What Is Chapter 4- Review (Day 1) Write the Letter for the Correct Answer

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iv.1 Exponential Functions

i .

$g(x)={0.875}^{\mathrm{10}}$ and $j(\mathrm{10})={1095.6}^{-2x}$ represent exponential functions.

3 .

About $1.548$ billion people; past the twelvemonth 2031, India'due south population will exceed China'south past about 0.001 billion, or i million people.

4 .

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

5 .

$f(x)=\mathrm{ii}{\left(\mathrm{1.five}\right)}^x$

6 .

$f(x)=\sqrt{\mathrm{ii}}{\left(\sqrt{\mathrm{two}}\right)}^{\mathrm{10}}.$ Answers may vary due to circular-off error. The answer should be very close to $1.4142{\left(\mathrm{i.4142}\right)}^{\mathrm{10}}.$

seven .

$y\approx 12\cdot {1.85}^x$

ten .

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

12 .

three.77E-26 (This is calculator notation for the number written as $\mathrm{iii.77}\times {\mathrm{ten}}^{-26}$ in scientific annotation. While the output of an exponential part is never zero, this number is so shut to zip that for all practical purposes we can accept zero as the respond.)

4.2 Graphs of Exponential Functions

one .

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

2 .

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

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 $y=0.$

six .

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

4.3 Logarithmic Functions

one .

1. ⓐ ${\log }_{\mathrm{ten}}\left(1,000,000\right)=\mathrm{vi}$ is equivalent to ${10}^{\mathrm{six}}=\mathrm{i},000,000$
2. ⓑ ${\log }_5\left(25\right)=\mathrm{ii}$ is equivalent to $5^2=25$

2 .

1. ⓐ $3^2=9$ is equivalent to ${\log }_3(\mathrm{ix})=2$
2. ⓑ ${\mathrm{five}}^{\mathrm{three}}=125$ is equivalent to ${\log }_5(125)=3$
3. ⓒ $2^{-1}=\frac{1}{2}$ is equivalent to ${\text{log}}_{\mathrm{ii}}(\frac{\mathrm{ane}}{\mathrm{ii}})=-\mathrm{i}$

three .

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

four .

${\log }_{\mathrm{two}}\left(\frac{1}{32}\right)=-5$

five .

$\log (\mathrm{i},000,000)=6$

6 .

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

7 .

The deviation in magnitudes was nigh $\mathrm{3.929.}$

viii .

It is non possible to accept the logarithm of a negative number in the set of real numbers.

four.four Graphs of Logarithmic Functions

3 .

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

v .

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

6 .

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

seven .

The domain is $\left(2,\infty \right),$ the range is $\left(-\infty ,\infty \right),$ and the vertical asymptote is $\mathrm{10}=\mathrm{two.}$

viii .

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

eleven .

$f(x)=2\ln (x+\mathrm{iii})-1$

iv.five Logarithmic Properties

i .

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

2 .

${\log }_3\left(x+3\right)-{\log }_{\mathrm{three}}\left(\mathrm{ten}-1\right)-{\log }_{\mathrm{three}}\left(\mathrm{ten}-2\right)$

6 .

$\mathrm{ii}\log \mathrm{10}+3\log y-\mathrm{four}\log z$

8 .

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

ix .

$\log \left(\frac{3\cdot 5}{\mathrm{iv}\cdot 6}\right);$ tin can too be written $\log \left(\frac{5}{8}\right)$ by reducing the fraction to lowest terms.

10 .

$\log \left(\frac{\mathrm{v}{\left(x-1\right)}^{\mathrm{iii}}\sqrt{x}}{\left(7\mathrm{ten}-1\right)}\right)$

11 .

$\log \frac{x^{12}{\left(x+5\right)}^{\mathrm{iv}}}{{\left(2x+\mathrm{three}\right)}^4};$ this answer could also be written $\log {\left(\frac{{\mathrm{ten}}^3\left(x+5\right)}{\left(\mathrm{two}\mathrm{ten}+3\right)}\right)}^4.$

12 .

The pH increases by nearly 0.301.

14 .

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

4.6 Exponential and Logarithmic Equations

iv .

The equation has no solution.

5 .

$\mathrm{ten}=\frac{\ln \mathrm{three}}{\ln \left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}$

vi .

