Chapter 6 Functions Review Questions and Excersices #5

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6.i Exponential Functions

1.

$\mathrm{yard}(\mathrm{10})={0.875}^x$ and $j(x)={\mathrm{1095.half\; dozen}}^{-2x}$ correspond exponential functions.

3.

About $\mathrm{i.548}$ billion people; by the yr 2031, India'south population volition exceed Cathay'south by about 0.001 billion, or 1 million people.

four.

$\left(0,129\right)$ and $\left(2,236\right);N(t)=129{\left(\text{1}\text{.3526}\right)}^t$

5.

$f(x)=2{\left(\mathrm{one.5}\right)}^x$

6.

$f(x)=\sqrt{2}{\left(\sqrt{2}\right)}^x.$ Answers may vary due to round-off error. The answer should be very close to $1.4142{\left(\mathrm{i.4142}\right)}^x.$

7.

$y\approx 12\cdot {1.85}^{\mathrm{ten}}$

ten.

$e^{-\mathrm{0.five}}\approx 0.60653$

12.

3.77E-26 (This is calculator notation for the number written equally $3.77\times {10}^{-26}$ in scientific notation. While the output of an exponential function is never goose egg, this number is so close to zero that for all practical purposes we can accept cypher as the answer.)

half-dozen.2 Graphs of Exponential Functions

1.

The domain is $\left(-\infty ,\infty \right);$ the range is $\left(0,\infty \right);$ the horizontal asymptote is $y=0.$

ii.

The domain is $\left(-\infty ,\infty \right);$ the range is $\left(3,\infty \right);$ the horizontal asymptote is $y=3.$

4.

The domain is $\left(-\infty ,\infty \right);$ the range is $\left(0,\infty \right);$ the horizontal asymptote is $y=0.$

5.

The domain is $\left(-\infty ,\infty \right);$ the range is $\left(0,\infty \right);$ the horizontal asymptote is $y=0.$

vi.

$f(x)=-\frac{\mathrm{i}}{3}{e}^x-\mathrm{ii};$ the domain is $\left(-\infty ,\infty \right);$ the range is $\left(-\infty ,\mathrm{-two}\right);$ the horizontal asymptote is $y=\mathrm{-2.}$

half dozen.iii Logarithmic Functions

1.

1. ⓐ ${\log }_{\mathrm{ten}}\left(1,000,000\right)=\mathrm{vi}$ is equivalent to ${10}^6=1,000,000$
2. ⓑ ${\log }_{\mathrm{v}}\left(25\right)=2$ is equivalent to $5^2=25$

2.

1. ⓐ $3^{\mathrm{two}}=9$ is equivalent to ${\log }_3(\mathrm{ix})=\mathrm{ii}$
2. ⓑ ${\mathrm{v}}^{\mathrm{three}}=125$ is equivalent to ${\log }_{\mathrm{v}}(125)=3$
3. ⓒ $2^{-\mathrm{i}}=\frac{1}{2}$ is equivalent to ${\text{log}}_2(\frac{\mathrm{one}}{\mathrm{two}})=-\mathrm{i}$

3.

${\log }_{121}\left(11\right)=\frac{1}{2}$ (recalling that $\sqrt{121}={(121)}^{\frac{\mathrm{i}}{2}}=\mathrm{xi}$ )

4.

${\log }_{\mathrm{ii}}\left(\frac{\mathrm{ane}}{32}\right)=-5$

5.

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

6.

$\log \left(123\right)\approx 2.0899$

vii.

The difference in magnitudes was virtually $\mathrm{iii.929.}$

8.

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

6.4 Graphs of Logarithmic Functions

iii.

The domain is $\left(0,\infty \right),$ the range is $\left(-\infty ,\infty \right),$ and the vertical asymptote is $x=0.$

4.

The domain is $\left(-\mathrm{four},\infty \right),$ the range $\left(-\infty ,\infty \right),$ and the asymptote $\mathrm{10}=–4.$

5.

The domain is $\left(0,\infty \right),$ the range is $\left(-\infty ,\infty \right),$ and the vertical asymptote is $\mathrm{ten}=0.$

vi.

The domain is $\left(0,\infty \right),$ the range is $\left(-\infty ,\infty \right),$ and the vertical asymptote is $x=0.$

vii.

