Cpm Pre Calc Review and Preview Chapter 4

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4.one Exponential Functions

1 .

$g(\mathrm{10})={0.875}^x$ and $j(\mathrm{ten})={\mathrm{1095.half\; dozen}}^{-\mathrm{ii}\mathrm{ten}}$ represent exponential functions.

three .

Virtually $\mathrm{i.548}$ billion people; by the yr 2031, India's population volition exceed China'due south by nearly 0.001 billion, or 1 one thousand thousand people.

4 .

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

5 .

$f(\mathrm{ten})=2{\left(1.5\right)}^x$

vi .

$f(x)=\sqrt{2}{\left(\sqrt{\mathrm{two}}\right)}^{\mathrm{ten}}.$ Answers may vary due to circular-off mistake. The reply should be very close to $\mathrm{ane.4142}{\left(1.4142\right)}^x.$

vii .

$y\approx 12\cdot {1.85}^{\mathrm{10}}$

10 .

${\mathrm{due\; east}}^{-\mathrm{0.v}}\approx 0.60653$

12 .

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

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 $y=3.$

4 .

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

five .

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

six .

$f(\mathrm{10})=-\frac{\mathrm{ane}}{3}{e}^x-\mathrm{ii};$ the domain is $\left(-\infty ,\infty \right);$ the range is $\left(-\infty ,\mathrm{-2}\right);$ the horizontal asymptote is $y=\mathrm{-2.}$

4.three Logarithmic Functions

ane .

1. ⓐ ${\log }_{\mathrm{ten}}\left(1,000,000\right)=6$ is equivalent to ${10}^6=1,000,000$
2. ⓑ ${\log }_5\left(25\right)=\mathrm{ii}$ is equivalent to $5^2=25$

2 .

1. ⓐ ${\mathrm{three}}^2=\mathrm{ix}$ is equivalent to ${\log }_3(9)=2$
2. ⓑ $5^{\mathrm{three}}=125$ is equivalent to ${\log }_5(125)=3$
3. ⓒ ${\mathrm{two}}^{-\mathrm{ane}}=\frac{1}{2}$ is equivalent to ${\text{log}}_{\mathrm{ii}}(\frac{\mathrm{ane}}{\mathrm{two}})=-\mathrm{one}$

3 .

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

4 .

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

5 .

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

6 .

$\log \left(123\right)\approx \mathrm{two.0899}$

7 .

The difference in magnitudes was about $\mathrm{three.929.}$

viii .

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

four.4 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 $\left(-4,\infty \right),$ the range $\left(-\infty ,\infty \right),$ and the asymptote $x=–4.$

5 .

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 $x=0.$

vii .

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

viii .

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

xi .

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

4.5 Logarithmic Properties

one .

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

2 .

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

half-dozen .

$\mathrm{ii}\log x+\mathrm{iii}\log y-\mathrm{iv}\log z$

eight .

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

9 .

$\log \left(\frac{3\cdot 5}{4\cdot 6}\right);$ can also be written $\log \left(\frac{5}{\mathrm{eight}}\right)$ past reducing the fraction to everyman terms.

10 .

$\log \left(\frac{\mathrm{five}{\left(\mathrm{10}-1\right)}^3\sqrt{x}}{\left(\mathrm{vii}x-1\right)}\right)$

eleven .

$\log \frac{{\mathrm{10}}^{12}{\left(\mathrm{10}+5\right)}^4}{{\left(2x+3\right)}^{\mathrm{four}}};$ this answer could also be written $\log {\left(\frac{{\mathrm{ten}}^3\left(\mathrm{10}+5\right)}{\left(2x+3\right)}\right)}^{\mathrm{four}}.$

12 .

The pH increases past about 0.301.

14 .

$\frac{\ln 100}{\ln 5}\approx \frac{4.6051}{\mathrm{i.6094}}=2.861$

4.6 Exponential and Logarithmic Equations

4 .

The equation has no solution.

5 .

$\mathrm{10}=\frac{\ln 3}{\ln \left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}$

6 .

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

7 .

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

12 .

$\mathrm{10}=1$ or $x=-\mathrm{one}$

13 .

$t=703,800,000\times \frac{\ln (0.8)}{\ln (0.5)}\text{years}\approx 226,572,993\text{years}.$

four.7 Exponential and Logarithmic Models

one .

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

2 .

less than 230 years, 229.3157 to be exact

three .

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

6 .

Exponential. $y=2{\mathrm{due\; east}}^{0.5x}.$

7 .

$y=3{\mathrm{east}}^{\left(\ln 0.5\right)x}$

iv.viii Plumbing fixtures Exponential Models to Data

ane .

1. ⓐ The exponential regression model that fits these data is $y=522.88585984{\left(\mathrm{ane.19645256}\right)}^{\mathrm{10}}.$
2. ⓑ If spending continues at this rate, the graduate's credit card debt will be $4,499.38 after one twelvemonth.

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, virtually 171,000 games volition be sold in the year 2015.

3 .

1. ⓐ The logistic regression model that fits these data is $y=\frac{25.65665979}{1+6.113686306e^{-0.3852149008x}}.$
2. ⓑ If the population continues to grow at this rate, at that place will be about $\text{25,634}$ seals in 2020.
3. ⓒ To the nearest whole number, the conveying capacity is 25,657.

4.1 Section Exercises

ane .

