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13.2.4: Chapter 4

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    117279
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    Try It

    4.1 Exponential Functions

    1.

    g(x)=0.875xg(x)=0.875x and j(x)=1095.6−2xj(x)=1095.6−2x represent exponential functions.

    2.

    5.55565.5556

    3.

    About 1.5481.548 billion people; by the year 2031, India’s population will exceed China’s by about 0.001 billion, or 1 million people.

    4.

    (0,129)(0,129) and (2,236);N(t)=129(1.3526)t(2,236);N(t)=129(1.3526)t

    5.

    f(x)=2(1.5)xf(x)=2(1.5)x

    6.

    f(x)=2–√(2–√)x.f(x)=2(2)x. Answers may vary due to round-off error. The answer should be very close to 1.4142(1.4142)x.1.4142(1.4142)x.

    7.

    y≈12⋅1.85xy≈12⋅1.85x

    8.

    about $3,644,675.88

    9.

    $13,693

    10.

    e−0.5≈0.60653e−0.5≈0.60653

    11.

    $3,659,823.44

    12.

    3.77E-26 (This is calculator notation for the number written as 3.77×10−263.77×10−26 in scientific notation. While the output of an exponential function is never zero, this number is so close to zero that for all practical purposes we can accept zero as the answer.)

    4.2 Graphs of Exponential Functions

    1.

    The domain is (−∞,∞);(−∞,∞); the range is (0,∞);(0,∞); the horizontal asymptote is y=0.y=0.

    Graph of the increasing exponential function f(x) = 4^x with labeled points at (-1, 0.25), (0, 1), and (1, 4).2.

    The domain is (−∞,∞);(−∞,∞); the range is (3,∞);(3,∞); the horizontal asymptote is y=3.y=3.

    Graph of the function, f(x) = 2^(x-1)+3, with an asymptote at y=3. Labeled points in the graph are (-1, 3.25), (0, 3.5), and (1, 4).3.

    x≈−1.608x≈−1.608

    4.

    The domain is (−∞,∞);(−∞,∞); the range is (0,∞);(0,∞); the horizontal asymptote is y=0.y=0.

    Graph of the function, f(x) = (1/2)(4)^(x), with an asymptote at y=0. Labeled points in the graph are (-1, 0.125), (0, 0.5), and (1, 2).5.

    The domain is (−∞,∞);(−∞,∞); the range is (0,∞);(0,∞); the horizontal asymptote is y=0.y=0.

    Graph of the function, g(x) = -(1.25)^(-x), with an asymptote at y=0. Labeled points in the graph are (-1, 1.25), (0, 1), and (1, 0.8).6.

    f(x)=−13ex−2;f(x)=−13ex−2; the domain is (−∞,∞);(−∞,∞); the range is (−∞,−2);(−∞,−2); the horizontal asymptote is y=−2.y=−2.

    4.3 Logarithmic Functions

    1.

    1. ⓐlog10(1,000,000)=6log10(1,000,000)=6 is equivalent to 106=1,000,000106=1,000,000
    2. ⓑlog5(25)=2log5(25)=2 is equivalent to 52=2552=25

    2.

    1. ⓐ 32=932=9 is equivalent to log3(9)=2log3(9)=2
    2. ⓑ 53=12553=125 is equivalent to log5(125)=3log5(125)=3
    3. ⓒ 2−1=122−1=12 is equivalent to log2(12)=−1log2(12)=−1

    3.

    log121(11)=12log121(11)=12 (recalling that 121−−−√=(121)12=11121=(121)12=11 )

    4.

    log2(132)=−5log2(132)=−5

    5.

    log(1,000,000)=6log(1,000,000)=6

    6.

    log(123)≈2.0899log(123)≈2.0899

    7.

    The difference in magnitudes was about 3.929.3.929.

    8.

    It is not possible to take the logarithm of a negative number in the set of real numbers.

    4.4 Graphs of Logarithmic Functions

    1.

    (2,∞)(2,∞)

    2.

    (5,∞)(5,∞)

    3.Graph of f(x)=log_(1/5)(x) with labeled points at (1/5, 1) and (1, 0). The y-axis is the asymptote.

    The domain is (0,∞),(0,∞), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=0.x=0.

    4.Graph of two functions. The parent function is y=log_3(x), with an asymptote at x=0 and labeled points at (1, 0), and (3, 1).The translation function f(x)=log_3(x+4) has an asymptote at x=-4 and labeled points at (-3, 0) and (-1, 1).

    The domain is (−4,∞),(−4,∞), the range (−∞,∞),(−∞,∞), and the asymptote x=–4.x=–4.

