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== 4.6#h(0.6em)Chapter Summary and Review


    
      
=== Key Concepts


      + If a quantity is multiplied by a constant factor, #math.equation(block: false, alt: "b")[$b$], in each time period, we say that it undergoes #strong[exponential growth] or #strong[decay] . The constant #math.equation(block: false, alt: "b")[$b$] is called the #strong[growth factor] if #math.equation(block: false, alt: "b greater than 1")[$b > 1$] and the #strong[decay factor] if #math.equation(block: false, alt: "0 less than b less than 1")[$0 < b < 1$].
+ Quantities that increase or decrease by a constant percent in each time period grow or decay exponentially.
+ #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 b to the power t")[$P ( t ) = P_(0) b^(t)$]models exponential growth and decay.

#math.equation(block: false, alt: "P sub 0 equals P open parenthesis 0 close parenthesis")[$P_(0) = P ( 0 )$] is the initial value of #math.equation(block: false, alt: "P")[$P$];

#math.equation(block: false, alt: "b")[$b$] is the growth or decay factor.

+ If #math.equation(block: false, alt: "b greater than 1")[$b > 1$], then #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$] is increasing, and #math.equation(block: false, alt: "b equals 1 plus r")[$b = 1 + r$], where #math.equation(block: false, alt: "r")[$r$] represents percent increase.
+ If #math.equation(block: false, alt: "0 less than b less than 1")[$0 < b < 1$], then #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$] is decreasing, and #math.equation(block: false, alt: "b equals 1 minus r")[$b = 1 − r$], where #math.equation(block: false, alt: "r")[$r$] represents percent decrease.
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Interest Compounded Annually]

The amount #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] accumulated (principal plus interest) in an account bearing interest compounded annually is

#math.equation(block: true, alt: "A open parenthesis t close parenthesis equals open parenthesis 1 plus r close parenthesis to the power t")[$A ( t ) = ( 1 + r )^(t)$]where

#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([], [], [#math.equation(block: false, alt: "P")[$P$]], [is the principal invested,]),
  [], [], [#math.equation(block: false, alt: "r")[$r$]], [is the interest rate,],
  [], [], [#math.equation(block: false, alt: "t")[$t$]], [is the time period, in years.],
))
]
+ In linear growth, a constant amount is #emph[added] to the output for each unit increase in the input. In exponential growth, the output is #emph[multiplied] by a constant factor for each unit increase in the input.
+ An #strong[exponential function] has the form
			#math.equation(block: true, alt: "f open parenthesis x close parenthesis equals a b to the power x , where b greater than 0 and b not equal to 1 , a not equal to 0")[$f ( x ) = a b^(x) , " " " " " where " " " " " b > 0 " " " " " and " b ≠ 1 , " " " " " " a ≠ 0$]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Properties of Exponential Functions, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a b to the power x , a greater than 0")[$f ( x ) = a b^(x) , " " " " a > 0$]]

+ Domain: all real numbers
+ Range: all positive numbers
+ If #math.equation(block: false, alt: "b greater than 1")[$b > 1$], the function is increasing and concave up; if #math.equation(block: false, alt: "0 less than b less than 1")[$0 < b < 1$], the function is decreasing and concave up.
+ The #math.equation(block: false, alt: "y")[$y$]-intercept is #math.equation(block: false, alt: "open parenthesis a , 0 close parenthesis")[$( a , 0 )$]. There is no #math.equation(block: false, alt: "x")[$x$]-intercept
]
+ The graphs of exponential functions can be transformed by shifts, stretches, and reflections.
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Reflections of Graphs]

+ The graph of #math.equation(block: false, alt: "y equals minus f open parenthesis x close parenthesis")[$y = − f ( x )$] is the reflection of the graph of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] about the #math.equation(block: false, alt: "x")[$x$]-axis.
+ The graph of #math.equation(block: false, alt: "y equals f open parenthesis minus x close parenthesis")[$y = f ( − x )$] is the reflection of the graph of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] about the #math.equation(block: false, alt: "y")[$y$]-axis.
]
+ Exponential functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a b to the power x")[$f ( x ) = a b^(x)$] have different properties than power functions #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals k x to the power p")[$f ( x ) = k x^(p)$].
+ We can solve #strong[exponential equations] by writing both sides with the same base and equating the exponents.
+ We can use graphs to find approximate solutions to exponential equations.
+ We use logrithms 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 ⁡ 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")[$b > 0$], #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 ⁡ x 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 10 logarithms are called #strong[common logarithms], and #math.equation(block: false, alt: "log ⁡ x")[$log x$] means #math.equation(block: false, alt: "log sub 10 ⁡ x")[$log_(10) x$].
+ #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.
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Properties of Logarithms]

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

+ #math.equation(block: false, alt: "log sub b ⁡ open parenthesis x y close parenthesis equals log sub b ⁡ x plus log sub b ⁡ y")[$log_(b) ( x y ) = log_(b) x + log_(b) y$]
+ #math.equation(block: false, alt: "log sub b ⁡ the fraction x over y equals log sub b ⁡ x minus log sub b ⁡ y")[$log_(b) display(frac(x, y)) = log_(b) x − log_(b) y$]
+ #math.equation(block: false, alt: "log sub b ⁡ x to the power k equals k log sub b ⁡ x")[$log_(b) x^(k) = k log_(b) x$]
]
+ We can use the properties of logarithms to solve exponential equations with any base.
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Compounded Interest]

