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== 5.5#h(0.6em)Chapter Summary and Review


    
      
=== Key Concepts


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

If the inverse of a function #math.equation(block: false, alt: "f")[$f$] is also a function, then the inverse is denoted by the symbol #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$], and

#math.equation(block: true, alt: "f to the power minus 1 open parenthesis b close parenthesis equals a if and only if f open parenthesis a close parenthesis equals b")[$f^(− 1) ( b ) = a " " " " " if and only if " " " " " f ( a ) = b$]
]
+ We can make a table of values for the inverse function, #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$], by interchanging the columns of a table for #math.equation(block: false, alt: "f")[$f$].
+ If a function is defined by a formula in the form #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$], we can find a formula for its inverse function by solving the equation for #math.equation(block: false, alt: "x")[$x$] to get #math.equation(block: false, alt: "x equals f to the power minus 1 open parenthesis y close parenthesis")[$x = f^(− 1) ( y )$].
+ The inverse function #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] undoes the effect of the function #math.equation(block: false, alt: "f")[$f$], that is, if we apply the inverse function to the output of #math.equation(block: false, alt: "f")[$f$], we return to the original input value.
+ If #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] is the inverse function for #math.equation(block: false, alt: "f")[$f$], then #math.equation(block: false, alt: "f")[$f$] is also the inverse function for #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]
+ The graphs of #math.equation(block: false, alt: "f")[$f$] and its inverse function are #strong[symmetric about the line #math.equation(block: false, alt: "y equals x")[$y = x$]].
+ #strong[Horizontal line test]: If no horizontal line intersects the graph of a function more than once, then the inverse is also a function.
+ A function that passes the horizontal line test is called #strong[one-to-one].
+ The inverse of a function #math.equation(block: false, alt: "f")[$f$] is also a function if and only if #math.equation(block: false, alt: "f")[$f$] is one-to-one.
+ We define the 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 ⁡ 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"))[
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")[$log_(b) ( b^(x) ) = x " " " " ", for all " x " " " " " and " " " " " " " b^(log_(b) ( x )) = x " " " " ", for " x > 0$]
]
+ #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 where 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.
]
+ The natural base is an irrational number called #math.equation(block: false, alt: "e")[$e$], where
		#math.equation(block: true, alt: "e approximately equals 2.71828182845")[$e ≈ 2.71828182845$]
+ The #strong[natural exponential function] is the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals e to the power x")[$f ( x ) = e^(x)$]. The #strong[natural log function] is the function #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals ln ⁡ open parenthesis x close parenthesis equals log sub e ⁡ open parenthesis x close parenthesis")[$g ( x ) = ln ( x ) = log_(e) ( x )$].
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Conversion Formulas for Natural Logs]

#math.equation(block: true, alt: "y equals ln ⁡ open parenthesis x close parenthesis if and only if e to the power y equals x")[$bold(italic(y)) = bold(ln) ( bold(italic(x)) ) " " " " " if and only if " " " " " bold(italic(e))^(bold(italic(y))) = bold(italic(x))$]
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Properties of Natural Logarithms]

If #math.equation(block: false, alt: "x , y greater than 0")[$x , y > 0$], then

+ #math.equation(block: false, alt: "ln ⁡ open parenthesis x y close parenthesis equals ln ⁡ open parenthesis x close parenthesis plus ln ⁡ open parenthesis y close parenthesis")[$ln ( x y ) = ln ( x ) + ln ( y )$]
+ #math.equation(block: false, alt: "ln ⁡ open parenthesis the fraction x over y close parenthesis equals ln ⁡ open parenthesis x close parenthesis minus ln ⁡ open parenthesis y close parenthesis")[$ln ( display(frac(x, y)) ) = ln ( x ) − ln ( y )$]
+ #math.equation(block: false, alt: "ln ⁡ open parenthesis x to the power k close parenthesis equals k ln ⁡ open parenthesis x close parenthesis")[$ln ( x^(k) ) = k ln ( x )$]

also

#math.equation(block: true, alt: "ln ⁡ open parenthesis e to the power x close parenthesis equals x , for all x and e to the power ln ⁡ open parenthesis x close parenthesis equals x , for x greater than 0")[$ln ( e^(x) ) = x ", for all " x " and " e^(ln ( x )) = x ", for " x > 0$]
]
+ We use the natural logarithm to solve exponential equations with base #math.equation(block: false, alt: "e")[$e$].
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Exponential Growth and Decay]