    5.Graph of two functions. The parent function is y=log_2(x), with an asymptote at x=0 and labeled points at (1, 0), and (2, 1).The translation function f(x)=log_2(x)+2 has an asymptote at x=0 and labeled points at (0.25, 0) and (0.5, 1).

    The domain is (0,∞),(0,∞), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=0.x=0.

    6.Graph of two functions. The parent function is y=log_4(x), with an asymptote at x=0 and labeled points at (1, 0), and (4, 1).The translation function f(x)=(1/2)log_4(x) has an asymptote at x=0 and labeled points at (1, 0) and (16, 1).

    The domain is (0,∞),(0,∞), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=0.x=0.

    7.Graph of f(x)=3log(x-2)+1 with an asymptote at x=2.

    The domain is (2,∞),(2,∞), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=2.x=2.

    8.Graph of f(x)=-log(-x) with an asymptote at x=0.

    The domain is (−∞,0),(−∞,0), the range is (−∞,∞),(−∞,∞), and the vertical asymptote is x=0.x=0.

    9.

    x≈3.049x≈3.049

    10.

    x=1x=1

    11.

    f(x)=2ln(x+3)−1f(x)=2ln(x+3)−1

    4.5 Logarithmic Properties

    1.

    logb2+logb2+logb2+logbk=3logb2+logbklogb2+logb2+logb2+logbk=3logb2+logbk

    2.

    log3(x+3)−log3(x−1)−log3(x−2)log3(x+3)−log3(x−1)−log3(x−2)

    3.

    2lnx2lnx

    4.

    −2ln(x)−2ln(x)

    5.

    log316log316

    6.

    2logx+3logy−4logz2logx+3logy−4logz

    7.

    23lnx23lnx

    8.

    12ln(x−1)+ln(2x+1)−ln(x+3)−ln(x−3)12ln(x−1)+ln(2x+1)−ln(x+3)−ln(x−3)

    9.

    log(3⋅54⋅6);log(3⋅54⋅6); can also be written log(58)log(58) by reducing the fraction to lowest terms.

    10.

    log(5(x−1)3x√(7x−1))log(5(x−1)3x(7x−1))

    11.

    logx12(x+5)4(2x+3)4;logx12(x+5)4(2x+3)4; this answer could also be written log(x3(x+5)(2x+3))4.log(x3(x+5)(2x+3))4.

    12.

    The pH increases by about 0.301.

    13.

    ln8ln0.5ln8ln0.5

    14.

    ln100ln5≈4.60511.6094=2.861ln100ln5≈4.60511.6094=2.861

    4.6 Exponential and Logarithmic Equations

    1.

    x=−2x=−2

    2.

    x=−1x=−1

    3.

    x=12x=12

    4.

    The equation has no solution.

    5.

    x=ln3ln(23/)x=ln3ln(23)

    6.

    t=2ln(113)t=2ln(113) or ln(113)2ln(113)2

    7.

    t=ln(12√)=−12ln(2)t=ln(12)=−12ln(2)

    8.

    x=ln2x=ln2

    9.

    x=e4x=e4

    10.

    x=e5−1x=e5−1

    11.

    x≈9.97x≈9.97

    12.

    x=1x=1 or x=−1x=−1

    13.

    t=703,800,000×ln(0.8)ln(0.5)years ≈226,572,993years.t=703,800,000×ln(0.8)ln(0.5)years ≈226,572,993years.

    4.7 Exponential and Logarithmic Models

    1.

    f(t)=A0e−0.0000000087tf(t)=A0e−0.0000000087t

    2.

    less than 230 years, 229.3157 to be exact

    3.

    f(t)=A0eln23tf(t)=A0eln23t

    4.

    6.026 hours

    5.

    895 cases on day 15

    6.

    Exponential. y=2e0.5x.y=2e0.5x.

    7.

    y=3e(ln0.5)xy=3e(ln0.5)x

    4.8 Fitting Exponential Models to Data

    1.

    1. ⓐ The exponential regression model that fits these data is y=522.88585984(1.19645256)x.y=522.88585984(1.19645256)x.
    2. ⓑ If spending continues at this rate, the graduate’s credit card debt will be $4,499.38 after one year.

    2.

    1. ⓐ The logarithmic regression model that fits these data is y=141.91242949+10.45366573ln(x)y=141.91242949+10.45366573ln(x)
    2. ⓑ If sales continue at this rate, about 171,000 games will be sold in the year 2015.

    3.