The amount #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] accumulated (principal plus interest) in an account bearing interest compounded #math.equation(block: false, alt: "n")[$n$] times annually is

#math.equation(block: true, alt: "A open parenthesis t close parenthesis equals open parenthesis 1 plus the fraction r over n close parenthesis to the power n t")[$A ( t ) = attach(( 1 + display(frac(r, n)) ), t: n t)$]where

#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([], [], [#math.equation(block: false, alt: "P")[$P$]], [is the principal invested,]),
  [], [], [#math.equation(block: false, alt: "r")[$r$]], [is the interest rate,],
  [], [], [#math.equation(block: false, alt: "t")[$t$]], [is the time period, in years.],
))
]
+ We can use the ratio method to fit an exponential function through two points.#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[To find an exponential function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a b to the power x")[$f ( x ) = a b^(x)$] through two points:]

+ Use the coordinates of the points to write two equations in #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$].
+ Divide one equation by the other to eliminate #math.equation(block: false, alt: "a")[$a$].
+ Solve for #math.equation(block: false, alt: "b")[$b$].
+ Substitute #math.equation(block: false, alt: "b")[$b$] into either equation and solve for #math.equation(block: false, alt: "a")[$a$].
]
+ Every increasing exponential has a fixed #strong[doubling time]. Every decreasing exponential function has a fixed #strong[half-life].
+ If #math.equation(block: false, alt: "D")[$D$] is the doubling time for a population, its growth law can be written as #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 02 to the power t / D")[$P ( t ) = P_(0) 2^(t / D)$].
+ If #math.equation(block: false, alt: "H")[$H$] is the half-life for a quantity, its decay law can be written as #math.equation(block: false, alt: "Q open parenthesis t close parenthesis equals Q sub 0 open parenthesis 0.5 close parenthesis to the power t / H")[$Q ( t ) = Q_(0) ( 0.5 )^(t / H)$].
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Future Value of an Annuity]

If you make #math.equation(block: false, alt: "n")[$n$] payments per year for #math.equation(block: false, alt: "t")[$t$] years into an annuity that pays interest rate #math.equation(block: false, alt: "r")[$r$] compounded #math.equation(block: false, alt: "n")[$n$] times per year, the #strong[future value], #math.equation(block: false, alt: "F V")[$F V$], of the annuity is

#math.equation(block: true, alt: "F V equals the fraction P open bracket open parenthesis 1 plus the fraction r over n close parenthesis to the power n t minus 1 close bracket over the fraction r over n")[$F V = display(frac(P [ attach(( 1 + display(frac(r, n)) ), t: n t) − 1 ], display(frac(r, n))))$]where each payment is #math.equation(block: false, alt: "P")[$P$] dollars.
]
+ #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Present Value of an Annuity]

If you wish to receive #math.equation(block: false, alt: "n")[$n$] payments per year for #math.equation(block: false, alt: "t")[$t$] years from a fund that earns interest rate #math.equation(block: false, alt: "r")[$r$] compounded #math.equation(block: false, alt: "n")[$n$] times per year, the #strong[present value], #math.equation(block: false, alt: "P V")[$P V$], of the annuity must be

#math.equation(block: true, alt: "P V equals the fraction P open bracket 1 minus open parenthesis 1 plus the fraction r over n close parenthesis to the power minus n t close bracket over the fraction r over n")[$P V = frac(P [ 1 − attach(( 1 + display(frac(r, n)) ), t: − n t) ], display(frac(r, n)))$]where each payment is #math.equation(block: false, alt: "P")[$P$] dollars.
]


    
    
      
=== Chapter 4 Review Problems


      For Problems 1–4,


      + Write a function that describes exponential growth or decay.
+ Evaluate the function at the given values.


      
        The number of computer science degrees awarded by Monroe College has increased by a factor of #math.equation(block: false, alt: "1.5")[$1.5$] every #math.equation(block: false, alt: "5")[$5$] years since 1984. If the college granted #math.equation(block: false, alt: "8")[$8$] degrees in 1984, how many did it award in 1994? In 2005?


        
          + #math.equation(block: false, alt: "D equals 8 open parenthesis 1.5 close parenthesis to the power t / 5")[$D = 8 ( 1.5 )^(t / 5)$]
+ #math.equation(block: false, alt: "18")[$18$]; #math.equation(block: false, alt: "44")[$44$]


        
      
      
        The price of public transportation has been rising by #math.equation(block: false, alt: "10 %")[$10 upright(%)$] per year since #math.equation(block: false, alt: "1975")[$1975$]. If it cost \$#math.equation(block: false, alt: "0.25")[$0.25$] to ride the bus in #math.equation(block: false, alt: "1975")[$1975$], how much did it cost in #math.equation(block: false, alt: "1985")[$1985$]? How much will it cost in the year #math.equation(block: false, alt: "2010")[$2010$] if the current trend continues?


      
      
        A certain medication is eliminated from the body at a rate of #math.equation(block: false, alt: "15 %")[$15 upright(%)$] per hour. If an initial dose of #math.equation(block: false, alt: "100")[$100$] milligrams is taken at #math.equation(block: false, alt: "8")[$8$] a.m., how much is left at #math.equation(block: false, alt: "12")[$12$] noon? At #math.equation(block: false, alt: "6")[$6$] p.m.?