The domain is $\left(\mathrm{two},\infty \right),$ the range is $\left(-\infty ,\infty \right),$ and the vertical asymptote is $x=2.$

8.

The domain is $\left(-\infty ,0\right),$ the range is $\left(-\infty ,\infty \right),$ and the vertical asymptote is $\mathrm{10}=0.$

11.

$f(\mathrm{ten})=2\ln (x+3)-\mathrm{one}$

6.v Logarithmic Properties

1.

${\log }_b\mathrm{two}+{\log }_b\mathrm{ii}+{\log }_{b}2+{\log }_{b}k=3{\log }_{b}2+{\log }_b\mathrm{one\; thousand}$

2.

${\log }_{\mathrm{three}}\left(x+\mathrm{iii}\right)-{\log }_3\left(x-1\right)-{\log }_3\left(\mathrm{ten}-\mathrm{ii}\right)$

half dozen.

$\mathrm{two}\log x+3\log y-\mathrm{iv}\log z$

viii.

$\frac{\mathrm{i}}{2}\ln \left(\mathrm{ten}-1\right)+\ln \left(\mathrm{two}x+\mathrm{i}\right)-\ln \left(\mathrm{10}+3\right)-\ln \left(\mathrm{10}-3\right)$

9.

$\log \left(\frac{\mathrm{iii}\cdot 5}{4\cdot \mathrm{six}}\right);$ can as well be written $\log \left(\frac{5}{8}\right)$ past reducing the fraction to everyman terms.

10.

$\log \left(\frac{5{\left(x-1\right)}^3\sqrt{x}}{\left(\mathrm{seven}\mathrm{ten}-1\right)}\right)$

11.

$\log \frac{{\mathrm{ten}}^{12}{\left(x+\mathrm{five}\right)}^4}{{\left(\mathrm{two}\mathrm{10}+3\right)}^4};$ this answer could besides exist written $\log {\left(\frac{x^{\mathrm{three}}\left(x+\mathrm{five}\right)}{\left(2x+3\right)}\right)}^{\mathrm{iv}}.$

12.

The pH increases by most 0.301.

14.

$\frac{\ln 100}{\ln 5}\approx \frac{4.6051}{1.6094}=2.861$

6.6 Exponential and Logarithmic Equations

4.

The equation has no solution.

five.

$x=\frac{\ln \mathrm{three}}{\ln \left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\mathrm{iii}$}\right.\right)}$

6.

$t=\mathrm{two}\ln \left(\frac{\mathrm{eleven}}{\mathrm{iii}}\right)$ or $\ln {\left(\frac{\mathrm{xi}}{3}\right)}^{\mathrm{ii}}$

7.

$t=\ln \left(\frac{\mathrm{ane}}{\sqrt{2}}\right)=-\frac{1}{2}\ln \left(\mathrm{ii}\right)$

13.

$t=703,800,000\times \frac{\ln (0.8)}{\ln (\mathrm{0.five})}\text{years}\approx 226,572,993\text{years}.$

half-dozen.vii Exponential and Logarithmic Models

ane.

$f(t)=A_0{\mathrm{east}}^{-0.0000000087t}$

2.

less than 230 years, 229.3157 to be exact

3.

$f(t)=A_0{\mathrm{east}}^{\frac{\ln 2}{3}t}$

6.

Exponential. $y=2e^{\mathrm{0.v}\mathrm{10}}.$

7.

$y=\mathrm{three}{e}^{\left(\ln 0.5\right)x}$

6.eight Fitting Exponential Models to Information

i.

1. ⓐ The exponential regression model that fits these data is $y=522.88585984{\left(1.19645256\right)}^x.$
2. ⓑ If spending continues at this charge per unit, the graduate's credit menu debt will be $4,499.38 afterwards ane year.

2.

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

3.

1. ⓐ The logistic regression model that fits these information is $y=\frac{25.65665979}{\mathrm{ane}+\mathrm{half-dozen.113686306}{e}^{-0.3852149008x}}.$
2. ⓑ If the population continues to grow at this rate, there will be about $\text{25,634}$ seals in 2020.
3. ⓒ To the nearest whole number, the conveying capacity is 25,657.

6.one Section Exercises

one.