    1. ⓐ The logistic regression model that fits these data is y=25.656659791+6.113686306e−0.3852149008x.y=25.656659791+6.113686306e−0.3852149008x.
    2. ⓑ If the population continues to grow at this rate, there will be about 25,634 25,634  seals in 2020.
    3. ⓒ To the nearest whole number, the carrying capacity is 25,657.

    4.1 Section Exercises

    1.

    Linear functions have a constant rate of change. Exponential functions increase based on a percent of the original.

    3.

    When interest is compounded, the percentage of interest earned to principal ends up being greater than the annual percentage rate for the investment account. Thus, the annual percentage rate does not necessarily correspond to the real interest earned, which is the very definition of nominal.

    5.

    exponential; the population decreases by a proportional rate. .

    7.

    not exponential; the charge decreases by a constant amount each visit, so the statement represents a linear function. .

    9.

    The forest represented by the function B(t)=82(1.029)t.B(t)=82(1.029)t.

    11.

    After t=20t=20 years, forest A will have 4343 more trees than forest B.

    13.

    Answers will vary. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, but because forest B actually grows at a faster rate, the population will eventually become larger than forest A and will remain that way as long as the population growth models hold. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other environmental and biological factors.

    15.

    exponential growth; The growth factor, 1.06,1.06, is greater than 1.1.

    17.

    exponential decay; The decay factor, 0.97,0.97, is between 00 and 1.1.

    19.

    f(x)=2000(0.1)xf(x)=2000(0.1)x

    21.

    f(x)=(16)−35(16)x5≈2.93(0.699)xf(x)=(16)−35(16)x5≈2.93(0.699)x

    23.

    Linear

    25.

    Neither

    27.

    Linear

    29.

    $10,250$10,250

    31.

    $13,268.58$13,268.58

    33.

    P=A(t)⋅(1+rn)−ntP=A(t)⋅(1+rn)−nt

    35.

    $4,572.56$4,572.56

    37.

    4%4%

    39.

    continuous growth; the growth rate is greater than 0.0.

    41.

    continuous decay; the growth rate is less than 0.0.

    43.

    $669.42$669.42

    45.

    f(−1)=−4f(−1)=−4

    47.

    f(−1)≈−0.2707f(−1)≈−0.2707

    49.

    f(3)≈483.8146f(3)≈483.8146

    51.

    y=3⋅5xy=3⋅5x

    53.

    y≈18⋅1.025xy≈18⋅1.025x

    55.

    y≈0.2⋅1.95xy≈0.2⋅1.95x

    57.

    APY=A(t)−aa=a(1+r365)365(1)−aa=a[(1+r365)365−1]a=(1+r365)365−1;APY=A(t)−aa=a(1+r365)365(1)−aa=a[ (1+r365)365−1 ]a=(1+r365)365−1; I(n)=(1+rn)n−1I(n)=(1+rn)n−1

    59.

    Let ff be the exponential decay function f(x)=a⋅(1b)xf(x)=a⋅(1b)x such that b>1.b>1. Then for some number n>0,n>0, f(x)=a⋅(1b)x=a(b−1)x=a((en)−1)x=a(e−n)x=a(e)−nx.f(x)=a⋅(1b)x=a(b−1)x=a((en)−1)x=a(e−n)x=a(e)−nx.

    61.

    47,62247,622 fox

    63.

    1.39%;1.39%; $155,368.09$155,368.09

    65.

    $35,838.76$35,838.76

    67.

    $82,247.78;$82,247.78; $449.75$449.75

    4.2 Section Exercises

    1.

    An asymptote is a line that the graph of a function approaches, as xx either increases or decreases without bound. The horizontal asymptote of an exponential function tells us the limit of the function’s values as the independent variable gets either extremely large or extremely small.

    3.

    g(x)=4(3)−x;g(x)=4(3)−x; y-intercept: (0,4);(0,4); Domain: all real numbers; Range: all real numbers greater than 0.0.

    5.

    g(x)=−10x+7;g(x)=−10x+7; y-intercept: (0,6);(0,6); Domain: all real numbers; Range: all real numbers less than 7.7.

    7.

    g(x)=2(14)x;g(x)=2(14)x; y-intercept: (0,2);(0,2); Domain: all real numbers; Range: all real numbers greater than 0.0.

    9.Graph of two functions, g(-x)=-2(0.25)^(-x) in blue and g(x)=-2(0.25)^x in orange.

    y-intercept: (0,−2)(0,−2)

    11.

    Graph of three functions, g(x)=3(2)^(x) in blue, h(x)=3(4)^(x) in green, and f(x)=3(1/4)^(x) in orange.

    13.

    B

    15.

    A

    17.

    E

    19.

    D

    21.