        
          + #math.equation(block: false, alt: "M equals 100 open parenthesis 0.85 close parenthesis to the power t")[$M = 100 ( 0.85 )^(t)$]
+ #math.equation(block: false, alt: "52.2")[$52.2$] mg; #math.equation(block: false, alt: "19.7")[$19.7$] mg


        
      
      
        After the World Series, sales of T-shirts and other baseball memorabilia decline #math.equation(block: false, alt: "30 %")[$30 upright(%)$] per week. If \$#math.equation(block: false, alt: "200 , 000")[$200 , 000$] worth of souvenirs were sold during the Series, how much will be sold #math.equation(block: false, alt: "4")[$4$] weeks later? After #math.equation(block: false, alt: "6")[$6$] weeks?


      
      For Problems 5-8, use the laws of exponents to simplify.


      
        #math.equation(block: true, alt: "open parenthesis 4 n to the power x plus 5 close parenthesis squared")[$( 4 n^(x + 5) )^(2)$]


        
          #math.equation(block: true, alt: "16 n to the power 2 x plus 10")[$16 n^(2 x + 10)$]


        
      
      
        #math.equation(block: true, alt: "9 to the power x times 3 to the power x minus 3")[$9^(x) ⋅ 3^(x − 3)$]


      
      
        #math.equation(block: true, alt: "the fraction m to the power x plus 2 over m to the power 2 x plus 4")[$display(frac(m^(x + 2), m^(2 x + 4)))$]


        
          #math.equation(block: true, alt: "the fraction 1 over m to the power x plus 2")[$display(frac(1, m^(x + 2)))$]


        
      
      
        #math.equation(block: true, alt: "the cube root of 8 to the power 2 x plus 1 times 8 to the power x minus 2")[$root(3, 8^(2 x + 1) ⋅ 8^(x − 2))$]


      
      For Problems 5-8, find a growth or decay law for the function.


      
        #figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]]),
  [#math.equation(block: false, alt: "g open parenthesis t close parenthesis")[$g ( t )$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "13.6")[$13.6$]], [#math.equation(block: false, alt: "11.56")[$11.56$]], [#math.equation(block: false, alt: "9.83")[$9.83$]],
))


        
          #math.equation(block: true, alt: "g open parenthesis t close parenthesis equals 16 open parenthesis 0.85 close parenthesis to the power t")[$g ( t ) = 16 ( 0.85 )^(t)$]


        
      
      
        #figure(table(
  columns: 5, align: left, inset: 6pt,
  table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]]),
  [#math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "19.2")[$19.2$]], [#math.equation(block: false, alt: "30.72")[$30.72$]], [#math.equation(block: false, alt: "49.15")[$49.15$]],
))


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


        
          #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 500 open parenthesis the fraction 1 over 5 close parenthesis to the power x")[$f ( x ) = 500 attach(( display(frac(1, 5)) ), t: x)$]


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


      
      
        The president's approval rating increased by #math.equation(block: false, alt: "12 %")[$12 upright(%)$] and then decreased by #math.equation(block: false, alt: "15 %")[$15 upright(%)$]. What was the net change in his approval rating?


        
          #math.equation(block: false, alt: "4.8 %")[$4.8 upright(%)$] loss


        
      
      
        The number of students at Salt Creek Elementary School fell by #math.equation(block: false, alt: "18 %")[$18 upright(%)$] last year but increased by #math.equation(block: false, alt: "26 %")[$26 upright(%)$] this year. What was the net change in the number of students?


      
      
        Enviroco's stock is growing exponentially in value and increased by #math.equation(block: false, alt: "33.8 %")[$33.8 upright(%)$] over the past #math.equation(block: false, alt: "5")[$5$] years. What was its annual rate of increase?


        
          #math.equation(block: false, alt: "6 %")[$6 upright(%)$] loss


        
      
      
        Sales of the software package Home Accountant 3.0 fell exponentially when the new version came out, decreasing by #math.equation(block: false, alt: "60 %")[$60 upright(%)$] over the past 3 months. What was the monthly rate of decrease?


      
      For Problems 17–20,


      + Graph the function.
+ List all intercepts and asymptotes.
+ Give the range of the function on the domain #math.equation(block: false, alt: "open bracket minus 3 , 3 close bracket")[$[ − 3 , 3 ]$].


      
        #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals 6 open parenthesis 1.2 close parenthesis to the power t")[$f ( t ) = 6 ( 1.2 )^(t)$]


        
          + #figure(figph[growth], alt: "growth", caption: none)
+ #math.equation(block: false, alt: "y")[$y$]-intercept #math.equation(block: false, alt: "open parenthesis 0 , 6 close parenthesis")[$( 0 , 6 )$]; asymptote: #math.equation(block: false, alt: "y equals 0")[$y = 0$]
+ #math.equation(block: false, alt: "open bracket 3.472 , 10.368 close bracket")[$[ 3.472 , 10.368 ]$]


        
      
      
        #math.equation(block: true, alt: "g open parenthesis t close parenthesis equals 35 open parenthesis 0.6 close parenthesis to the power minus t")[$g ( t ) = 35 ( 0.6 )^(− t)$]