The function

#math.equation(block: true, alt: "P open parenthesis t close parenthesis equals P sub 0 e to the power k t")[$bold(italic(P)) ( bold(italic(t)) ) = bold(italic(P))_(0) bold(italic(e))^(bold(italic(k)) bold(italic(t)))$]describes exponential growth if #math.equation(block: false, alt: "k greater than 0")[$k > 0$], and exponential decay if #math.equation(block: false, alt: "k less than 0")[$k < 0$].
]
+ #strong[Continuous compounding]: The amount accumulated in an account after #math.equation(block: false, alt: "t")[$t$] years at interest rate #math.equation(block: false, alt: "r")[$r$] compounded continuously is given by
			#math.equation(block: true, alt: "A open parenthesis t close parenthesis equals P e to the power r t")[$A ( t ) = P e^(r t)$]
		where #math.equation(block: false, alt: "P")[$P$] is the principal invested.
+ A #strong[log scale] is useful for plotting values that vary greatly in magnitude. We plot the log of the variable, instead of the variable itself.
+ A log scale is a #strong[multiplicative scale]: Each increment of equal length on the scale indicates that the value is multiplied by an equal amount.
+ The pH value of a substance is defined by the formula
			#math.equation(block: true, alt: "pH equals minus log sub 10 ⁡ open parenthesis open bracket H to the power plus close bracket close parenthesis")[$"pH" = − log_(10) ( [ H^(+) ] )$]
			where #math.equation(block: false, alt: "open bracket H to the power plus close bracket")[$[ H^(+) ]$] denotes the concentration of hydrogen ions in the substance.
+ The loudness of a sound is measured in decibels, #math.equation(block: false, alt: "D")[$D$], by
			#math.equation(block: true, alt: "D equals 10 log sub 10 ⁡ open parenthesis the fraction I over 10 to the power minus 12 close parenthesis")[$D = 10 log_(10) ( frac(I, 10^(− 12)) )$]
			where #math.equation(block: false, alt: "I")[$I$] is the intensity of its sound waves (in watts per square meter).
+ The Richter magnitude, #math.equation(block: false, alt: "M")[$M$], of an earthquake is given by
			#math.equation(block: true, alt: "M equals log sub 10 ⁡ open parenthesis the fraction A over A sub 0 close parenthesis")[$M = log_(10) ( frac(A, A_(0)) )$]
			where #math.equation(block: false, alt: "A")[$A$] is the amplitude of its seismographic trace and #math.equation(block: false, alt: "A sub 0")[$A_(0)$] is the amplitude of the smallest detectable earthquake.
+ A #emph[difference] of #math.equation(block: false, alt: "K")[$K$] units on a logarithmic scale corresponds to a #emph[factor] of #math.equation(block: false, alt: "10 to the power K")[$10^(K)$] units in the value of the variable.


    
    
      
=== Chapter 5 Review Problems


      For Problems 1–4, make a table of values for the inverse function.


      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x cubed plus x plus 1")[$f ( x ) = x^(3) + x + 1$]


        
          #figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "11")[$11$]]),
  [#math.equation(block: false, alt: "x equals f to the power minus 1 open parenthesis y close parenthesis")[$x = f^(− 1) ( y )$]], [#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: true, alt: "g open parenthesis x close parenthesis equals x plus 6 the cube root of x")[$g ( x ) = x + 6 root(3, x)$]


      
      
        #math.equation(block: true, alt: "g open parenthesis w close parenthesis equals the fraction 1 plus w over w minus 3")[$g ( w ) = display(frac(1 + w, w − 3))$]


        
          #figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction minus 1 over 3")[$frac(− 1, 3)$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]]),
  [#math.equation(block: false, alt: "w equals g to the power minus 1 open parenthesis y close parenthesis")[$w = g^(− 1) ( y )$]], [#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: true, alt: "f open parenthesis n close parenthesis equals the fraction n over 1 plus n")[$f ( n ) = display(frac(n, 1 + n))$]


      
      For Problems 5–6, use the graph to find the function values.