    C

    23.

    Graph of two functions, f(x)=(1/2)(4)^(x) in blue and -f(x)=(-1/2)(4)^x in orange.

    25.

    Graph of two functions, -f(x)=(4)(2)^(x)-2 in blue and f(x)=(-4)(2)^x+1 in orange.

    27.Graph of h(x)=2^(x)+3.

    Horizontal asymptote: h(x)=3;h(x)=3; Domain: all real numbers; Range: all real numbers strictly greater than 3.3.

    29.

    As x→∞x→∞ , f(x)→−∞f(x)→−∞ ;
    As x→−∞x→−∞ , f(x)→−1f(x)→−1

    31.

    As x→∞x→∞ , f(x)→2f(x)→2 ;
    As x→−∞x→−∞ , f(x)→∞f(x)→∞

    33.

    f(x)=4x−3f(x)=4x−3

    35.

    f(x)=4x−5f(x)=4x−5

    37.

    f(x)=4−xf(x)=4−x

    39.

    y=−2x+3y=−2x+3

    41.

    y=−2(3)x+7y=−2(3)x+7

    43.

    g(6)=800+13≈800.3333g(6)=800+13≈800.3333

    45.

    h(−7)=−58h(−7)=−58

    47.

    x≈−2.953x≈−2.953

    49.

    x≈−0.222x≈−0.222

    51.

    The graph of G(x)=(1b)xG(x)=(1b)x is the refelction about the y-axis of the graph of F(x)=bx;F(x)=bx; For any real number b>0b>0 and function f(x)=bx,f(x)=bx, the graph of (1b)x(1b)x is the the reflection about the y-axis, F(−x).F(−x).

    53.

    The graphs of g(x)g(x) and h(x)h(x) are the same and are a horizontal shift to the right of the graph of f(x);f(x); For any real number n, real number b>0,b>0, and function f(x)=bx,f(x)=bx, the graph of (1bn)bx(1bn)bx is the horizontal shift f(x−n).f(x−n).

    4.3 Section Exercises

    1.

    A logarithm is an exponent. Specifically, it is the exponent to which a base bb is raised to produce a given value. In the expressions given, the base bb has the same value. The exponent, y,y, in the expression byby can also be written as the logarithm, logbx,logbx, and the value of xx is the result of raising bb to the power of y.y.

    3.

    Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation by=x,by=x, and then properties of exponents can be applied to solve for x.x.

    5.

    The natural logarithm is a special case of the logarithm with base bb in that the natural log always has base e.e. Rather than notating the natural logarithm as loge(x),loge(x), the notation used is ln(x).ln(x).

    7.

    ac=bac=b

    9.

    xy=64xy=64

    11.

    15b=a15b=a

    13.

    13a=14213a=142

    15.

    en=wen=w

    17.

    logc(k)=dlogc(k)=d

    19.

    log19y=xlog19y=x

    21.

    logn(103)=4logn(103)=4

    23.

    logy(39100)=xlogy(39100)=x

    25.

    ln(h)=kln(h)=k

    27.

    x=2−3=18x=2−3=18

    29.

    x=33=27x=33=27

    31.

    x=912=3x=912=3

    33.

    x=6−3=1216x=6−3=1216

    35.

    x=e2x=e2

    37.

    3232

    39.

    1.061.06

    41.

    14.12514.125

    43.

    1212

    45.

    44

    47.

    −3−3

    49.

    −12−12

    51.

    00

    53.

    1010

    55.

    2.7082.708

    57.

    0.1510.151

    59.

    No, the function has no defined value for x=0.x=0. To verify, suppose x=0x=0 is in the domain of the function f(x)=log(x).f(x)=log(x). Then there is some number nn such that n=log(0).n=log(0). Rewriting as an exponential equation gives: 10n=0,10n=0, which is impossible since no such real number nn exists. Therefore, x=0x=0 is not the domain of the function f(x)=log(x).f(x)=log(x).

    61.

    Yes. Suppose there exists a real number xx such that lnx=2.lnx=2. Rewriting as an exponential equation gives x=e2,x=e2, which is a real number. To verify, let x=e2.x=e2. Then, by definition, ln(x)=ln(e2)=2.ln(x)=ln(e2)=2.

    63.

    No; ln(1)=0,ln(1)=0, so ln(e1.725)ln(1)ln(e1.725)ln(1) is undefined.

    65.

    22

    4.4 Section Exercises

    1.

    Since the functions are inverses, their graphs are mirror images about the line y=x.y=x. So for every point (a,b)(a,b) on the graph of a logarithmic function, there is a corresponding point (b,a)(b,a) on the graph of its inverse exponential function.

