should be to write #math.equation(block: false, alt: "4 x equals log ⁡ open parenthesis 20 close parenthesis")[$4 x = log ( 20 )$]; #math.equation(block: false, alt: "x equals the fraction log ⁡ 20 over 4")[$x = frac(log 20, 4)$].


        
      
      
        #math.equation(block: true, alt: "12 plus 6 to the power x, equals 42; 6 to the power x, equals 30; x, equals 5, open parenthesis I n c o r r e c t ! close parenthesis")[$12 + 6^(x) & = 42 &  &  \ 6^(x) & = 30 \ x & = 5 &  & ( I n c o r r e c t ! )$]
      
      
        The population of the state of California increased during the years 1990 to 2000 according to the formula


        #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 29 , 760 , 021 open parenthesis 10 close parenthesis to the power 0.0056 t")[$P ( t ) = 29 , 760 , 021 ( 10 )^(0.0056 t)$]
        where #math.equation(block: false, alt: "t")[$t$] is measured in years since #math.equation(block: false, alt: "1990")[$1990$].


        + What was the population in #math.equation(block: false, alt: "2000")[$2000$]?
+ Assuming the same rate of growth, estimate the population of California in the years #math.equation(block: false, alt: "2010")[$2010$], #math.equation(block: false, alt: "2015")[$2015$], and #math.equation(block: false, alt: "2020")[$2020$].
+ When did the population of California reach #math.equation(block: false, alt: "35 , 000 , 000")[$35 , 000 , 000$]?
+ When should the population reach #math.equation(block: false, alt: "40 , 000 , 000")[$40 , 000 , 000$]?
+ Graph the function #math.equation(block: false, alt: "P")[$P$] with a suitable domain and range, then verify your answers to parts (a) through (d).


        
          + #math.equation(block: false, alt: "33 , 855 , 812")[$33 , 855 , 812$]
+ #math.equation(block: false, alt: "38 , 515 , 295")[$38 , 515 , 295$]; #math.equation(block: false, alt: "41 , 080 , 265")[$" " 41 , 080 , 265$]; #math.equation(block: false, alt: "43 , 816 , 051")[$" " 43 , 816 , 051$]
+ #math.equation(block: false, alt: "2002")[$2002$]
+ #math.equation(block: false, alt: "2012")[$2012$]
+ #figure(figph[exponential growth], alt: "exponential growth", caption: none)


        
      
      
        The population of the state of New York increased during the years 1990 to 2000 according to the formula


        #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 17 , 990 , 455 open parenthesis 10 close parenthesis to the power 0.0023 t")[$P ( t ) = 17 , 990 , 455 ( 10 )^(0.0023 t)$]
        where #math.equation(block: false, alt: "t")[$t$] is measured in years since #math.equation(block: false, alt: "1990")[$1990$].


        + What was the population in #math.equation(block: false, alt: "2000")[$2000$]?
+ Assuming the same rate of growth, estimate the population of New York in the years #math.equation(block: false, alt: "2010")[$2010$], #math.equation(block: false, alt: "2015")[$2015$], and #math.equation(block: false, alt: "2020")[$2020$].
+ When did the population of New York reach #math.equation(block: false, alt: "20 , 000 , 000")[$20 , 000 , 000$]?
+ When should the population reach #math.equation(block: false, alt: "30 , 000 , 000")[$30 , 000 , 000$]?
+ Graph the function #math.equation(block: false, alt: "P")[$P$] with a suitable domain and range, then verify your answers to parts (a) through (d).