$t=\mathrm{two}\ln \left(\frac{11}{3}\right)$ or $\ln {\left(\frac{11}{3}\right)}^2$

vii .

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

12 .

$\mathrm{10}=\mathrm{one}$ or $\mathrm{ten}=-\mathrm{i}$

thirteen .

$t=703,800,000\times \frac{\ln (\mathrm{0.viii})}{\ln (0.5)}\text{years}\approx 226,572,993\text{years}.$

four.7 Exponential and Logarithmic Models

1 .

$f(t)=A_0{e}^{-0.0000000087t}$

2 .

less than 230 years, 229.3157 to exist exact

3 .

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

6 .

Exponential. $y=\mathrm{ii}{e}^{0.5\mathrm{10}}.$

seven .

$y=\mathrm{three}{e}^{\left(\ln \mathrm{0.v}\right)x}$

4.8 Plumbing fixtures Exponential Models to Information

one .

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'south credit bill of fare debt will be $4,499.38 after one 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, nearly 171,000 games volition be sold in the year 2015.

3 .

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

4.i Section Exercises

1 .

Linear functions have a abiding rate of change. Exponential functions increase based on a pct of the original.

iii .

When interest is compounded, the percentage of interest earned to master ends up being greater than the annual percentage rate for the investment account. Thus, the almanac percentage charge per unit does non necessarily correspond to the real involvement earned, which is the very definition of nominal.

five .

exponential; the population decreases by a proportional charge per unit. .

7 .

not exponential; the accuse decreases by a constant corporeality each visit, and then the statement represents a linear function. .

nine .

The forest represented by the function $B(t)=82{(\mathrm{i.029})}^t.$

eleven .

Later on $t=20$ 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 woods A volition increasingly exceed forest B, but because forest B actually grows at a faster rate, the population volition eventually become larger than forest A and will remain that way as long every bit 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.

fifteen .

exponential growth; The growth factor, $1.06,$ is greater than $1.$

17 .

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

19 .

$f(\mathrm{ten})=2000{(0.1)}^x$

21 .

$f(x)={\left(\frac{\mathrm{ane}}{\mathrm{six}}\right)}^{-\frac{3}{\mathrm{five}}}{\left(\frac{\mathrm{one}}{6}\right)}^{\frac{\mathrm{ten}}{\mathrm{v}}}\approx 2.93{\left(0.699\right)}^x$

31 .

$\$\mathrm{xiii},268.58$

33 .

$P=A(t)\cdot {\left(1+\frac{r}{n}\right)}^{-nt}$

39 .

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

41 .

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

47 .

$f(-1)\approx -0.2707$

49 .

$f(3)\approx 483.8146$

53 .

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

55 .

$y\approx 0.2\cdot {\mathrm{i.95}}^{\mathrm{10}}$

57 .

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

59 .

Let $f$ be the exponential decay role $f(x)=a\cdot {\left(\frac{1}{b}\right)}^x$ such that $b>1.$ Then for some number $\mathrm{north}>0,$ $f(x)=a\cdot {\left(\frac{1}{b}\right)}^{\mathrm{10}}=a{\left(b^{-\mathrm{one}}\right)}^x=a{\left({\left(e^n\right)}^{-\mathrm{one}}\right)}^{\mathrm{ten}}=a{\left(e^{-n}\right)}^x=a{\left(\mathrm{due\; east}\right)}^{-\mathrm{north}x}.$

63 .

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

65 .

$\$35,838.76$

67 .

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

4.two Section Exercises

one .

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 part tells usa the limit of the part'south values equally the independent variable gets either extremely large or extremely small.

3 .

$\mathrm{thousand}(x)=4{\left(3\right)}^{-\mathrm{ten}};$ y-intercept: $(0,4);$ Domain: all real numbers; Range: all real numbers greater than $0.$

5 .

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

7 .

$g(x)=2{\left(\frac{\mathrm{one}}{4}\right)}^x;$ y-intercept: $\left(0,\text{2}\right);$ Domain: all existent numbers; Range: all real numbers greater than $0.$

9 .

y-intercept: $(0,-2)$

27 .

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

29 .