    3.

    Shifting the function right or left and reflecting the function about the y-axis will affect its domain.

    5.

    No. A horizontal asymptote would suggest a limit on the range, and the range of any logarithmic function in general form is all real numbers.

    7.

    Domain: (−∞,12);(−∞,12); Range: (−∞,∞)(−∞,∞)

    9.

    Domain: (−174,∞);(−174,∞); Range: (−∞,∞)(−∞,∞)

    11.

    Domain: (5,∞);(5,∞); Vertical asymptote: x=5x=5

    13.

    Domain: (−13,∞);(−13,∞); Vertical asymptote: x=−13x=−13

    15.

    Domain: (−3,∞);(−3,∞); Vertical asymptote: x=−3x=−3

    17.

    Domain: (37,∞)(37,∞) ;
    Vertical asymptote: x=37x=37 ; End behavior: as x→(37)+,f(x)→−∞x→(37)+,f(x)→−∞ and as x→∞,f(x)→∞x→∞,f(x)→∞

    19.

    Domain: (−3,∞)(−3,∞) ; Vertical asymptote: x=−3x=−3 ;
    End behavior: as x→−3+x→−3+ , f(x)→−∞f(x)→−∞ and as x→∞x→∞ , f(x)→∞f(x)→∞

    21.

    Domain: (1,∞);(1,∞); Range: (−∞,∞);(−∞,∞); Vertical asymptote: x=1;x=1; x-intercept: (54,0);(54,0); y-intercept: DNE

    23.

    Domain: (−∞,0);(−∞,0); Range: (−∞,∞);(−∞,∞); Vertical asymptote: x=0;x=0; x-intercept: (−e2,0);(−e2,0); y-intercept: DNE

    25.

    Domain: (0,∞);(0,∞); Range: (−∞,∞);(−∞,∞); Vertical asymptote: x=0;x=0; x-intercept: (e3,0);(e3,0); y-intercept: DNE

    27.

    B

    29.

    C

    31.

    B

    33.

    C

    35.

    Graph of two functions, g(x) = log_(1/2)(x) in orange and f(x)=log(x) in blue.

    37.

    Graph of two functions, g(x) = ln(1/2)(x) in orange and f(x)=e^(x) in blue.

    39.

    C

    41.

    Graph of f(x)=log_2(x+2).

    43.

    Graph of f(x)=ln(-x).

    45.

    Graph of g(x)=log(6-3x)+1.

    47.

    f(x)=log2(−(x−1))f(x)=log2(−(x−1))

    49.

    f(x)=3log4(x+2)f(x)=3log4(x+2)

    51.

    x=2x=2

    53.

    x≈2.303x≈2.303

    55.

    x≈−0.472x≈−0.472

    57.

    The graphs of f(x)=log12(x)f(x)=log12(x) and g(x)=−log2(x)g(x)=−log2(x) appear to be the same; Conjecture: for any positive base b≠1,b≠1, logb(x)=−log1b(x).logb(x)=−log1b(x).

    59.

    Recall that the argument of a logarithmic function must be positive, so we determine where x+2x−4>0x+2x−4>0 . From the graph of the function f(x)=x+2x−4,f(x)=x+2x−4, note that the graph lies above the x-axis on the interval (−∞,−2)(−∞,−2) and again to the right of the vertical asymptote, that is (4,∞).(4,∞). Therefore, the domain is (−∞,−2)∪(4,∞).(−∞,−2)∪(4,∞).

    8904f41612ee5574cc00b1bec1f9607457806014

    4.5 Section Exercises

    1.

    Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, logb(x1n)=1nlogb(x).logb(x1n)=1nlogb(x).

    3.

    logb(2)+logb(7)+logb(x)+logb(y)logb(2)+logb(7)+logb(x)+logb(y)

    5.

    logb(13)−logb(17)logb(13)−logb(17)

    7.

    −kln(4)−kln(4)

    9.

    ln(7xy)ln(7xy)

    11.

    logb(4)logb(4)

    13.

    logb(7)logb(7)

    15.

    15log(x)+13log(y)−19log(z)15log(x)+13log(y)−19log(z)

    17.

    32log(x)−2log(y)32log(x)−2log(y)

    19.

    83log(x)+143log(y)83log(x)+143log(y)

    21.

    ln(2x7)ln(2x7)

    23.

    log(xz3y√)log(xz3y)

    25.

    log7(15)=ln(15)ln(7)log7(15)=ln(15)ln(7)

    27.

    log11(5)=log5(5)log5(11)=1blog11(5)=log5(5)log5(11)=1b

    29.

    log11(611)=log5(611)log5(11)=log5(6)−log5(11)log5(11)=a−bb=ab−1log11(611)=log5(611)log5(11)=log5(6)−log5(11)log5(11)=a−bb=ab−1

    31.