      
      
        The absolute magnitude, #math.equation(block: false, alt: "M")[$M$], of a star is a measurement of its brightness. For example, our Sun, not a particularly bright star, has magnitude #math.equation(block: false, alt: "M equals 4.83")[$M = 4.83$]. The magnitude in turn is a measure of the luminosity, #math.equation(block: false, alt: "L")[$L$], or amount of light energy emitted by the star, where


        #math.equation(block: true, alt: "L equals L sub 010 to the power minus 0.4 M")[$L = L_(0) 10^(− 0.4 M)$]
        + The luminosity of a star is measured in solar units, so that our Sun has luminosity #math.equation(block: false, alt: "L equals 1")[$L = 1$]. Use the values of #math.equation(block: false, alt: "L")[$L$] and #math.equation(block: false, alt: "M")[$M$] for the Sun to calculate a value of #math.equation(block: false, alt: "L sub 0")[$L_(0)$] in the equation above.
+ Is luminosity an increasing or decreasing function of magnitude? Graph the function on the domain #math.equation(block: false, alt: "open bracket minus 3 , 3 close bracket")[$[ − 3 , 3 ]$]. What is its range on that domain?
+ The luminosity of Sirius is #math.equation(block: false, alt: "22.5")[$22.5$] times that of the Sun, or #math.equation(block: false, alt: "L equals 22.5")[$L = 22.5$]. Calculate the magnitude of Sirius.
+ If two stars differ in magnitude by #math.equation(block: false, alt: "5")[$5$], what is the ratio of their luminosities?
+ A decrease in magnitude by #math.equation(block: false, alt: "1")[$1$] corresponds to an increase in luminosity by what factor? Give an exact value and an approximation to four decimal places.
+ Normal stars have magnitudes between #math.equation(block: false, alt: "minus 10")[$− 10$] and #math.equation(block: false, alt: "19")[$19$]. What range of luminosities do stars exhibit?


        
          + #math.equation(block: false, alt: "85.5")[$85.5$]
+ Decreasing; range: #math.equation(block: false, alt: "open bracket 5.4 , 1355.2 close bracket")[$[ 5.4 , 1355.2 ]$]#figure(figph[exponential decay], alt: "exponential decay", caption: none)
+ #math.equation(block: false, alt: "1.45")[$1.45$]
+ #math.equation(block: false, alt: "the fraction 1 over 100")[$display(frac(1, 100))$]
+ #math.equation(block: false, alt: "10 to the power 0.4 approximately equals 2.5119")[$10^(0.4) ≈ 2.5119$]
+ #math.equation(block: false, alt: "2.15 times 10 to the power minus 6")[$2.15 × 10^(− 6)$] to #math.equation(block: false, alt: "855 , 067")[$855 , 067$]


        
      
      
        The loudness of a sound is a consequence of its intensity, #math.equation(block: false, alt: "I")[$I$], or the amount of energy it generates, in watts per square meter. The intensity is related to the decibel level, #math.equation(block: false, alt: "D")[$D$], which is another measure of loudness, by


        #math.equation(block: true, alt: "I equals 10 to the power minus 12 plus D / 10")[$I = 10^(− 12 + D / 10)$]
        + Is intensity an increasing or decreasing function of decibel level? The faintest sound a healthy human can hear is #math.equation(block: false, alt: "0")[$0$] decibels. What is the intensity of a #math.equation(block: false, alt: "0")[$0$] decibel sound?
+ A whisper produces an energy intensity of #math.equation(block: false, alt: "10 to the power minus 9")[$10^(− 9)$] watts per square meter. What is the decibel level of a whisper?
+ If two sounds differ in loudness by #math.equation(block: false, alt: "10")[$10$] decibels, what is the ratio of their intensities?
+ An increase in loudness of #math.equation(block: false, alt: "1")[$1$] decibel produces a just noticeable difference to the human ear. By what factor does the intensity increase?
+ Sounds of #math.equation(block: false, alt: "130")[$130$] decibels are at the threshold of pain for people. What is the range of the intensity function on the domain #math.equation(block: false, alt: "open bracket 0 , 130 close bracket")[$[ 0 , 130 ]$]?