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


        + #math.equation(block: false, alt: "P to the power minus 1 open parenthesis 350 close parenthesis")[$P^(− 1) ( 350 )$]
+ #math.equation(block: false, alt: "P to the power minus 1 open parenthesis 100 close parenthesis")[$P^(− 1) ( 100 )$]


        
          + #math.equation(block: false, alt: "P to the power minus 1 open parenthesis 350 close parenthesis equals 40")[$P^(− 1) ( 350 ) = 40$]
+ #math.equation(block: false, alt: "P to the power minus 1 open parenthesis 100 close parenthesis equals 0")[$P^(− 1) ( 100 ) = 0$]


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


        + #math.equation(block: false, alt: "H to the power minus 1 open parenthesis 200 close parenthesis")[$H^(− 1) ( 200 )$]
+ #math.equation(block: false, alt: "H to the power minus 1 open parenthesis 75 close parenthesis")[$H^(− 1) ( 75 )$]


      
      For Problems 7–12,


      + Find a formula for the inverse #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$] of each function.
+ Graph the function and its inverse on the same set of axes, along with the graph of #math.equation(block: false, alt: "y equals x")[$y = x$].


      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x plus 4")[$f ( x ) = x + 4$]


        
          + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals x minus 4")[$f^(− 1) ( x ) = x − 4$]
+ #figure(figph[line and inverse], alt: "line and inverse", caption: none)


        
      
      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction x minus 2 over 4")[$f ( x ) = display(frac(x − 2, 4))$]


      
      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x cubed minus 1")[$f ( x ) = x^(3) − 1$]


        
          + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the cube root of x plus 1")[$f^(− 1) ( x ) = root(3, x + 1)$]
+ #figure(figph[cubic and inverse], alt: "cubic and inverse", caption: none)


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


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


        
          + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis x close parenthesis equals the fraction 1 over x minus 2")[$f^(− 1) ( x ) = display(frac(1, x − 2))$]
+ #figure(figph[translated reciprocal and inverse], alt: "translated reciprocal and inverse", caption: none)


        
      
      
        #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the cube root of x minus 2")[$f ( x ) = root(3, x) − 2$]


      
      
        If #math.equation(block: false, alt: "F open parenthesis t close parenthesis equals the fraction 3 over 4 t plus 2")[$F ( t ) = display(frac(3, 4)) t + 2$], find #math.equation(block: false, alt: "F to the power minus 1 open parenthesis 2 close parenthesis")[$F^(− 1) ( 2 )$].


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


        
      
      
        If #math.equation(block: false, alt: "G open parenthesis x close parenthesis equals the fraction 1 over x minus 4")[$G ( x ) = display(frac(1, x)) − 4$], find #math.equation(block: false, alt: "G to the power minus 1 open parenthesis 3 close parenthesis")[$G^(− 1) ( 3 )$].


      
      
        The table shows the revenue, #math.equation(block: false, alt: "R")[$R$], from sales of the Miracle Mop as a function of the number of dollars spent on advertising, #math.equation(block: false, alt: "A")[$A$]. Let #math.equation(block: false, alt: "f")[$f$] be the name of the function defined by the table, so #math.equation(block: false, alt: "R equals f open parenthesis A close parenthesis")[$R = f ( A )$].


        #figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "A")[$A$] (thousands #linebreak() of dollars) #linebreak()], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "150")[$150$]], [#math.equation(block: false, alt: "200")[$200$]], [#math.equation(block: false, alt: "250")[$250$]], [#math.equation(block: false, alt: "300")[$300$]]),
  [#math.equation(block: false, alt: "R")[$R$] (thousands #linebreak() of dollars) #linebreak()], [#math.equation(block: false, alt: "250")[$250$]], [#math.equation(block: false, alt: "280")[$280$]], [#math.equation(block: false, alt: "300")[$300$]], [#math.equation(block: false, alt: "310")[$310$]], [#math.equation(block: false, alt: "315")[$315$]],
))


        + Evaluate #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 300 close parenthesis")[$f^(− 1) ( 300 )$]. Explain its meaning in this context.
+ Write two equations to answer the following question, one using #math.equation(block: false, alt: "f")[$f$] and one using #math.equation(block: false, alt: "f to the power minus 1")[$f^(− 1)$]: How much should we spend on advertising to generate revenue of #math.equation(block: false, alt: "$ 250 , 000")[$upright(\$) 250 , 000$]?