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


        
          + #figure(figph[growth], alt: "growth", caption: none)
+ #math.equation(block: false, alt: "x")[$x$]-intercept #math.equation(block: false, alt: "open parenthesis the fraction log ⁡ 3 over log ⁡ 2 , 0 close parenthesis")[$( frac(log 3, log 2) , 0 )$]; #math.equation(block: false, alt: "y")[$y$]-intercept #math.equation(block: false, alt: "open parenthesis 0 , minus 2 close parenthesis")[$( 0 , − 2 )$]; asymptote: #math.equation(block: false, alt: "y equals minus 3")[$y = − 3$]
+ #math.equation(block: false, alt: "open bracket minus 2.875 , 5 close bracket")[$[ − 2.875 , 5 ]$]


        
      
      
        #math.equation(block: true, alt: "R open parenthesis x close parenthesis equals 2 to the power x plus 3")[$R ( x ) = 2^(x + 3)$]


      
      For Problems 21-24, solve the equation.


      
        #math.equation(block: true, alt: "3 to the power x plus 2 equals 9 to the power 1 / 3")[$3^(x + 2) = 9^(1 / 3)$]


        
          #math.equation(block: true, alt: "the fraction minus 4 over 3")[$display(frac(− 4, 3))$]


        
      
      
        #math.equation(block: true, alt: "2 to the power x minus 1 equals 8 to the power minus 2 x")[$2^(x − 1) = 8^(− 2 x)$]


      
      
        #math.equation(block: true, alt: "4 to the power 2 x plus 1 equals 8 to the power x minus 3")[$4^(2 x + 1) = 8^(x − 3)$]


        
          #math.equation(block: true, alt: "minus 11")[$− 11$]


        
      
      
        #math.equation(block: true, alt: "3 to the power x squared minus 4 equals 27")[$3^(x^(2) − 4) = 27$]


      
      For Problems 25–28,


      + Graph both functions in the same window. Are they equivalent?
+ Justify your answer to part (a) algebraically.


      
        #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 5 open parenthesis 2 to the power t / 8 close parenthesis , Q open parenthesis t close parenthesis equals 5 open parenthesis 1.0905 close parenthesis to the power t")[$P ( t ) = 5 ( 2^(t / 8) ) , " " " " " " Q ( t ) = 5 ( 1.0905 )^(t)$]


        
          + #figure(figph[two growth functions], alt: "two growth functions", caption: none)

 #linebreak() Not (quite) equivalent
+ #math.equation(block: false, alt: "2 to the power 1 / 8 approximately equals 1.090507733 greater than 1.0905")[$2^(1 / 8) ≈ 1.090507733 > 1.0905$]


        
      
      
        #math.equation(block: true, alt: "M open parenthesis x close parenthesis equals 4 open parenthesis 3 to the power x / 5 close parenthesis , N open parenthesis x close parenthesis equals 4 open parenthesis 1.2457 close parenthesis to the power x")[$M ( x ) = 4 ( 3^(x / 5) ) , " " " " " " N ( x ) = 4 ( 1.2457 )^(x)$]


      
      
        #math.equation(block: true, alt: "H open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 3 close parenthesis to the power x minus 2 , G open parenthesis x close parenthesis equals 9 open parenthesis the fraction 1 over 3 close parenthesis to the power x")[$H ( x ) = attach(( display(frac(1, 3)) ), t: x − 2) , " " " " " " G ( x ) = 9 attach(( display(frac(1, 3)) ), t: x)$]


        
          + #figure(figph[two growth functions], alt: "two growth functions", caption: none)

 #linebreak() Equivalent
+ #math.equation(block: false, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power x minus 2 equals open parenthesis the fraction 1 over 3 close parenthesis to the power x times open parenthesis the fraction 1 over 3 close parenthesis to the power minus 2 equals open parenthesis the fraction 1 over 3 close parenthesis to the power x times 9")[$attach(( display(frac(1, 3)) ), t: x − 2) = attach(( display(frac(1, 3)) ), t: x) ⋅ attach(( display(frac(1, 3)) ), t: − 2) = attach(( display(frac(1, 3)) ), t: x) ⋅ 9$]


        
      
      
        #math.equation(block: true, alt: "F open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 2 close parenthesis to the power 2 x minus 3 , L open parenthesis x close parenthesis equals 8 open parenthesis the fraction 1 over 4 close parenthesis to the power x")[$F ( x ) = attach(( display(frac(1, 2)) ), t: 2 x − 3) , " " " " " " L ( x ) = 8 attach(( display(frac(1, 4)) ), t: x)$]


      
      For Problems 29–32, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x")[$f ( x ) = 2^(x)$].


      + Write a formula for the function.
+ Use transformations to sketch the graph, indicating any intercepts and asymptotes.