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

3.

When interest is compounded, the pct of interest earned to principal ends up being greater than the almanac percentage rate for the investment business relationship. Thus, the annual pct rate does not necessarily represent 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 past a abiding corporeality each visit, so the statement represents a linear function. .

nine.

The woods represented by the role $B(t)=82{(\mathrm{ane.029})}^t.$

11.

After $t=\mathrm{xx}$ years, forest A will have $43$ 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 really grows at a faster rate, the population volition somewhen become larger than forest A and will remain that fashion equally 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.

xv.

exponential growth; The growth gene, $1.06,$ is greater than $\mathrm{i.}$

17.

exponential decay; The decay gene, $0.97,$ is between $0$ and $1.$

19.

$f(x)=2000{(\mathrm{0.one})}^{\mathrm{10}}$

21.

$f(x)={\left(\frac{\mathrm{one}}{6}\right)}^{-\frac{3}{5}}{\left(\frac{1}{\mathrm{half\; dozen}}\right)}^{\frac{\mathrm{10}}{5}}\approx 2.93{\left(0.699\right)}^{\mathrm{10}}$

31.

$\$13,268.58$

33.

$P=A(t)\cdot {\left(\mathrm{one}+\frac{r}{n}\right)}^{-nt}$

39.

continuous growth; the growth rate is greater than $0.$

41.

continuous disuse; the growth rate is less than $0.$

47.

$f(-1)\approx -0.2707$

49.

$f(3)\approx 483.8146$

53.

$y\approx \mathrm{eighteen}\cdot {1.025}^{\mathrm{ten}}$

55.

$y\approx 0.2\cdot {1.95}^x$

57.

$\text{APY}=\frac{A(t)-a}{a}=\frac{a{\left(1+\frac{r}{365}\right)}^{365(\mathrm{ane})}-a}{a}=\frac{a\left[{\left(\mathrm{one}+\frac{r}{365}\right)}^{365}-1\right]}{a}={\left(1+\frac{r}{365}\right)}^{365}-\mathrm{one};$ $I(\mathrm{north})={\left(1+\frac{r}{\mathrm{due\; north}}\right)}^n-1$

59.

Allow $f$ be the exponential disuse function $f(x)=a\cdot {\left(\frac{1}{b}\right)}^{\mathrm{10}}$ such that $b>\mathrm{one.}$ And then for some number $\mathrm{due\; north}>0,$ $f(\mathrm{ten})=a\cdot {\left(\frac{1}{b}\right)}^{\mathrm{ten}}=a{\left(b^{-1}\right)}^{\mathrm{ten}}=a{\left({\left(e^n\right)}^{-\mathrm{i}}\right)}^x=a{\left(e^{-n}\right)}^x=a{\left(e\right)}^{-\mathrm{northward}x}.$

63.

$\mathrm{i.39}\%;$ $\$155,368.09$

65.

$\$35,838.76$

67.

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

6.ii Department Exercises

i.

An asymptote is a line that the graph of a function approaches, as $x$ either increases or decreases without bound. The horizontal asymptote of an exponential function tells u.s.a. the limit of the part's values as the independent variable gets either extremely big or extremely small.

iii.

$g(\mathrm{10})=\mathrm{iv}{\left(3\right)}^{-x};$ y-intercept: $(0,\mathrm{four});$ Domain: all real numbers; Range: all real numbers greater than $0.$

5.

$g(\mathrm{ten})=-{10}^x+7;$ y-intercept: $\left(0,6\right);$ Domain: all existent numbers; Range: all real numbers less than $7.$

vii.

$g(x)=2{\left(\frac{\mathrm{i}}{\mathrm{four}}\right)}^{\mathrm{ten}};$ y-intercept: $\left(0,\text{2}\right);$ Domain: all real numbers; Range: all real numbers greater than $0.$

9.

y-intercept: $(0,-\mathrm{ii})$

27.

Horizontal asymptote: $h(\mathrm{10})=3;$ Domain: all real numbers; Range: all real numbers strictly greater than $3.$

29.

As $x\to \infty$ , $f\left(x\right)\to -\infty$ ;
As $x\to -\infty$ , $f\left(x\right)\to -\mathrm{i}$

31.