As $x\to \infty$ , $f\left(x\right)\to -\infty$ ;
Every bit $x\to -\infty$ , $f\left(x\right)\to -1$

31 .

As $x\to \infty$ , $f\left(x\right)\to 2$ ;
Every bit $x\to -\infty$ , $f\left(x\right)\to \infty$

33 .

$f\left(\mathrm{10}\right)=4^{\mathrm{ten}}-3$

35 .

$f(x)={\mathrm{iv}}^{\mathrm{10}-\mathrm{v}}$

37 .

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

39 .

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

41 .

$y=-2{\left(\mathrm{iii}\right)}^x+\mathrm{vii}$

43 .

$g(6)=800+\frac{1}{3}\approx 800.3333$

51 .

The graph of $G(\mathrm{10})={\left(\frac{1}{b}\right)}^{\mathrm{10}}$ is the refelction near the y-centrality of the graph of $F(x)=b^x;$ For any real number $b>0$ and office $f(x)=b^{\mathrm{ten}},$ the graph of ${\left(\frac{\mathrm{one}}{b}\right)}^{\mathrm{ten}}$ is the the reflection about the y-axis, $F(-x).$

53 .

The graphs of $g(\mathrm{ten})$ and $h(\mathrm{10})$ are the same and are a horizontal shift to the right of the graph of $f(x);$ For any real number north, real number $b>0,$ and office $f(\mathrm{10})=b^{\mathrm{ten}},$ the graph of $\left(\frac{1}{b^{\mathrm{north}}}\right)b^{\mathrm{ten}}$ is the horizontal shift $f(x-n).$

iv.iii Section Exercises

one .

A logarithm is an exponent. Specifically, it 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 also be written as the logarithm, ${\log }_{b}x,$ and the value of $x$ is the upshot of raising $b$ to the ability of $y.$

3 .

Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can exist converted to the exponential equation $b^y=x,$ and then properties of exponents can exist applied to solve for $\mathrm{ten}.$

5 .

The natural logarithm is a special example of the logarithm with base $b$ in that the natural log ever has base of operations $e.$ Rather than notating the natural logarithm as ${\log }_e\left(x\right),$ the notation used is $\ln \left(x\right).$

17 .

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

nineteen .

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

21 .

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

23 .

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

27 .

$x={\mathrm{two}}^{-3}=\frac{\mathrm{ane}}{8}$

29 .

$x={\mathrm{iii}}^3=27$

31 .

$x={\mathrm{ix}}^{\frac{1}{2}}=3$

33 .

$x=6^{-3}=\frac{1}{216}$

59 .

No, the function has no defined value for $x=0.$ To verify, suppose $x=0$ is in the domain of the function $f(x)=\log (\mathrm{ten}).$ Then there is some number $n$ such that $n=\log (0).$ Rewriting as an exponential equation gives: ${\mathrm{ten}}^n=0,$ which is incommunicable since no such real number $\mathrm{northward}$ exists. Therefore, $x=0$ is not the domain of the role $f(\mathrm{10})=\log (\mathrm{ten}).$

61 .

Yes. Suppose there exists a existent number $x$ such that $\ln x=2.$ Rewriting as an exponential equation gives $\mathrm{10}=e^2,$ which is a real number. To verify, let $x={\mathrm{east}}^2.$ Then, past definition, $\ln \left(\mathrm{ten}\right)=\ln \left({\mathrm{due\; east}}^2\right)=\mathrm{two.}$

63 .

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

4.4 Department Exercises

1 .

Since the functions are inverses, their graphs are mirror images about the line $y=\mathrm{10}.$ So for every point $(a,b)$ on the graph of a logarithmic office, at that place is a corresponding point $(b,a)$ on the graph of its changed exponential role.

3 .

Shifting the part correct or left and reflecting the part nearly the y-axis volition bear upon its domain.

5 .

No. A horizontal asymptote would advise a limit on the range, and the range of any logarithmic office in full general grade is all real numbers.

7 .

Domain: $\left(-\infty ,\frac{1}{2}\right);$ Range: $\left(-\infty ,\infty \right)$

9 .

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

xi .

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

thirteen .

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

15 .

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

17 .