      
        #math.equation(block: true, alt: "y equals 4 plus f open parenthesis x plus 1 close parenthesis")[$y = 4 + f ( x + 1 )$]


        
          + #math.equation(block: false, alt: "y equals 4 plus 2 to the power x plus 1")[$y = 4 + 2^(x + 1)$]
+ Shift the graph of #math.equation(block: false, alt: "f 1")[$f " " " " 1$]  unit left, #math.equation(block: false, alt: "4")[$4$] units up.#figure(figph[translated growth], alt: "translated growth", caption: none)


        
      
      
        #math.equation(block: true, alt: "y equals minus 3 plus f open parenthesis x minus 2 close parenthesis")[$y = − 3 + f ( x − 2 )$]


      
      
        #math.equation(block: true, alt: "y equals 6 minus 3 f open parenthesis x close parenthesis")[$y = 6 − 3 f ( x )$]


        
          + #math.equation(block: false, alt: "y equals 6 minus 3 times 2 to the power x")[$y = 6 − 3 ⋅ 2^(x)$]
+ Scale vertically by #math.equation(block: false, alt: "3")[$3$], reflect about #math.equation(block: false, alt: "x")[$x$]-axis, shift #math.equation(block: false, alt: "6")[$6$] units up.#figure(figph[transformed growth], alt: "transformed growth", caption: none)


        
      
      
        #math.equation(block: true, alt: "y equals 10 minus 4 f open parenthesis x close parenthesis")[$y = 10 − 4 f ( x )$]


      
      In Problems 33–36, we compare power and exponential functions. Let


      #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 4 x to the power 1.5 , g open parenthesis x close parenthesis equals 4 open parenthesis 1.5 close parenthesis to the power x")[$f ( x ) = 4 x^(1.5) , " " " " " " g ( x ) = 4 ( 1.5 )^(x)$]
      
        Graph both functions in the window #math.equation(block: false, alt: "Xmin equals 0")[$"Xmin" = 0$], #math.equation(block: false, alt: "Xmax equals 10")[$"Xmax" = 10$], #math.equation(block: false, alt: "Ymin equals 0")[$"Ymin" = 0$], #math.equation(block: false, alt: "Ymax equals 120")[$"Ymax" = 120$]. Which function grows more rapidly for large values of #math.equation(block: false, alt: "x")[$x$]?


        
          #figure(figph[power and exponential], alt: "power and exponential", caption: none)


          #math.equation(block: false, alt: "g")[$g$] eventually grows faster.


        
      
      
        Estimate the solutions of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals g open parenthesis x close parenthesis")[$f ( x ) = g ( x )$]. For what values of #math.equation(block: false, alt: "x")[$x$] is #math.equation(block: false, alt: "f open parenthesis x close parenthesis greater than g open parenthesis x close parenthesis")[$f ( x ) > g ( x )$]?


      
      
        When #math.equation(block: false, alt: "x")[$x$] doubles from #math.equation(block: false, alt: "2")[$2$] to #math.equation(block: false, alt: "4")[$4$], #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] grows by a factor of \_\_\_\_, and #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$] grows by a factor of \_\_\_\_.


        
          #math.equation(block: true, alt: "2 to the power 1.5 approximately equals 2.83")[$2^(1.5) ≈ 2.83$]; #math.equation(block: true, alt: "2.25")[$2.25$]


        
      
      
        What is the range of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] on the domain #math.equation(block: false, alt: "open bracket 0 , 100 close bracket")[$[ 0 , 100 ]$]? What is the range of #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$] on the same domain?


      
      
        "Within belts of uniform moisture conditions and comparable vegetation, the organic matter content of soil decreases exponentially with increasing temperature." Data indicate that the organic content doubles with each #math.equation(block: false, alt: "10 degrees")[$10^(∘)$]C decrease in temperature. Write a formula for this function, stating clearly what each variable represents. (Source: Leopold, Wolman, Gordon, and Miller, 1992)


        
          #math.equation(block: false, alt: "M equals M sub 0 open parenthesis 2 close parenthesis to the power t / 10")[$M = M_(0) ( 2 )^(t / 10)$], where #math.equation(block: false, alt: "M")[$M$] is the organic content, #math.equation(block: false, alt: "M sub 0")[$M_(0)$] is the organic content at #math.equation(block: false, alt: "0 degrees")[$0^(∘)$]C, and #math.equation(block: false, alt: "t")[$t$] is the temperature in #math.equation(block: false, alt: "∘")[$∘$] Celsius.


        
      
      
        In 1951, a study of barley yields under diverse soil conditions led to the formula


        #math.equation(block: true, alt: "Y equals c V to the power a G to the power b")[$Y = c V^(a) G^(b)$]
        where #math.equation(block: false, alt: "V")[$V$] is a soil texture rating, #math.equation(block: false, alt: "G")[$G$] is a drainage rating, and #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$] are constants. In fields with similar drainage systems, the formula gives barley yields, #math.equation(block: false, alt: "Y")[$Y$], as a function of #math.equation(block: false, alt: "V")[$V$], the soil texture. What type of function is it? If it is an increasing function, what can you say about #math.equation(block: false, alt: "a")[$a$]? (Source: Briggs and Courtney, 1985)


      
      For Problems 39-44, find the logarithm.


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


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


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


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


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


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


      
      
        #math.equation(block: true, alt: "log sub 10 ⁡ 10 to the power minus 3")[$log_(10) 10^(− 3)$]


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


        
      
      
        #math.equation(block: true, alt: "log sub 10 ⁡ 0.0001")[$log_(10) 0.0001$]


      
      For Problems 45-46, write the equation in logarithmic form.