      
      The atmospheric pressure decreases with altitude above the surface of the Earth. For Problems 53–58, use the function


      #math.equation(block: true, alt: "P open parenthesis h close parenthesis equals 30 open parenthesis 10 close parenthesis to the power minus 0.09 h")[$P ( h ) = 30 ( 10 )^(− 0.09 h)$]
      where altitude, #math.equation(block: false, alt: "h")[$h$], is given in miles and atmospheric pressure, #math.equation(block: false, alt: "P")[$P$], in inches of mercury. Graph this function in the window


      #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 9.4; Ymin, equals 0, Ymax equals 30")[$"Xmin" & = 0 &  & "Xmax" = 9.4 \ "Ymin" & = 0 &  & "Ymax" = 30$]
      Solve the problems below algebraically, and verify with your graph.


      
        The elevation of Mount Everest, the highest mountain in the world, is #math.equation(block: false, alt: "29 , 028")[$29 , 028$] feet. What is the atmospheric pressure at the top? Hint: #math.equation(block: false, alt: "1")[$1$] mile #math.equation(block: false, alt: "equals 5280")[$= 5280$] feet


        
          #math.equation(block: false, alt: "9.60")[$9.60$] in


        
      
      
        The elevation of Mount McKinley, the highest mountain in the United States, is #math.equation(block: false, alt: "20 , 320")[$20 , 320$] feet. What is the atmospheric pressure at the top?


      
      
        How high above sea level is the atmospheric pressure #math.equation(block: false, alt: "20.2")[$20.2$] inches of mercury?


        
          #math.equation(block: false, alt: "1.91")[$1.91$] mi


        
      
      
        How high above sea level is the atmospheric pressure #math.equation(block: false, alt: "16.1")[$16.1$] inches of mercury?


      
      
        Find the height above sea level at which the atmospheric pressure is equal to one-half the pressure at sea level. Hint: What is the altitude at sea level?


        
          #math.equation(block: false, alt: "3.34")[$3.34$] mi


        
      
      
        Find the height above sea level at which the atmospheric pressure is equal to one-fourth the pressure at sea level. Hint: What is the altitude at sea level?


      
      For Problems 59–66, simplify the expression.


      
        #math.equation(block: true, alt: "log sub 2 ⁡ open parenthesis log sub 4 ⁡ open parenthesis 16 close parenthesis close parenthesis")[$log_(2) ( log_(4) ( 16 ) )$]


        
          #math.equation(block: true, alt: "1")[$1$]


        
      
      
        #math.equation(block: true, alt: "log sub 5 ⁡ open parenthesis log sub 5 ⁡ open parenthesis 5 close parenthesis close parenthesis")[$log_(5) ( log_(5) ( 5 ) )$]


      
      
        #math.equation(block: true, alt: "log sub 10 ⁡ open bracket log sub 3 ⁡ open parenthesis log sub 5 ⁡ open parenthesis 125 close parenthesis close parenthesis close bracket")[$log_(10) [ log_(3) ( log_(5) ( 125 ) ) ]$]


        
          #math.equation(block: true, alt: "0")[$0$]


        
      
      
        #math.equation(block: true, alt: "log sub 10 ⁡ open bracket log sub 2 ⁡ open parenthesis log sub 3 ⁡ open parenthesis 9 close parenthesis close parenthesis close bracket")[$log_(10) [ log_(2) ( log_(3) ( 9 ) ) ]$]


      
      
        #math.equation(block: true, alt: "log sub 2 ⁡ open bracket log sub 2 ⁡ open parenthesis log sub 3 ⁡ open parenthesis 81 close parenthesis close parenthesis close bracket")[$log_(2) [ log_(2) ( log_(3) ( 81 ) ) ]$]


        
          #math.equation(block: true, alt: "1")[$1$]


        
      
      
        #math.equation(block: true, alt: "log sub 4 ⁡ open bracket log sub 2 ⁡ open parenthesis log sub 3 ⁡ open parenthesis 81 close parenthesis close parenthesis close bracket")[$log_(4) [ log_(2) ( log_(3) ( 81 ) ) ]$]


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


        
          #math.equation(block: true, alt: "0")[$0$]


        
      
      
        #math.equation(block: true, alt: "log sub b ⁡ open parenthesis log sub a ⁡ open parenthesis a to the power b close parenthesis close parenthesis")[$log_(b) ( log_(a) ( a^(b) ) )$]