Domain: $(\frac{3}{7},\infty )$ ;
Vertical asymptote: $x=\frac{3}{7}$ ; End beliefs: as $x\to {\left(\frac{\mathrm{iii}}{7}\right)}^{+},f(x)\to -\infty$ and as $x\to \infty ,f(x)\to \infty$

19 .

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

21 .

Domain: $\left(1,\infty \right);$ Range: $\left(-\infty ,\infty \right);$ Vertical asymptote: $x=\mathrm{i};$ 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: $\mathrm{ten}=0;$ x-intercept: $\left(-{\mathrm{due\; east}}^{\mathrm{two}},0\right);$ y-intercept: DNE

25 .

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

47 .

$f(\mathrm{ten})={\log }_2(-(x-\mathrm{i}))$

49 .

$f(x)=\mathrm{three}{\log }_4(x+\mathrm{two})$

57 .

The graphs of $f(x)={\log }_{\frac{\mathrm{i}}{2}}\left(x\right)$ and $\mathrm{chiliad}(\mathrm{10})=-{\log }_2\left(x\right)$ appear to be the same; Theorize: for whatever positive base $b\ne \mathrm{ane},$ ${\log }_b\left(x\right)=-{\log }_{\frac{\mathrm{ane}}{b}}\left(\mathrm{ten}\right).$

59 .

Think that the argument of a logarithmic office must be positive, and then we determine where $\frac{x+2}{\mathrm{ten}-4}>0$ . From the graph of the part $f\left(\mathrm{10}\right)=\frac{x+2}{x-\mathrm{four}},$ note that the graph lies above the x-axis on the interval $\left(-\infty ,-2\right)$ and again to the right of the vertical asymptote, that is $\left(4,\infty \right).$ Therefore, the domain is $\left(-\infty ,-2\right)\cup \left(4,\infty \right).$

4.five Section Exercises

1 .

Any root expression can be rewritten as an expression with a rational exponent then that the power rule tin can be applied, making the logarithm easier to calculate. Thus, ${\log }_b\left(x^{\frac{\mathrm{one}}{\mathrm{north}}}\right)=\frac{\mathrm{i}}{n}{\log }_b(x).$

three .

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

five .

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

13 .

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

fifteen .

$\mathrm{fifteen}\log (x)+13\log (y)-19\log (z)$

17 .

$\frac{\mathrm{three}}{2}\log (x)-\mathrm{two}\log (y)$

19 .

$\frac{\mathrm{eight}}{\mathrm{three}}\log (x)+\frac{14}{3}\log (y)$

21 .

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

23 .

$\log \left(\frac{x{z}^3}{\sqrt{y}}\right)$

25 .

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

27 .

${\log }_{\mathrm{eleven}}\left(\mathrm{v}\right)=\frac{{\log }_5\left(5\right)}{{\log }_5\left(11\right)}=\frac{\mathrm{ane}}{b}$

29 .

${\log }_{11}\left(\frac{\mathrm{six}}{\mathrm{eleven}}\right)=\frac{{\log }_5\left(\frac{6}{\mathrm{xi}}\right)}{{\log }_{\mathrm{five}}\left(\mathrm{xi}\right)}=\frac{{\log }_{\mathrm{v}}\left(\mathrm{half\; dozen}\right)-{\log }_{\mathrm{five}}\left(\mathrm{xi}\right)}{{\log }_{\mathrm{five}}\left(11\right)}=\frac{a-b}{b}=\frac{a}{b}-\mathrm{ane}$

39 .

$\mathrm{ten}=4;$ Past the caliber rule: ${\log }_{\mathrm{half\; dozen}}\left(x+2\right)-{\log }_6\left(x-\mathrm{three}\right)={\log }_6\left(\frac{\mathrm{ten}+\mathrm{two}}{x-3}\right)=\mathrm{ane.}$

Rewriting as an exponential equation and solving for $\mathrm{ten}:$

$$\begin{array}{ll}{6}^1\hfill & =\frac{x+\mathrm{two}}{\mathrm{ten}-\mathrm{iii}}\hfill \\ 0\hfill & =\frac{x+2}{\mathrm{10}-3}-6\hfill \\ 0\hfill & =\frac{\mathrm{ten}+\mathrm{two}}{\mathrm{10}-3}-\frac{6\left(x-3\right)}{\left(x-\mathrm{iii}\right)}\hfill \\ 0\hfill & =\frac{x+\mathrm{ii}-\mathrm{half-dozen}x+\mathrm{xviii}}{x-\mathrm{three}}\hfill \\ 0\hfill & =\frac{x-4}{x-3}\hfill \\ \text{}\mathrm{10}\hfill & =4\hfill \end{array}$$