      
        #math.equation(block: true, alt: "0.3 to the power minus 2 equals x plus 1")[$0.3^(− 2) = x + 1$]


        
          #math.equation(block: true, alt: "log sub 0.3 ⁡ open parenthesis x plus 1 close parenthesis equals minus 2")[$log_(0.3) ( x + 1 ) = − 2$]


        
      
      
        #math.equation(block: true, alt: "4 to the power 0.3 t equals 3 N sub 0")[$4^(0.3 t) = 3 N_(0)$]


      
      For Problems 47-50, solve.


      
        #math.equation(block: true, alt: "4 times 10 to the power 1.3 x equals 20.4")[$4 ⋅ 10^(1.3 x) = 20.4$]


        
          #math.equation(block: true, alt: "the fraction log ⁡ 5.1 over 1.3 approximately equals 0.5433")[$display(frac(log 5.1, 1.3)) ≈ 0.5433$]


        
      
      
        #math.equation(block: true, alt: "127 equals 2 open parenthesis 10 to the power 0.5 x close parenthesis minus 17.3")[$127 = 2 ( 10^(0.5 x) ) − 17.3$]


      
      
        #math.equation(block: true, alt: "3 open parenthesis 10 to the power minus 0.7 x close parenthesis plus 6.1 equals 9")[$3 ( 10^(− 0.7 x) ) + 6.1 = 9$]


        
          #math.equation(block: true, alt: "the fraction log ⁡ open parenthesis 2.9 / 3 close parenthesis over minus 0.7 approximately equals 0.21")[$display(frac(log ( 2.9 / 3 ), − 0.7)) ≈ 0.21$]


        
      
      
        #math.equation(block: true, alt: "40 open parenthesis 1 minus 10 to the power minus 1.2 x close parenthesis equals 30")[$40 ( 1 − 10^(− 1.2 x) ) = 30$]


      
      For Problems 51-54, write the expression in terms of simpler logarithms. (Assume that all variables and variable expressions denote positive real numbers.)


      
        #math.equation(block: true, alt: "log sub b ⁡ open parenthesis the fraction x y to the power 1 / 3 over z squared close parenthesis")[$log_(b) ( display(frac(x y^(1 / 3), z^(2))) )$]


        
          #math.equation(block: true, alt: "log sub b ⁡ x plus the fraction 1 over 3 log sub b ⁡ y minus 2 log sub b ⁡ z")[$log_(b) x + display(frac(1, 3)) log_(b) y − 2 log_(b) z$]


        
      
      
        #math.equation(block: true, alt: "log sub b ⁡ the square root of the fraction L squared over 2 R")[$log_(b) sqrt(display(frac(L^(2), 2 R)))$]


      
      
        #math.equation(block: true, alt: "log sub 10 ⁡ open parenthesis x the cube root of the fraction x over y close parenthesis")[$log_(10) ( x root(3, display(frac(x, y))) )$]


        
          #math.equation(block: true, alt: "the fraction 4 over 3 log ⁡ x minus the fraction 1 over 3 log ⁡ y")[$display(frac(4, 3)) log x − display(frac(1, 3)) log y$]


        
      
      
        #math.equation(block: true, alt: "log sub 10 ⁡ the square root of open parenthesis s minus a close parenthesis open parenthesis s minus g close parenthesis squared")[$log_(10) sqrt(( s − a ) ( s − g )^(2))$]


      
      For Problems 55-58, write the expression as a single logarithm with coefficient #math.equation(block: false, alt: "1")[$1$].


      
        #math.equation(block: true, alt: "the fraction 1 over 3 open parenthesis log sub 10 ⁡ x minus 2 log sub 10 ⁡ y close parenthesis")[$display(frac(1, 3)) ( log_(10) x − 2 log_(10) y )$]


        
          #math.equation(block: true, alt: "log ⁡ the cube root of the fraction x over y squared")[$log root(3, display(frac(x, y^(2))))$]


        
      
      
        #math.equation(block: true, alt: "the fraction 1 over 2 log sub 10 ⁡ open parenthesis 3 x close parenthesis minus the fraction 2 over 3 log sub 10 ⁡ y")[$display(frac(1, 2)) log_(10) ( 3 x ) − display(frac(2, 3)) log_(10) y$]


      
      
        #math.equation(block: true, alt: "the fraction 1 over 3 log sub 10 ⁡ 8 minus 2 open parenthesis log sub 10 ⁡ 8 minus log sub 10 ⁡ 2 close parenthesis")[$display(frac(1, 3)) log_(10) 8 − 2 ( log_(10) 8 − log_(10) 2 )$]


        
          #math.equation(block: true, alt: "log ⁡ the fraction 1 over 8")[$log display(frac(1, 8))$]


        
      
      
        #math.equation(block: true, alt: "the fraction 1 over 2 open parenthesis log sub 10 ⁡ 9 plus 2 log sub 10 ⁡ 4 close parenthesis plus 2 log sub 10 ⁡ 5")[$display(frac(1, 2)) ( log_(10) 9 + 2 log_(10) 4 ) + 2 log_(10) 5$]


      
      For Problems 59–62, solve the equation by using base 10 logarithms.