        
          + #math.equation(block: false, alt: "f to the power minus 1 open parenthesis 300 close parenthesis equals 200")[$f^(− 1) ( 300 ) = 200$]: #math.equation(block: false, alt: "$ 200 , 000")[$upright(\$) 200 , 000$] in advertising results in #math.equation(block: false, alt: "$ 300 , 000")[$upright(\$) 300 , 000$] in revenue.
+ #math.equation(block: false, alt: "f open parenthesis A close parenthesis equals 250")[$f ( A ) = 250$] or #math.equation(block: false, alt: "A equals f to the power minus 1 open parenthesis 250 close parenthesis")[$A = f^(− 1) ( 250 )$]


        
      
      
        The table shows the systolic blood pressure, #math.equation(block: false, alt: "S")[$S$], of a patient as a function of the dosage, #math.equation(block: false, alt: "d")[$d$], of medication he receives. Let #math.equation(block: false, alt: "g")[$g$] be the name of the function defined by the table, so #math.equation(block: false, alt: "S equals g open parenthesis d close parenthesis")[$S = g ( d )$].


        #figure(table(
  columns: 6, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "d")[$d$] (mg)], [#math.equation(block: false, alt: "190")[$190$]], [#math.equation(block: false, alt: "195")[$195$]], [#math.equation(block: false, alt: "200")[$200$]], [#math.equation(block: false, alt: "210")[$210$]], [#math.equation(block: false, alt: "220")[$220$]]),
  [#math.equation(block: false, alt: "S")[$S$] (mm Hg)], [#math.equation(block: false, alt: "220")[$220$]], [#math.equation(block: false, alt: "200")[$200$]], [#math.equation(block: false, alt: "190")[$190$]], [#math.equation(block: false, alt: "185")[$185$]], [#math.equation(block: false, alt: "183")[$183$]],
))


        + Evaluate #math.equation(block: false, alt: "g to the power minus 1 open parenthesis 200 close parenthesis")[$g^(− 1) ( 200 )$]. Explain its meaning in this context.
+ Write two equations to answer the following question, one using #math.equation(block: false, alt: "g")[$g$] and one using #math.equation(block: false, alt: "g to the power minus 1")[$g^(− 1)$]: What dosage results in systolic blood pressure of #math.equation(block: false, alt: "220")[$220$]?


      
      For Problems 17–24, write the equation in exponential form.


      
        #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis 0.001 close parenthesis equals z")[$log_(10) ( 0.001 ) = z$]


        
          #math.equation(block: true, alt: "10 to the power z equals 0.001")[$10^(z) = 0.001$]


        
      
      
        #math.equation(block: true, alt: "log sub 3 ⁡ open parenthesis 20 close parenthesis equals t")[$log_(3) ( 20 ) = t$]


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


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


        
      
      
        #math.equation(block: true, alt: "log sub 5 ⁡ open parenthesis 3 close parenthesis equals 6 minus 2 p")[$log_(5) ( 3 ) = 6 − 2 p$]


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


        
          #math.equation(block: true, alt: "b cubed equals 3 x plus 1")[$b^(3) = 3 x + 1$]


        
      
      
        #math.equation(block: true, alt: "log sub m ⁡ open parenthesis 8 close parenthesis equals 4 t")[$log_(m) ( 8 ) = 4 t$]


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


        
          #math.equation(block: true, alt: "n to the power p minus 1 equals q")[$n^(p − 1) = q$]


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


      
      For Problems 25–28, simplify.


      
        #math.equation(block: true, alt: "10 to the power log ⁡ open parenthesis 6 n close parenthesis")[$10^(log ( 6 n ))$]


        
          #math.equation(block: true, alt: "6 n")[$6 n$]


        
      
      
        #math.equation(block: true, alt: "log ⁡ 100 to the power x")[$log 100^(x)$]


      
      
        #math.equation(block: true, alt: "log sub 2 ⁡ 4 to the power x plus 3")[$log_(2) 4^(x + 3)$]


        
          #math.equation(block: true, alt: "2 x plus 6")[$2 x + 6$]


        
      
      
        #math.equation(block: true, alt: "3 to the power 2 log sub 3 ⁡ open parenthesis t close parenthesis")[$3^(2 log_(3) ( t ))$]


      
      For Problems 29–36, solve.