Checking, we find that ${\log }_{\mathrm{half\; dozen}}\left(\mathrm{four}+\mathrm{two}\right)-{\log }_6\left(4-\mathrm{iii}\right)={\log }_6\left(6\right)-{\log }_6\left(1\right)$ is defined, then $\mathrm{10}=4.$

41 .

Let $b$ and $n$ be positive integers greater than $\mathrm{one.}$ So, by the modify-of-base formula, ${\log }_b\left(\mathrm{north}\right)=\frac{{\log }_{\mathrm{north}}\left(n\right)}{{\log }_n\left(b\right)}=\frac{1}{{\log }_n\left(b\right)}.$

four.6 Section Exercises

1 .

Decide start if the equation can be rewritten so that each side uses the same base. If and so, the exponents tin can be set equal to each other. If the equation cannot exist 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 ane-to-1 property can be used if both sides of the equation tin be rewritten equally a single logarithm with the same base of operations. 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 exist used when each side of the equation cannot be rewritten as a single logarithm with the same base.

15 .

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

17 .

$\mathrm{chiliad}=-\frac{\ln \left(38\right)}{3}$

19 .

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

23 .

$x=\frac{\ln \left(\frac{3}{5}\right)-3}{8}$

29 .

${10}^{-2}=\frac{\mathrm{ane}}{100}$

51 .

$x=\mathrm{ix}$

53 .

$x=\frac{{\mathrm{due\; east}}^2}{3}\approx \mathrm{2.v}$

55 .

$x=-5$

57 .

$\mathrm{10}=\frac{e+\mathrm{ten}}{4}\approx \mathrm{3.two}$

59 .

No solution

61 .

$x=\frac{11}{5}\approx 2.2$

63 .

$x=\frac{101}{11}\approx 9.2$

65 .

about $\$27,710.24$

67 .

most five years

69 .

$\frac{\ln (17)}{5}\approx 0.567$

71 .

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

75 .

$x\approx -\text{44655}.\text{7143}$

79 .

$t=\ln \left({\left(\frac{y}{A}\right)}^{\frac{\mathrm{ane}}{m}}\right)$

81 .

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

4.7 Section Exercises

1 .

Half-life is a mensurate of decay 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 one-half of the initial amount of that substance or quantity to decay.

iii .

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

5 .

An guild of magnitude is the nearest power of x by which a quantity exponentially grows. It is as well an guess position on a logarithmic calibration; 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 most ${\mathrm{x}}^{\text{ii}}$ times, or ii orders of magnitude greater, than the mass of Earth.

7 .

$f(0)\approx \mathrm{16.vii};$ The amount initially present is about 16.7 units.

11 .

exponential; $f(x)={\mathrm{ane.ii}}^x$

13 .

logarithmic

15 .

logarithmic

23 .

$\mathrm{four}$ half-lives; $8.18$ minutes

25 .

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

27 .

Let $y=b^x$ for some non-negative existent number $b$ such that $b\ne 1.$ Then,

$\begin{array}{l}\ln (y)=\ln (b^{\mathrm{10}})\hfill \\ \ln (y)=\mathrm{10}\ln (b)\hfill \\ e^{\ln (y)}={\mathrm{eastward}}^{x\ln (b)}\hfill \\ y=e^{\mathrm{ten}\ln (b)}\hfill \end{array}$

29 .

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

33 .

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

35 .

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

37 .

$f(t)=1350e^{(0.03466t)};$ after 3 hours: $P(180)\approx 691,200$

39 .

$f(t)=256e^{(0.068110t)};$ doubling time: nearly $\mathrm{ten}$ minutes

43 .

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

45 .

most $\text{113}$ minutes

47 .

$\log \left(x\right)=\mathrm{one.five};x\approx 31.623$

49 .

MMS magnitude: $5.82$

4.8 Section Exercises

i .

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

3 .