As $x\to \infty$ , $f\left(\mathrm{ten}\right)\to 2$ ;
As $x\to -\infty$ , $f\left(x\right)\to \infty$

33.

$f\left(x\right)=4^x-3$

35.

$f(x)={\mathrm{iv}}^{x-5}$

37.

$f\left(\mathrm{10}\right)=4^{-x}$

39.

$y=-2^{\mathrm{ten}}+3$

41.

$y=-2{\left(\mathrm{three}\right)}^{\mathrm{10}}+7$

43.

$k(\mathrm{half-dozen})=800+\frac{\mathrm{one}}{3}\approx 800.3333$

51.

The graph of $G(x)={\left(\frac{1}{b}\right)}^x$ is the refelction most the y-centrality of the graph of $F(x)=b^x;$ For whatever real number $b>0$ and function $f(\mathrm{ten})=b^x,$ the graph of ${\left(\frac{\mathrm{ane}}{b}\right)}^x$ is the the reflection nearly the y-axis, $F(-x).$

53.

The graphs of $g(x)$ and $h(x)$ are the aforementioned and are a horizontal shift to the correct of the graph of $f(x);$ For any existent number due north, real number $b>0,$ and office $f(x)=b^x,$ the graph of $\left(\frac{1}{b^n}\right)b^x$ is the horizontal shift $f(x-\mathrm{north}).$

6.3 Section Exercises

one.

A logarithm is an exponent. Specifically, information technology is the exponent to which a base $b$ is raised to produce a given value. In the expressions given, the base $b$ has the same value. The exponent, $y,$ in the expression $b^y$ can likewise be written every bit the logarithm, ${\log }_{b}x,$ and the value of $x$ is the result of raising $b$ to the ability of $y.$

3.

Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation $b^y=x,$ and and so properties of exponents can be applied to solve for $x.$

5.

The natural logarithm is a special case of the logarithm with base $b$ in that the natural log always has base $e.$ Rather than notating the natural logarithm as ${\log }_e\left(x\right),$ the annotation used is $\ln \left(\mathrm{10}\right).$

17.

${\text{log}}_c(m)=d$

19.

${\log }_{19}y=x$

21.

${\log }_n\left(103\right)=4$

23.

${\log }_y\left(\frac{39}{100}\right)=x$

27.

$\mathrm{10}=2^{-3}=\frac{1}{8}$

29.

$\mathrm{ten}=3^3=27$

31.

$\mathrm{10}=9^{\frac{1}{2}}=3$

33.

$x=6^{-3}=\frac{\mathrm{ane}}{216}$

59.

No, the role has no divers value for $\mathrm{10}=0.$ To verify, suppose $\mathrm{ten}=0$ is in the domain of the function $f(\mathrm{10})=\log (x).$ And then at that place is some number $n$ such that $n=\log (0).$ Rewriting as an exponential equation gives: ${\mathrm{x}}^n=0,$ which is impossible since no such real number $n$ exists. Therefore, $x=0$ is non the domain of the role $f(\mathrm{10})=\log (x).$

61.

Yes. Suppose there exists a real number $x$ such that $\ln x=\mathrm{ii.}$ Rewriting as an exponential equation gives $x=e^2,$ which is a existent number. To verify, allow $\mathrm{ten}={\mathrm{east}}^2.$ Then, by definition, $\ln \left(x\right)=\ln \left(e^2\right)=2.$

63.

No; $\ln \left(\mathrm{ane}\right)=0,$ so $\frac{\ln \left({\mathrm{eastward}}^{1.725}\right)}{\ln \left(1\right)}$ is undefined.

6.4 Section Exercises

one.

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

3.

Shifting the function correct or left and reflecting the function nearly 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 role in general form is all real numbers.

vii.

Domain: $\left(-\infty ,\frac{\mathrm{i}}{2}\right);$ Range: $\left(-\infty ,\infty \right)$

9.

Domain: $\left(-\frac{17}{4},\infty \right);$ Range: $\left(-\infty ,\infty \right)$

11.

Domain: $\left(5,\infty \right);$ Vertical asymptote: $\mathrm{ten}=\mathrm{v}$

13.

Domain: $\left(-\frac{\mathrm{i}}{3},\infty \right);$ Vertical asymptote: $x=-\frac{1}{3}$

15.