      
        #math.equation(block: true, alt: "log sub 3 ⁡ open parenthesis the fraction 1 over 3 close parenthesis equals y")[$log_(3) ( display(frac(1, 3)) ) = y$]


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


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


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


        
          #math.equation(block: true, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]


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


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


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


        
      
      
        #math.equation(block: true, alt: "log sub b ⁡ open parenthesis 9 close parenthesis equals the fraction 1 over 2")[$log_(b) ( 9 ) = display(frac(1, 2))$]


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


        
          #math.equation(block: true, alt: "the fraction minus 15 over 8")[$display(frac(− 15, 8))$]


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


      
      For Problems 37–40, solve.


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


        
          #math.equation(block: true, alt: "the fraction 9 over 4")[$display(frac(9, 4))$]


        
      
      
        #math.equation(block: true, alt: "log sub 2 ⁡ open parenthesis x plus 2 close parenthesis minus log sub 2 ⁡ open parenthesis 3 close parenthesis equals 6")[$log_(2) ( x + 2 ) − log_(2) ( 3 ) = 6$]


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


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


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


      
      For Problems 41–46, solve.


      
        #math.equation(block: true, alt: "e to the power x equals 4.7")[$e^(x) = 4.7$]


        
          #math.equation(block: true, alt: "x approximately equals 1.548")[$x ≈ 1.548$]


        
      
      
        #math.equation(block: true, alt: "e to the power x equals 0.5")[$e^(x) = 0.5$]


      
      
        #math.equation(block: true, alt: "ln ⁡ open parenthesis x close parenthesis equals 6.02")[$ln ( x ) = 6.02$]


        
          #math.equation(block: true, alt: "x approximately equals 411.58")[$x ≈ 411.58$]


        
      
      
        #math.equation(block: true, alt: "ln ⁡ open parenthesis x close parenthesis equals minus 1.4")[$ln ( x ) = − 1.4$]


      
      
        #math.equation(block: true, alt: "4.73 equals 1.2 e to the power 0.6 x")[$4.73 = 1.2 e^(0.6 x)$]


        
          #math.equation(block: true, alt: "x approximately equals 2.286")[$x ≈ 2.286$]


        
      
      
        #math.equation(block: true, alt: "1.75 equals 0.3 e to the power minus 1.2 x")[$1.75 = 0.3 e^(− 1.2 x)$]


      
      For Problems 47–50, simplify.


      
        #math.equation(block: true, alt: "e to the power ln ⁡ open parenthesis x close parenthesis / 2")[$e^(ln ( x ) / 2)$]


        
          #math.equation(block: true, alt: "the square root of x")[$sqrt(x)$]


        
      
      
        #math.equation(block: true, alt: "ln ⁡ open parenthesis open parenthesis the fraction 1 over e close parenthesis close parenthesis to the power 2 n")[$ln attach(( ( display(frac(1, e)) ) ), t: 2 n)$]


      
      
        #math.equation(block: true, alt: "ln ⁡ open parenthesis the fraction e to the power k over e cubed close parenthesis")[$ln ( display(frac(e^(k), e^(3))) )$]


        
          #math.equation(block: true, alt: "k minus 3")[$k − 3$]


        
      
      
        #math.equation(block: true, alt: "e to the power ln ⁡ open parenthesis e plus x close parenthesis")[$e^(ln ( e + x ))$]


      
      
        In 1970, the population of New York City was #math.equation(block: false, alt: "7 , 894 , 862")[$7 , 894 , 862$]. In 1980, the population had fallen to #math.equation(block: false, alt: "7 , 071 , 639")[$7 , 071 , 639$].


        + Write an exponential function using base #math.equation(block: false, alt: "e")[$e$] for the population of New York over that decade.
+ By what percent did the population decline annually?


        
          + #math.equation(block: false, alt: "P equals 7 , 894 , 862 e to the power minus 0.011 t")[$P = 7 , 894 , 862 e^(− 0.011 t)$]
+ #math.equation(block: false, alt: "1.095 %")[$1.095 upright(%)$]


        
      
      
        In 1990, the population of New York City was #math.equation(block: false, alt: "7 , 322 , 564")[$7 , 322 , 564$]. In 2000, the population was #math.equation(block: false, alt: "8 , 008 , 278")[$8 , 008 , 278$].