    33

    33.

    2.813592.81359

    35.

    0.939130.93913

    37.

    −2.23266−2.23266

    39.

    x=4;x=4; By the quotient rule: log6(x+2)−log6(x−3)=log6(x+2x−3)=1.log6(x+2)−log6(x−3)=log6(x+2x−3)=1.

    Rewriting as an exponential equation and solving for x:x:

    610000​x=x+2x−3=x+2x−3−6=x+2x−3−6(x−3)(x−3)=x+2−6x+18x−3=x−4x−3=461=x+2x−30=x+2x−3−60=x+2x−3−6(x−3)(x−3)0=x+2−6x+18x−30=x−4x−3​x=4

    Checking, we find that log6(4+2)−log6(4−3)=log6(6)−log6(1)log6(4+2)−log6(4−3)=log6(6)−log6(1) is defined, so x=4.x=4.

    41.

    Let bb and nn be positive integers greater than 1.1. Then, by the change-of-base formula, logb(n)=logn(n)logn(b)=1logn(b).logb(n)=logn(n)logn(b)=1logn(b).

    4.6 Section Exercises

    1.

    Determine first if the equation can be rewritten so that each side uses the same base. If so, the exponents can be set equal to each other. If the equation cannot be rewritten so that each side uses the same base, then apply the logarithm to each side and use properties of logarithms to solve.

    3.

    The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.

    5.

    x=−13x=−13

    7.

    n=−1n=−1

    9.

    b=65b=65

    11.

    x=10x=10

    13.

    No solution

    15.

    p=log(178)−7p=log(178)−7

    17.

    k=−ln(38)3k=−ln(38)3

    19.

    x=ln(383)−89x=ln(383)−89

    21.

    x=ln12x=ln12

    23.

    x=ln(35)−38x=ln(35)−38

    25.

    no solution

    27.

    x=ln(3)x=ln(3)

    29.

    10−2=110010−2=1100

    31.

    n=49n=49

    33.

    k=136k=136

    35.

    x=9−e8x=9−e8

    37.

    n=1n=1

    39.

    No solution

    41.

    No solution

    43.

    x=±103x=±103

    45.

    x=10x=10

    47.

    x=0x=0

    49.

    x=34x=34

    51.

    x=9x=9

    Graph of log_9(x)-5=y and y=-4.53.

    x=e23≈2.5x=e23≈2.5

    Graph of ln(3x)=y and y=2.55.

    x=−5x=−5

    Graph of log(4)+log(-5x)=y and y=2.57.

    x=e+104≈3.2x=e+104≈3.2

    Graph of ln(4x-10)-6=y and y=-5.59.

    No solution

    Graph of log_11(-2x^2-7x)=y and y=log_11(x-2).61.

    x=115≈2.2x=115≈2.2

    Graph of log_9(3-x)=y and y=log_9(4x-8).63.

    x=10111≈9.2x=10111≈9.2

    Graph of 3/log_2(10)-log(x-9)=y and y=log(44).65.

    about $27,710.24$27,710.24

    Graph of f(x)=6500e^(0.0725x) with the labeled point at (20, 27710.24).67.

    about 5 years

    Graph of P(t)=1650e^(0.5x) with the labeled point at (5, 20000).69.

    ln(17)5≈0.567ln(17)5≈0.567

    71.

    x=log(38)+5log(3) 4log(3)≈2.078x=log(38)+5log(3) 4log(3)≈2.078

    73.

    x≈2.2401x≈2.2401

    75.

    x≈−44655.7143x≈−44655.7143

    77.

    about 5.835.83

    79.

    t=ln((yA)1k)t=ln((yA)1k)

    81.

    t=ln((T−TsT0−Ts)−1k)t=ln((T−TsT0−Ts)−1k)

    4.7 Section Exercises

    1.

    Half-life is a measure of decay and is thus associated with exponential decay models. The half-life of a substance or quantity is the amount of time it takes for half of the initial amount of that substance or quantity to decay.

    3.

    Doubling time is a measure of growth and is thus associated with exponential growth models. The doubling time of a substance or quantity is the amount of time it takes for the initial amount of that substance or quantity to double in size.

    5.