Regression assay is the process of finding an equation that all-time fits a given set of information 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 information points on the scatter graph tin can help decide which regression feature to use. Once this is determined, select the advisable regression analysis control from the STAT then CALC card.

5 .

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

11 .

$P(0)=22$ ; 175

15 .

y-intercept: $\left(0,\mathrm{xv}\right)$

19 .

about $6.8$ months.

27 .

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

33 .

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

35 .

When $f\left(\mathrm{ten}\right)=250,\mathrm{10}\approx \mathrm{three.6}$

37 .

$y=5.063+\mathrm{i.934}\text{log}\left(\mathrm{ten}\right)$

43 .

When $f\left(\mathrm{x}\right)\approx \mathrm{2.three}$

45 .

When $f\left(x\right)=8,x\approx 0.82$

47 .

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

55 .

When $f\left(x\right)=68,x\approx 4.9$

57 .

$f\left(\mathrm{ten}\right)=1.034341{\left(\mathrm{ane.281204}\right)}^{\mathrm{10}}$ ; $m\left(\mathrm{10}\right)=\mathrm{four.035510}$ ; the regression curves are symmetrical about $y=x$ , so it appears that they are inverse functions.

59 .

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

Review Exercises

one .

exponential decay; The growth factor, $0.825,$ is between $0$ and $1.$

iii .

$y=0.25{\left(3\right)}^x$

5 .

$\mathrm{\$}42,888.18$

vii .

continuous decay; the growth rate is negative.

9 .

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

11 .

$g(x)=\mathrm{seven}{\left(\mathrm{vi.5}\right)}^{-\mathrm{ten}};$ y-intercept: $(0,\text{7});$ Domain: all existent numbers; Range: all real numbers greater than $0.$

13 .

${17}^x=4913$

fifteen .

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

17 .

$x={64}^{\frac{1}{3}}=4$

19 .

$\log \left(\text{0}\text{.000001}\right)=-6$

21 .

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

25 .

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

27 .

${\text{log}}_8\left(65xy\right)$

29 .

$\ln \left(\frac{z}{\mathrm{10}y}\right)$

31 .

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

33 .

$\ln \left(2\right)+\ln \left(b\right)+\frac{\ln \left(b+\mathrm{ane}\right)-\ln \left(b-1\right)}{\mathrm{two}}$

35 .

${\log }_{\mathrm{vii}}\left(\frac{{\mathrm{five}}^3{w}^{\mathrm{six}}}{\sqrt[3]{u}}\right)$

37 .

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

45 .

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

51 .

nigh $5.45$ years

53 .

$f^{-1}\left(\mathrm{ten}\right)=\sqrt[\mathrm{iii}]{2^{\mathrm{four}x}-1}$

55 .

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

61 .

exponential

63 .

$y=4{\left(0.2\right)}^x;$ $y=4e^{\text{-1.609438}x}$

67 .

logarithmic; $y=16.68718-9.71860\ln (x)$

Practice Exam

5 .

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

7 .

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

9 .

$\mathrm{10}={\left(\frac{\mathrm{one}}{7}\right)}^2=\frac{\mathrm{i}}{49}$

11 .

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

13 .

Domain: $\mathrm{10}<3;$ Vertical asymptote: $x=\mathrm{three};$ Cease beliefs: $x\to 3^{-},f(\mathrm{10})\to -\infty$ and $x\to -\infty ,f(\mathrm{ten})\to \infty$

15 .

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

17 .

$3\ln \left(y\right)+2\ln \left(z\right)+\frac{\ln \left(x-4\right)}{\mathrm{three}}$

19 .

$x=\frac{\frac{\ln \left(1000\right)}{\ln \left(16\right)}+5}{\mathrm{iii}}\approx 2.497$

21 .

$a=\frac{\ln \left(4\right)+8}{\mathrm{ten}}$

29 .

$f(t)=112{\mathrm{east}}^{-.019792t};$ half-life: nearly $35$ days

31 .

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

33 .

logarithmic

35 .

exponential; $y=\mathrm{xv.10062}{\left(1.24621\right)}^x$

37 .

logistic; $y=\frac{18.41659}{1+7.54644e^{-0.68375\mathrm{10}}}$

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Source: https://openstax.org/books/precalculus/pages/chapter-4

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