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


        
          #math.equation(block: true, alt: "the fraction log ⁡ 63 over log ⁡ 3 approximately equals 3.77")[$display(frac(log 63, log 3)) ≈ 3.77$]


        
      
      
        #math.equation(block: true, alt: "4 times 2 to the power 1.2 x equals 64")[$4 ⋅ 2^(1.2 x) = 64$]


      
      
        #math.equation(block: true, alt: "1200 equals 24 times 6 to the power minus 0.3 x")[$1200 = 24 ⋅ 6^(− 0.3 x)$]


        
          #math.equation(block: true, alt: "the fraction log ⁡ 50 over minus 0.3 log ⁡ 6 approximately equals minus 7.278")[$display(frac(log 50, − 0.3 log 6)) ≈ − 7.278$]


        
      
      
        #math.equation(block: true, alt: "0.08 equals 12 times 3 to the power minus 1.5 x")[$0.08 = 12 ⋅ 3^(− 1.5 x)$]


      
      
        Solve #math.equation(block: false, alt: "N equals N sub 0 open parenthesis 10 to the power k t close parenthesis")[$N = N_(0) ( 10^(k t) )$] for #math.equation(block: false, alt: "t")[$t$].


        
          #math.equation(block: true, alt: "the fraction log ⁡ open parenthesis N / N sub 0 close parenthesis over k")[$display(frac(log ( N / N_(0) ), k))$]


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


      
      
        The population of Dry Gulch has been declining according to the function


        #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 3800 times 2 to the power minus t / 20")[$P ( t ) = 3800 ⋅ 2^(− t / 20)$]
        where #math.equation(block: false, alt: "t")[$t$] is the number of years since the town's heyday in #math.equation(block: false, alt: "1910")[$1910$].


        + What was the population of Dry Gulch in 1990?
+ In what year did the population dip below #math.equation(block: false, alt: "120")[$120$] people?


        
          + #math.equation(block: false, alt: "238")[$238$]
+ #math.equation(block: false, alt: "2010")[$2010$]


        
      
      
        The number of compact discs produced each year by Delta Discs is given by the function


        #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals 8000 times 3 to the power t / 4")[$N ( t ) = 8000 ⋅ 3^(t / 4)$]
        where #math.equation(block: false, alt: "t")[$t$] is the number of years since discs were introduced in #math.equation(block: false, alt: "1980")[$1980$].


        + How many discs did Delta produce in #math.equation(block: false, alt: "1989")[$1989$]?
+ In what year did Delta first produce over 2 million discs?


      
      
        + Write a formula for the cost of a camera #math.equation(block: false, alt: "t")[$t$] years from now if it costs \$#math.equation(block: false, alt: "90")[$90$] now and the inflation rate is #math.equation(block: false, alt: "6 %")[$6 upright(%)$] annually.
+ How much will the camera cost #math.equation(block: false, alt: "10")[$10$] months from now?
+ How long will it be before the camera costs \$120?


        
          + #math.equation(block: false, alt: "C equals 90 open parenthesis 1.06 close parenthesis to the power t")[$C = 90 ( 1.06 )^(t)$]
+ \$#math.equation(block: false, alt: "94.48")[$94.48$]
+ #math.equation(block: false, alt: "5")[$5$] years


        
      
      
        + Write a formula for the cost of a sofa #math.equation(block: false, alt: "t")[$t$] years from now if it costs \$#math.equation(block: false, alt: "1200")[$1200$] now and the inflation rate is #math.equation(block: false, alt: "8 %")[$8 upright(%)$] annually.
+ How much will the sofa cost #math.equation(block: false, alt: "20")[$20$] months from now?
+ How long will it be before the sofa costs \$#math.equation(block: false, alt: "1500")[$1500$]?


      
      
        Francine inherited \$#math.equation(block: false, alt: "5000")[$5000$] and plans to deposit the money in an account that compounds interest monthly.


        + If she can get #math.equation(block: false, alt: "5.5 %")[$5.5 upright(%)$] interest, how long will it take for the money to grow to #math.equation(block: false, alt: "$ 7500")[$\$ 7500$]?
+ What interest rate will she need if she would like the money to grow to #math.equation(block: false, alt: "$ 6000")[$\$ 6000$] in #math.equation(block: false, alt: "3")[$3$] years?


        
          + #math.equation(block: false, alt: "7.4")[$7.4$] years
+ #math.equation(block: false, alt: "6.1 %")[$6.1 upright(%)$]


        
      
      
        Delbert received a signing bonus of \$#math.equation(block: false, alt: "2500")[$2500$] and wants to invest the money in a certificate of deposit (CD) that compounds interest quarterly.


        + If the CD pays #math.equation(block: false, alt: "4.8 %")[$4.8 upright(%)$] interest, how long will it take his money to grow to #math.equation(block: false, alt: "$ 3000")[$\$ 3000$]?
+ What interest rate will he need if he would like the money to grow to #math.equation(block: false, alt: "$ 3000")[$\$ 3000$] in #math.equation(block: false, alt: "1")[$1$] year?


      
      For Problems 71-74, find an exponential growth or decay function that fits the data.