    An order of magnitude is the nearest power of ten by which a quantity exponentially grows. It is also an approximate position on a logarithmic scale; Sample response: Orders of magnitude are useful when making comparisons between numbers that differ by a great amount. For example, the mass of Saturn is 95 times greater than the mass of Earth. This is the same as saying that the mass of Saturn is about 102102 times, or 2 orders of magnitude greater, than the mass of Earth.

    7.

    f(0)≈16.7;f(0)≈16.7; The amount initially present is about 16.7 units.

    9.

    150

    11.

    exponential; f(x)=1.2xf(x)=1.2x

    13.

    logarithmic

    Graph of the question’s table.15.

    logarithmic

    Graph of the question’s table.17.

    Graph of P(t)=1000/(1+9e^(-0.6t))

    19.

    about 1.41.4 years

    21.

    about 7.37.3 years

    23.

    44 half-lives; 8.188.18 minutes

    25.

    M=23log(SS0)log(SS0)=32MSS0=103M2S=S0103M2M=23log(SS0)log(SS0)=32MSS0=103M2S=S0103M2

    27.

    Let y=bxy=bx for some non-negative real number bb such that b≠1.b≠1. Then,

    ln(y)=ln(bx)ln(y)=xln(b)eln(y)=exln(b) y=exln(b)ln(y)=ln(bx)ln(y)=xln(b)eln(y)=exln(b) y=exln(b)

    29.

    A=125e(−0.3567t);A≈43A=125e(−0.3567t);A≈43 mg

    31.

    about 6060 days

    33.

    A(t)=250e(−0.00822t);A(t)=250e(−0.00822t); half-life: about 8484 minutes

    35.

    r≈−0.0667,r≈−0.0667, So the hourly decay rate is about 6.67%6.67%

    37.

    f(t)=1350e(0.03466t);f(t)=1350e(0.03466t); after 3 hours: P(180)≈691,200P(180)≈691,200

    39.

    f(t)=256e(0.068110t);f(t)=256e(0.068110t); doubling time: about 1010 minutes

    41.

    about 8888 minutes

    43.

    T(t)=90e(−0.008377t)+75,T(t)=90e(−0.008377t)+75, where tt is in minutes.

    45.

    about 113113 minutes

    47.

    log(x)=1.5;x≈31.623log(x)=1.5;x≈31.623

    49.

    MMS magnitude: 5.825.82

    51.

    N(3)≈71N(3)≈71

    53.

    C

    4.8 Section Exercises

    1.

    Logistic models are best used for situations that have limited values. For example, populations cannot grow indefinitely since resources such as food, water, and space are limited, so a logistic model best describes populations.

    3.

    Regression analysis is the process of finding an equation that best fits a given set of data points. To perform a regression analysis on a graphing utility, first list the given points using the STAT then EDIT menu. Next graph the scatter plot using the STAT PLOT feature. The shape of the data points on the scatter graph can help determine which regression feature to use. Once this is determined, select the appropriate regression analysis command from the STAT then CALC menu.

    5.

    The y-intercept on the graph of a logistic equation corresponds to the initial population for the population model.

    7.

    C

    9.

    B

    11.

    P(0)=22P(0)=22 ; 175

    13.

    p≈2.67p≈2.67

    15.

    y-intercept: (0,15)(0,15)

    17.

    44 koi

    19.

    about 6.86.8 months.

    21.

    cfe27b9c7269463d6daa7e25befd71e44ea02355

    23.

    About 38 wolves

    25.

    About 8.7 years

    27.

    f(x)=776.682(1.426)xf(x)=776.682(1.426)x

    29.

    7da6a27df1883b0ff3fdc9046b8d5a0efec2b884

    31.

    9c7aff12909bba6713c4123c13cbd4aec4e2c1a8

    33.

    f(x)=731.92e−0.3038xf(x)=731.92e-0.3038x

    35.

    When f(x)=250,x≈3.6f(x)=250, x≈3.6

    37.

    y=5.063+1.934log(x)y=5.063+1.934log(x)

    39.

    4916912a4e70f3457357e26ac1f227676888bad2

    41.

    7e362f8eda808b9be85e0df7b11d344b43559095

    43.

    When f(10)≈2.3f(10)≈2.3

    45.

    When f(x)=8,x≈0.82f(x)=8, x≈0.82

    47.

    f(x)=25.0811+3.182e−0.545xf(x)=25.0811+3.182e−0.545x

    49.

    About 25

    51.

    f7bccc628befaf37a99c29fb9c9c12c6236e195f

    53.

    eef9631ae7d5d0f14ff3280a5f40aa5790217753

    55.