Domain: $\left(-\mathrm{iii},\infty \right);$ Vertical asymptote: $x=-3$

17.

Domain: $(\frac{3}{\mathrm{seven}},\infty )$ ;
Vertical asymptote: $\mathrm{10}=\frac{3}{7}$ ; Stop behavior: equally $x\to {\left(\frac{3}{7}\right)}^{+},f(x)\to -\infty$ and equally $\mathrm{10}\to \infty ,f(x)\to \infty$

19.

Domain: $\left(-3,\infty \right)$ ; Vertical asymptote: $x=-3$ ;
Cease behavior: as $x\to -3^{+}$ , $f(\mathrm{10})\to -\infty$ and as $\mathrm{ten}\to \infty$ , $f(x)\to \infty$

21.

Domain: $\left(1,\infty \right);$ Range: $\left(-\infty ,\infty \right);$ Vertical asymptote: $x=1;$ x-intercept: $\left(\frac{5}{\mathrm{iv}},0\right);$ y-intercept: DNE

23.

Domain: $\left(-\infty ,0\right);$ Range: $\left(-\infty ,\infty \right);$ Vertical asymptote: $\mathrm{ten}=0;$ 10-intercept: $\left(-{\mathrm{east}}^{\mathrm{ii}},0\right);$ y-intercept: DNE

25.

Domain: $\left(0,\infty \right);$ Range: $\left(-\infty ,\infty \right);$ Vertical asymptote: $\mathrm{ten}=0;$ x-intercept: $\left(e^3,0\right);$ y-intercept: DNE

47.

$f(\mathrm{10})={\log }_{\mathrm{two}}(-(\mathrm{10}-1))$

49.

$f(x)=3{\log }_4(\mathrm{ten}+\mathrm{ii})$

57.

The graphs of $f(x)={\log }_{\frac{\mathrm{i}}{2}}\left(\mathrm{10}\right)$ and $\mathrm{grand}(x)=-{\log }_{\mathrm{ii}}\left(\mathrm{ten}\right)$ announced to exist the same; Conjecture: for whatever positive base of operations $b\ne 1,$ ${\log }_b\left(x\right)=-{\log }_{\frac{\mathrm{ane}}{b}}\left(x\right).$

59.

Recall that the statement of a logarithmic part must be positive, so nosotros make up one's mind where $\frac{x+2}{\mathrm{ten}-4}>0$ . From the graph of the role $f\left(x\right)=\frac{x+\mathrm{ii}}{\mathrm{ten}-4},$ note that the graph lies above the 10-axis on the interval $\left(-\infty ,-2\right)$ and once more to the right of the vertical asymptote, that is $\left(4,\infty \right).$ Therefore, the domain is $\left(-\infty ,-\mathrm{ii}\right)\cup \left(\mathrm{iv},\infty \right).$

6.five Section Exercises

one.

Any root expression tin exist rewritten as an expression with a rational exponent so that the power dominion tin can exist applied, making the logarithm easier to calculate. Thus, ${\log }_b\left({\mathrm{10}}^{\frac{1}{\mathrm{due\; north}}}\right)=\frac{1}{n}{\log }_b(x).$

three.

${\log }_b\left(\mathrm{two}\right)+{\log }_b\left(7\right)+{\log }_b\left(x\right)+{\log }_b\left(y\right)$

5.

${\log }_b\left(\mathrm{xiii}\right)-{\log }_b\left(17\right)$

xiii.

${\text{log}}_b\left(7\right)$

15.

$15\log (x)+13\log (y)-19\log (z)$

17.

$\frac{3}{2}\log (x)-2\log (y)$

19.

$\frac{\mathrm{viii}}{3}\log (\mathrm{10})+\frac{14}{\mathrm{iii}}\log (y)$

21.

$\ln (2x^{\mathrm{vii}})$

23.

$\log \left(\frac{\mathrm{10}{z}^{\mathrm{iii}}}{\sqrt{y}}\right)$

25.

${\log }_{\mathrm{vii}}\left(\mathrm{xv}\right)=\frac{\ln \left(15\right)}{\ln \left(7\right)}$

27.

${\log }_{11}\left(\mathrm{five}\right)=\frac{{\log }_5\left(5\right)}{{\log }_5\left(11\right)}=\frac{1}{b}$

29.