        + Write an exponential function using base #math.equation(block: false, alt: "e")[$e$] for the population of New York over that decade.
+ By what percent did the population increase annually?


      
      
        You deposit \$#math.equation(block: false, alt: "1000")[$1000$] in a savings account paying #math.equation(block: false, alt: "5 %")[$5 upright(%)$] interest compounded continuously.


        + Find the amount in the account after #math.equation(block: false, alt: "7")[$7$] years.
+ How long will it take for the original principal to double?
+ Find a formula for the time #math.equation(block: false, alt: "t")[$t$] required for the amount to reach #math.equation(block: false, alt: "A")[$A$].


        
          + \$#math.equation(block: false, alt: "1419.07")[$1419.07$]
+ #math.equation(block: false, alt: "13.9")[$13.9$] years
+ #math.equation(block: false, alt: "t equals 20 ln ⁡ open parenthesis the fraction A over 1000 close parenthesis")[$t = 20 ln ( display(frac(A, 1000)) )$]


        
      
      
        The voltage, #math.equation(block: false, alt: "V")[$V$], across a capacitor in a certain circuit is given by the function


        #math.equation(block: true, alt: "V open parenthesis t close parenthesis equals 100 open parenthesis 1 minus e to the power minus 0.5 t close parenthesis")[$V ( t ) = 100 ( 1 − e^(− 0.5 t) )$]
        where #math.equation(block: false, alt: "t")[$t$] is the time in seconds.


        + Make a table of values and graph #math.equation(block: false, alt: "V open parenthesis t close parenthesis")[$V ( t )$] for #math.equation(block: false, alt: "t equals 0")[$t = 0$] to #math.equation(block: false, alt: "t equals 10")[$t = 10$].
+ Describe the graph. What happens to the voltage in the long run?
+ How much time must elapse (to the nearest hundredth of a second) for the voltage to reach #math.equation(block: false, alt: "75")[$75$] volts?


      
      
        Solve for #math.equation(block: false, alt: "t")[$t$]: #math.equation(block: false, alt: "y equals 12 e to the power minus k t plus 6")[$" " " " y = 12 e^(− k t) + 6$]


        
          #math.equation(block: true, alt: "t equals the fraction minus 1 over k ln ⁡ open parenthesis the fraction y minus 6 over 12 close parenthesis")[$t = display(frac(− 1, k)) ln ( display(frac(y − 6, 12)) )$]


        
      
      
        Solve for #math.equation(block: false, alt: "k")[$k$]: #math.equation(block: false, alt: "N equals N sub 0 plus 4 ln ⁡ open parenthesis k plus 10 close parenthesis")[$" " " " N = N_(0) + 4 ln ( k + 10 )$]


      
      
        Solve for #math.equation(block: false, alt: "M")[$M$]: #math.equation(block: false, alt: "Q equals the fraction 1 over t open parenthesis the fraction log ⁡ open parenthesis M close parenthesis over log ⁡ open parenthesis N close parenthesis close parenthesis")[$" " " " Q = display(frac(1, t)) ( display(frac(log ( M ), log ( N ))) )$]


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


        
      
      
        Solve for #math.equation(block: false, alt: "t")[$t$]: #math.equation(block: false, alt: "C sub H equals C sub L times 10 to the power k t")[$" " " " C_(H) = C_(L) ⋅ 10^(k) t$]


      
      
        Express #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 750 e to the power 0.32 t")[$P ( t ) = 750 e^(0.32 t)$] in the form #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 b to the power t")[$P ( t ) = P_(0) b^(t)$].


        
          #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 750 open parenthesis 1.3771 close parenthesis to the power t")[$P ( t ) = 750 ( 1.3771 )^(t)$]


        
      
      
        Express #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 80 e to the power minus 0.6 t")[$P ( t ) = 80 e^(− 0.6 t)$] in the form #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 b to the power t")[$P ( t ) = P_(0) b^(t)$].


      
      
        Express #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals 600 open parenthesis 0.4 close parenthesis to the power t")[$N ( t ) = 600 ( 0.4 )^(t)$] in the form #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals N sub 0 e to the power k t")[$N ( t ) = N_(0) e^(k t)$].