    When f(x)=68,x≈4.9f(x)=68, x≈4.9

    57.

    f(x)=1.034341(1.281204)xf(x)=1.034341(1.281204)x ; g(x)=4.035510g(x)=4.035510 ; the regression curves are symmetrical about y=xy=x , so it appears that they are inverse functions.

    59.

    f−1(x)=ln(a)−ln(cx−1)bf−1(x)=ln(a)-ln(cx-1)b

    Review Exercises

    1.

    exponential decay; The growth factor, 0.825,0.825, is between 00 and 1.1.

    3.

    y=0.25(3)xy=0.25(3)x

    5.

    $42,888.18$42,888.18

    7.

    continuous decay; the growth rate is negative.

    9.

    domain: all real numbers; range: all real numbers strictly greater than zero; y-intercept: (0, 3.5);

    Graph of f(x)=3.5(2^x)11.

    g(x)=7(6.5)−x;g(x)=7(6.5)−x; y-intercept: (0,7);(0,7); Domain: all real numbers; Range: all real numbers greater than 0.0.

    13.

    17x=491317x=4913

    15.

    logab=−25logab=−25

    17.

    x=6413=4x=6413=4

    19.

    log(0.000001)=−6log(0.000001)=−6

    21.

    ln(e−0.8648)=−0.8648ln(e−0.8648)=−0.8648

    23.

    Graph of g(x)=log(7x+21)-4.25.

    Domain: x>−5;x>−5; Vertical asymptote: x=−5;x=−5; End behavior: as x→−5+,f(x)→−∞x→−5+,f(x)→−∞ and as x→∞,f(x)→∞.x→∞,f(x)→∞.

    27.

    log8(65xy)log8(65xy)

    29.

    ln(zxy)ln(zxy)

    31.

    logy(12)logy(12)

    33.

    ln(2)+ln(b)+ln(b+1)−ln(b−1)2ln(2)+ln(b)+ln(b+1)−ln(b−1)2

    35.

    log7(v3w6u√3)log7(v3w6u3)

    37.

    x=log(125)log(5)+1712=53x=log(125)log(5)+1712=53

    39.

    x=−3x=−3

    41.

    no solution

    43.

    no solution

    45.

    x=ln(11)x=ln(11)

    47.

    a=e4−3a=e4−3

    49.

    x=±95x=±95

    51.

    about 5.455.45 years

    53.

    f−1(x)=24x−1−−−−−−√3f−1(x)=24x−13

    55.

    f(t)=300(0.83)t;f(t)=300(0.83)t;
    f(24)≈3.43  gf(24)≈3.43  g

    57.

    about 4545 minutes

    59.

    about 8.58.5 days

    61.

    exponential

    Graph of the table’s values.63.

    y=4(0.2)x;y=4(0.2)x; y=4e-1.609438xy=4e-1.609438x

    65.

    about 7.27.2 days

    67.

    logarithmic; y=16.68718−9.71860ln(x)y=16.68718−9.71860ln(x)

    Graph of the table’s values.

    Practice Test

    1.

    About 13 dolphins.

    3.

    $1,947$1,947

    5.

    y-intercept: (0,5)(0,5)

    Graph of f(-x)=5(0.5)^-x in blue and f(x)=5(0.5)^x in orange.7.

    8.5a=614.1258.5a=614.125

    9.

    x=(17)2=149x=(17)2=149

    11.

    ln(0.716)≈−0.334ln(0.716)≈−0.334

    13.

    Domain: x<3;x<3; Vertical asymptote: x=3;x=3; End behavior: x→3−,f(x)→−∞x→3−,f(x)→−∞ and x→−∞,f(x)→∞x→−∞,f(x)→∞

    15.

    logt(12)logt(12)

    17.

    3ln(y)+2ln(z)+ln(x−4)33ln(y)+2ln(z)+ln(x−4)3

    19.

    x=ln(1000)ln(16)+53≈2.497x=ln(1000)ln(16)+53≈2.497

    21.

    a=ln(4)+810a=ln(4)+810

    23.

    no solution

    25.

    x=ln(9)x=ln(9)

    27.

    x=±33√2x=±332

    29.

    f(t)=112e−.019792t;f(t)=112e−.019792t; half-life: about 3535 days

    31.

    T(t)=36e−0.025131t+35;T(60)≈43oFT(t)=36e−0.025131t+35;T(60)≈43oF

    33.

    logarithmic

    Graph of the table’s values.35.

    exponential; y=15.10062(1.24621)xy=15.10062(1.24621)x

    Graph of the table’s values.37.

    logistic; y=18.416591+7.54644e−0.68375xy=18.416591+7.54644e−0.68375x

    Graph of the table’s values.


    13.2.4: Chapter 4 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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