      
        #math.equation(block: true, alt: "f open parenthesis 2 close parenthesis equals 1714 , f open parenthesis 4 close parenthesis equals 1836")[$f ( 2 ) = 1714 , " " " " " " f ( 4 ) = 1836$]


        
          #math.equation(block: true, alt: "f open parenthesis x close parenthesis approximately equals 1600 open parenthesis 1.035 close parenthesis to the power x")[$f ( x ) ≈ 1600 ( 1.035 )^(x)$]


        
      
      
        #math.equation(block: true, alt: "g open parenthesis 1 close parenthesis equals 10 , 665 , g open parenthesis 6 close parenthesis equals 24 , 920")[$g ( 1 ) = 10 , 665 , " " " " " " g ( 6 ) = 24 , 920$]


      
      
        #math.equation(block: true, alt: "g open parenthesis 1 close parenthesis equals 45 , g open parenthesis 5 close parenthesis equals 0.00142")[$g ( 1 ) = 45 , " " " " " " g ( 5 ) = 0.00142$]


        
          #math.equation(block: true, alt: "g open parenthesis x close parenthesis approximately equals 600 open parenthesis 0.075 close parenthesis to the power x")[$g ( x ) ≈ 600 ( 0.075 )^(x)$]


        
      
      
        #math.equation(block: true, alt: "f open parenthesis 2 close parenthesis equals 17 , 464 , f open parenthesis 5 close parenthesis equals 16.690")[$f ( 2 ) = 17 , 464 , " " " " " " f ( 5 ) = 16.690$]


      
      
        The population of Sweden is growing at #math.equation(block: false, alt: "0.1 %")[$0.1 upright(%)$] annually.


        + What is the doubling time for Sweden's population?
+ In #math.equation(block: false, alt: "2005")[$2005$], the population of Sweden was #math.equation(block: false, alt: "9")[$9$] million. At the current rate of growth, how long will it take the population to reach #math.equation(block: false, alt: "10")[$10$] million?


        
          + #math.equation(block: false, alt: "the fraction log ⁡ 2 over log ⁡ 1.001 approximately equals 693")[$display(frac(log 2, log 1.001)) ≈ 693$] years
+ #math.equation(block: false, alt: "105")[$105$] years


        
      
      
        The bacteria #emph[E. sakazakii] is found in powdered infant formula and has a doubling time of #math.equation(block: false, alt: "4.98")[$4.98$] hours even if kept chilled to #math.equation(block: false, alt: "50 degrees")[$50^(∘)$]F.


        + What is the hourly growth rate for #emph[E. sakazakii?]
+ How long would it take a colony of #emph[E. sakazakii] to increase by #math.equation(block: false, alt: "50 %")[$50 upright(%)$]?


      
      
        Manganese-53 decays to chromium-53 with a half-life of #math.equation(block: false, alt: "3.7")[$3.7$] million years and is used to estimate the age of meteorites. What is the decay rate of manganese-53, with time expressed in millions of years?


        
          #math.equation(block: true, alt: "17 %")[$17 upright(%)$]


        
      
      
        The cold medication pseudoephedrine decays at a rate of #math.equation(block: false, alt: "5.95 %")[$5.95 upright(%)$] per hour in the body. What is the half-life of pseudoephedrine?


      
      
        You would like to buy a house with a #math.equation(block: false, alt: "20")[$20$]-year mortgage for \$#math.equation(block: false, alt: "300 , 000")[$300 , 000$], at an interest rate of #math.equation(block: false, alt: "6.25 %")[$6.25 upright(%)$], compounded monthly. Use the formula for the present value of an annuity to calculate your monthly payment.


        
          \$#math.equation(block: false, alt: "2192.78")[$2192.78$]


        
      
      
        Rosalie's retirement fund pays #math.equation(block: false, alt: "7 %")[$7 upright(%)$] interest compounded monthly. Use the formula for the future value of an annuity to calculate how much should she contribute monthly in order to have \$#math.equation(block: false, alt: "500 , 000")[$500 , 000$] in #math.equation(block: false, alt: "25")[$25$] years.


      
      
        An eccentric millionaire offers you a summer job for the month of June. She will pay you #math.equation(block: false, alt: "2")[$2$] cents for your first day of work and will double your wages every day thereafter. (Assume that you work every day, including weekends.)


        + Make a table showing your wages on each day. Do you see a pattern?
+ Write a function that gives your wages in terms of the number of days you have worked.
+ How much will you make on June 15? On June 30?


        
          + #figure(table(
  columns: 8, align: left, inset: 6pt,
  table.header([Day], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "⋯")[$⋯$]], [#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "⋯")[$⋯$]], [#math.equation(block: false, alt: "30")[$30$]]),
  [Wage (cent)], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "⋯")[$⋯$]], [#math.equation(block: false, alt: "2 to the power t")[$2^(t)$]], [#math.equation(block: false, alt: "⋯")[$⋯$]], [#math.equation(block: false, alt: "2 to the power 30")[$2^(30)$]],
))
+ #math.equation(block: false, alt: "W open parenthesis t close parenthesis equals 2 to the power t")[$W ( t ) = 2^(t)$] cents
+ \$#math.equation(block: false, alt: "327.68")[$327.68$]; \$#math.equation(block: false, alt: "10 , 737 , 418.24")[$10 , 737 , 418.24$]


        
      
      
        The king of Persia offered one of his subjects anything he desired in return for services rendered. The subject requested that the king give him an amount of grain calculated as follows: Place one grain of wheat on the first square of a chessboard, two grains on the second square, four grains on the third square, and so on, until the entire chessboard is covered.


        + Make a table showing the number of grains of wheat on each square of the chessboard.
+ Write a function for the amount of wheat on each square.
+ How many grains of wheat should be placed on the last (64th) square?