${\log }_{\mathrm{eleven}}\left(\frac{6}{\mathrm{xi}}\right)=\frac{{\log }_5\left(\frac{6}{11}\right)}{{\log }_5\left(11\right)}=\frac{{\log }_5\left(\mathrm{half-dozen}\right)-{\log }_5\left(\mathrm{xi}\right)}{{\log }_5\left(\mathrm{eleven}\right)}=\frac{a-b}{b}=\frac{a}{b}-1$

39.

$x=4;$ Past the caliber rule: ${\log }_{\mathrm{half\; dozen}}\left(x+\mathrm{two}\right)-{\log }_6\left(\mathrm{10}-3\right)={\log }_{\mathrm{half-dozen}}\left(\frac{x+\mathrm{two}}{\mathrm{10}-3}\right)=1.$

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

$$\begin{array}{ll}{\mathrm{vi}}^1\hfill & =\frac{x+\mathrm{two}}{x-\mathrm{three}}\hfill \\ 0\hfill & =\frac{\mathrm{10}+2}{\mathrm{10}-\mathrm{iii}}-6\hfill \\ 0\hfill & =\frac{x+2}{x-\mathrm{three}}-\frac{6\left(\mathrm{10}-3\right)}{\left(\mathrm{ten}-3\right)}\hfill \\ 0\hfill & =\frac{x+\mathrm{two}-\mathrm{vi}x+18}{\mathrm{10}-\mathrm{iii}}\hfill \\ 0\hfill & =\frac{\mathrm{10}-\mathrm{four}}{x-3}\hfill \\ \text{}x\hfill & =\mathrm{iv}\hfill \end{array}$$

Checking, we notice that ${\log }_6\left(\mathrm{iv}+2\right)-{\log }_6\left(\mathrm{iv}-3\right)={\log }_6\left(\mathrm{half-dozen}\right)-{\log }_6\left(\mathrm{i}\right)$ is divers, then $x=4.$

41.

Let $b$ and $\mathrm{due\; north}$ be positive integers greater than $1.$ Then, by the alter-of-base of operations formula, ${\log }_b\left(\mathrm{northward}\right)=\frac{{\log }_n\left(n\right)}{{\log }_n\left(b\right)}=\frac{\mathrm{ane}}{{\log }_n\left(b\right)}.$

6.vi Section Exercises

ane.

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

3.

The one-to-1 property can exist used if both sides of the equation can be rewritten as a single logarithm with the same base of operations. If and so, the arguments tin can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one belongings cannot be used when each side of the equation cannot exist rewritten equally a single logarithm with the same base.

15.

$p=\log \left(\frac{17}{8}\right)-7$

17.

$\mathrm{yard}=-\frac{\ln \left(38\right)}{\mathrm{iii}}$

19.

$\mathrm{ten}=\frac{\ln \left(\frac{38}{3}\right)-8}{9}$

23.

$\mathrm{ten}=\frac{\ln \left(\frac{3}{5}\right)-3}{\mathrm{eight}}$

29.

${10}^{-\mathrm{ii}}=\frac{1}{100}$

51.

$\mathrm{10}=9$

53.

$x=\frac{{\mathrm{eastward}}^{\mathrm{two}}}{\mathrm{iii}}\approx \mathrm{two.5}$

55.

$x=-5$

57.

$\mathrm{10}=\frac{e+\mathrm{ten}}{4}\approx 3.2$

59.

No solution

61.

$x=\frac{11}{5}\approx \mathrm{ii.2}$

63.

$x=\frac{101}{11}\approx \mathrm{9.two}$

65.

nigh $\$27,710.24$

67.

about v years

69.

$\frac{\ln (17)}{\mathrm{v}}\approx 0.567$

71.

$\mathrm{ten}=\frac{\log \left(38\right)+5\log \left(\mathrm{iii}\right)}{\mathrm{four}\log \left(3\right)}\approx 2.078$

75.

$\mathrm{10}\approx -\text{44655}.\text{7143}$

79.

$t=\ln \left({\left(\frac{y}{A}\right)}^{\frac{1}{g}}\right)$

81.