        
          #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals 600 e to the power minus 0.9163 t")[$N ( t ) = 600 e^(− 0.9163 t)$]


        
      
      
        Express #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals 100 open parenthesis 1.06 close parenthesis to the power t")[$N ( t ) = 100 ( 1.06 )^(t)$] in the form #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals N sub 0 e to the power k t")[$N ( t ) = N_(0) e^(k t)$].


      
      
        Plot the values on a log scale.


        #figure(table(
  columns: 5, align: left, inset: 6pt,
  [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0.04")[$0.04$]], [#math.equation(block: false, alt: "45")[$45$]], [#math.equation(block: false, alt: "1200")[$1200$]], [#math.equation(block: false, alt: "560 , 000")[$560 , 000$]],
))


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


        
      
      
        Plot the values on a log scale.


        #figure(table(
  columns: 5, align: left, inset: 6pt,
  [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0.0007")[$0.0007$]], [#math.equation(block: false, alt: "0.8")[$0.8$]], [#math.equation(block: false, alt: "3.2")[$3.2$]], [#math.equation(block: false, alt: "2500")[$2500$]],
))


      
      
        The graph describes a network of streams near Santa Fe, New Mexico. It shows the number of streams of a given order, which is a measure of their size. Use the graph to estimate the number of streams of orders #math.equation(block: false, alt: "3")[$3$], #math.equation(block: false, alt: "4")[$4$], #math.equation(block: false, alt: "8")[$8$], and #math.equation(block: false, alt: "9")[$9$]. (Source: Leopold, Wolman, and Miller)


        #figure(figph[stream order on semi-log scale], alt: "stream order on semi-log scale", caption: none)


        
          Order #math.equation(block: false, alt: "3")[$3$]: #math.equation(block: false, alt: "17 , 000")[$17 , 000$]; Order #math.equation(block: false, alt: "4")[$4$]: #math.equation(block: false, alt: "5000")[$5000$]; Order #math.equation(block: false, alt: "8")[$8$]: #math.equation(block: false, alt: "40")[$40$]; Order #math.equation(block: false, alt: "9")[$9$]: #math.equation(block: false, alt: "11")[$11$]


        
      
      
        Large animals use oxygen more efficiently when running than small animals do. The graph shows the amount of oxygen various animals use, per gram of their body weight, to run #math.equation(block: false, alt: "1")[$1$] kilometer. Estimate the body mass and oxygen use for a kangaroo rat, a dog, and a horse. (Source: Schmidt-Neilsen, 1972)


        #figure(figph[A log-log scatter plot with both axes on a logarithmic scale (horizontal from 0.01 to 1000, vertical from 0.01 to 10). A downward-sloping straight line fits six red data points labeled by animal: white mouse, kangaroo rat, ground squirrel, white rat, dog, and horse. A straight line on logarithmic axes indicates a power-law relationship.], alt: "A log-log scatter plot with both axes on a logarithmic scale (horizontal from 0.01 to 1000, vertical from 0.01 to 10). A downward-sloping straight line fits six red data points labeled by animal: white mouse, kangaroo rat, ground squirrel, white rat, dog, and horse. A straight line on logarithmic axes indicates a power-law relationship.", caption: none)


      
      
        The pH of an unknown substance is #math.equation(block: false, alt: "6.3")[$6.3$]. What is its hydrogen ion concentration?


        
          #math.equation(block: true, alt: "5 times 10 to the power minus 7")[$5 × 10^(− 7)$]


        
      
      
        The noise of a leaf blower was measured at #math.equation(block: false, alt: "110")[$110$] decibels. What was the intensity of the sound waves?


      
      
        A refrigerator produces #math.equation(block: false, alt: "50")[$50$] decibels of noise, and a vacuum cleaner produces #math.equation(block: false, alt: "85")[$85$] decibels. How much more intense are the sound waves from a vacuum cleaner than those from a refrigerator?


        
          #math.equation(block: true, alt: "3160")[$3160$]


        
      
      
        In 2004, a magnitude #math.equation(block: false, alt: "9.0")[$9.0$] earthquake struck Sumatra in Indonesia. How much more powerful was this quake than the 1906 San Francisco earthquake of magnitude #math.equation(block: false, alt: "8.3")[$8.3$]?