$t=\ln \left({\left(\frac{T-T_s}{T_0-T_s}\right)}^{-\frac{1}{\mathrm{thou}}}\right)$

half-dozen.7 Section Exercises

1.

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

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.

five.

An guild of magnitude is the nearest power of 10 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 peachy amount. For example, the mass of Saturn is 95 times greater than the mass of Earth. This is the same as proverb that the mass of Saturn is about ${10}^{\text{2}}$ times, or 2 orders of magnitude greater, than the mass of Globe.

7.

$f(0)\approx 16.7;$ The amount initially present is about 16.7 units.

11.

exponential; $f(\mathrm{10})={1.2}^{\mathrm{ten}}$

thirteen.

logarithmic

15.

logarithmic

23.

$4$ one-half-lives; $8.18$ minutes

25.

$\begin{array}{l}M=\frac{2}{\mathrm{three}}\log \left(\frac{S}{{\mathrm{South}}_0}\right)\hfill \\ \log \left(\frac{S}{S_0}\right)=\frac{3}{\mathrm{ii}}\mathrm{1000}\hfill \\ \frac{\mathrm{South}}{{\mathrm{Southward}}_0}={\mathrm{x}}^{\frac{\mathrm{three}\mathrm{1000}}{\mathrm{two}}}\hfill \\ S={\mathrm{Southward}}_0{10}^{\frac{3G}{2}}\hfill \end{array}$

27.

Allow $y=b^x$ for some not-negative real number $b$ such that $b\ne 1.$ Then,

$\begin{array}{l}\ln (y)=\ln (b^x)\hfill \\ \ln (y)=x\ln (b)\hfill \\ {\mathrm{east}}^{\ln (y)}={\mathrm{east}}^{x\ln (b)}\hfill \\ y=e^{x\ln (b)}\hfill \end{array}$

29.

$A=125e^{\left(-0.3567t\right)};A\approx 43$ mg

33.

$A(t)=250e^{(-0.00822t)};$ half-life: about $\text{84}$ minutes

35.

$r\approx -0.0667,$ So the hourly decay rate is near $6.67\%$

37.

$f(t)=1350{\mathrm{east}}^{(0.03466t)};$ after three hours: $P(180)\approx 691,200$

39.

$f(t)=256e^{(0.068110t)};$ doubling time: near $10$ minutes

43.

$T(t)=90e^{(-0.008377t)}+75,$ where $t$ is in minutes.

45.

about $\text{113}$ minutes

47.

$\log \left(\mathrm{10}\right)=\mathrm{ane.5};x\approx 31.623$

49.

MMS magnitude: $5.82$

vi.8 Section Exercises

i.

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

three.

Regression analysis is the process of finding an equation that all-time fits a given ready of data points. To perform a regression assay on a graphing utility, beginning list the given points using the STAT and so EDIT carte du jour. Next graph the scatter plot using the STAT PLOT characteristic. The shape of the information points on the scatter graph can help determine which regression feature to use. One time this is determined, select the appropriate regression analysis control from the STAT then CALC bill of fare.

5.

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

xi.

$P(0)=22$ ; 175

15.

y-intercept: $\left(0,15\right)$

19.

about $\mathrm{six.8}$ months.

27.

$f\left(\mathrm{ten}\right)=776.682{\left(1.426\right)}^{\mathrm{10}}$

33.

$f\left(\mathrm{10}\right)={731.92e}^{-0.3038\mathrm{10}}$

35.

When $f\left(x\right)=250,x\approx 3.6$

37.

$y=\mathrm{five.063}+\mathrm{i.934}\text{log}\left(x\right)$

43.

When $f\left(\mathrm{ten}\right)\approx 2.3$

45.

When $f\left(\mathrm{10}\right)=8,x\approx 0.82$

47.

$f(x)=\frac{25.081}{1+3.182e^{-0.545x}}$

55.

When $f\left(\mathrm{ten}\right)=68,x\approx \mathrm{4.ix}$

57.

$f\left(\mathrm{10}\right)=1.034341{\left(1.281204\right)}^x$ ; $g\left(x\right)=4.035510$ ; the regression curves are symmetrical about $y=x$ , and so information technology appears that they are inverse functions.

59.

$f^{-1}\left(x\right)=\frac{\text{ln}(a)-\text{ln}(\frac{c}{x}-\mathrm{ane})}{b}$

Review Exercises

1.

exponential decay; The growth gene, $0.825,$ is between $0$ and $\mathrm{ane.}$

3.

$y=0.25{\left(\mathrm{iii}\right)}^x$

7.

continuous decay; the growth charge per unit is negative.

ix.

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

11.

$g(x)=7{\left(\mathrm{6.five}\right)}^{-\mathrm{10}};$ y-intercept: $(0,\text{vii});$ Domain: all real numbers; Range: all existent numbers greater than $0.$

13.

${17}^x=4913$

15.

${\log }_{a}b=-\frac{2}{5}$

17.

$x={64}^{\frac{\mathrm{i}}{3}}=\mathrm{iv}$

19.

$\log \left(\text{0}\text{.000001}\right)=-\mathrm{half\; dozen}$

21.

$\ln \left(e^{-0.8648}\right)=-0.8648$

25.

Domain: $x>-\mathrm{v};$ Vertical asymptote: $x=-\mathrm{v};$ End behavior: as $x\to -5^{+},f(x)\to -\infty$ and as $\mathrm{10}\to \infty ,f(x)\to \infty .$

27.

${\text{log}}_{\mathrm{viii}}\left(65xy\right)$

29.

$\ln \left(\frac{z}{xy}\right)$

31.

${\text{log}}_y\left(12\right)$

33.

$\ln \left(\mathrm{two}\right)+\ln \left(b\right)+\frac{\ln \left(b+1\right)-\ln \left(b-\mathrm{one}\right)}{\mathrm{two}}$

35.

${\log }_7\left(\frac{v^{\mathrm{three}}{w}^6}{\sqrt[\mathrm{three}]{u}}\right)$

37.

$\mathrm{ten}=\frac{\frac{\log \left(125\right)}{\log \left(\mathrm{five}\right)}+17}{12}=\frac{5}{3}$

45.

$x=\ln \left(\mathrm{eleven}\right)$

51.

virtually $5.45$ years

53.

$f^{-\mathrm{one}}\left(x\right)=\sqrt[\mathrm{three}]{{\mathrm{ii}}^{\mathrm{iv}x}-1}$

55.

$f(t)=300{\left(0.83\right)}^t;$
$f(24)\approx \mathrm{three.43}g$

61.

exponential

63.

$y=\mathrm{iv}{\left(0.2\right)}^x;$ $y=4e^{\text{-i.609438}x}$

67.

logarithmic; $y=\mathrm{sixteen.68718}-9.71860\ln (\mathrm{ten})$

Practice Test

5.

y-intercept: $(0,\text{5})$

7.

${\mathrm{eight.5}}^a=614.125$

9.

$\mathrm{10}={\left(\frac{1}{\mathrm{seven}}\right)}^2=\frac{1}{49}$

xi.

$\ln \left(0.716\right)\approx -0.334$

thirteen.

Domain: $\mathrm{10}<3;$ Vertical asymptote: $\mathrm{10}=\mathrm{iii};$ End behavior: $x\to {\mathrm{iii}}^{-},f(\mathrm{10})\to -\infty$ and $x\to -\infty ,f(x)\to \infty$

15.

${\log }_t\left(12\right)$

17.

$3\ln \left(y\right)+2\ln \left(z\right)+\frac{\ln \left(x-\mathrm{four}\right)}{\mathrm{three}}$

19.

$x=\frac{\frac{\ln \left(\mathrm{yard}\right)}{\ln \left(16\right)}+\mathrm{v}}{\mathrm{iii}}\approx \mathrm{two.497}$

21.

$a=\frac{\ln \left(\mathrm{iv}\right)+8}{10}$

29.

$f(t)=112e^{-.019792t};$ one-half-life: about $35$ days

31.

$T(t)=36e^{-0.025131t}+35;T\left(60\right)\approx {43}^{\text{o}}\text{F}$

33.

logarithmic

35.

exponential; $y=\mathrm{fifteen.10062}{\left(\mathrm{ane.24621}\right)}^x$

37.

logistic; $y=\frac{\mathrm{xviii.41659}}{1+7.54644e^{-0.68375x}}$

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Source: https://openstax.org/books/college-algebra/pages/chapter-6

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