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

three .

When interest is compounded, the pct of involvement earned to main ends upwards being greater than the annual percent rate for the investment account. Thus, the almanac pct charge per unit does not necessarily represent to the real interest earned, which is the very definition of nominal.

5 .

exponential; the population decreases past a proportional rate. .

7 .

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

9 .

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

eleven .

Afterward $t=20$ years, wood A will accept $43$ more trees than forest B.

13 .

Answers will vary. Sample response: For a number of years, the population of wood A volition increasingly exceed forest B, but because woods B actually grows at a faster rate, the population will somewhen become larger than wood 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.

fifteen .

exponential growth; The growth factor, $\mathrm{one.06},$ is greater than $\mathrm{ane.}$

17 .

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

19 .

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

21 .

$f(\mathrm{10})={\left(\frac{1}{6}\right)}^{-\frac{3}{5}}{\left(\frac{1}{\mathrm{six}}\right)}^{\frac{\mathrm{10}}{5}}\approx \mathrm{ii.93}{\left(0.699\right)}^x$

31 .

$\$\mathrm{xiii},268.58$

33 .

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

39 .

continuous growth; the growth charge per unit is greater than $0.$

41 .

continuous disuse; the growth charge per unit is less than $0.$

47 .

$f(-1)\approx -0.2707$

49 .

$f(3)\approx 483.8146$

53 .

$y\approx 18\cdot {\mathrm{one.025}}^{\mathrm{ten}}$

55 .

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

57 .

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

59 .

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

63 .

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

65 .

$\$35,838.76$

67 .

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

four.2 Section Exercises

1 .

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 role tells u.s.a. the limit of the function's values equally the independent variable gets either extremely big or extremely small.

3 .

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

v .

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

7 .

$\mathrm{grand}(\mathrm{10})=\mathrm{ii}{\left(\frac{\mathrm{one}}{\mathrm{four}}\right)}^x;$ y-intercept: $\left(0,\text{2}\right);$ Domain: all real numbers; Range: all real numbers greater than $0.$

9 .

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

27 .

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

29 .

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

31 .

Equally $x\to \infty$ , $f\left(x\right)\to \mathrm{ii}$ ;
Every bit $\mathrm{10}\to -\infty$ , $f\left(\mathrm{10}\right)\to \infty$

33 .

$f\left(x\right)={\mathrm{four}}^x-3$

35 .

$f(\mathrm{ten})=4^{x-\mathrm{five}}$

37 .

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

39 .

$y=-2^x+3$

41 .

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

43 .

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

51 .

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

53 .

The graphs of $\mathrm{thou}(x)$ and $h(\mathrm{10})$ are the aforementioned and are a horizontal shift to the correct of the graph of $f(x);$ For whatsoever real number north, existent number $b>0,$ and office $f(\mathrm{10})=b^x,$ the graph of $\left(\frac{\mathrm{one}}{b^{\mathrm{north}}}\right)b^x$ is the horizontal shift $f(x-\mathrm{due\; north}).$

4.3 Section Exercises

1 .

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 aforementioned value. The exponent, $y,$ in the expression $b^y$ tin can also be written as the logarithm, ${\log }_{b}x,$ and the value of $\mathrm{10}$ is the event of raising $b$ to the power of $y.$

3 .

Since the equation of a logarithm is equivalent to an exponential equation, the logarithm tin can be converted to the exponential equation $b^y=x,$ and then backdrop of exponents tin 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 $\mathrm{east}.$ Rather than notating the natural logarithm every bit ${\log }_e\left(\mathrm{ten}\right),$ the notation used is $\ln \left(\mathrm{10}\right).$

17 .

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

19 .

${\log }_{19}y=\mathrm{ten}$

21 .

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

23 .

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

27 .

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

29 .

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

31 .

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

33 .

$x=6^{-\mathrm{three}}=\frac{\mathrm{i}}{216}$

59 .

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

61 .

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

63 .

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

four.4 Section Exercises

1 .

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

3 .

Shifting the role correct or left and reflecting the function about the y-axis will bear upon its domain.

5 .

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

7 .

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

9 .

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

11 .

Domain: $\left(5,\infty \right);$ Vertical asymptote: $\mathrm{10}=5$

thirteen .

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

15 .

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

17 .

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

19 .

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

21 .

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

23 .

Domain: $\left(-\infty ,0\right);$ Range: $\left(-\infty ,\infty \right);$ Vertical asymptote: $x=0;$ 10-intercept: $\left(-e^2,0\right);$ y-intercept: DNE

25 .

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

47 .

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

49 .

$f(x)=\mathrm{three}{\log }_4(x+2)$

57 .

The graphs of $f(x)={\log }_{\frac{1}{\mathrm{ii}}}\left(x\right)$ and $g(\mathrm{ten})=-{\log }_2\left(x\right)$ appear to be the same; Conjecture: for any positive base $b\ne 1,$ ${\log }_b\left(\mathrm{ten}\right)=-{\log }_{\frac{\mathrm{ane}}{b}}\left(x\right).$

59 .

Recall that the argument of a logarithmic part must be positive, and so we determine where $\frac{\mathrm{ten}+2}{\mathrm{10}-4}>0$ . From the graph of the function $f\left(x\right)=\frac{\mathrm{ten}+2}{x-\mathrm{iv}},$ notation that the graph lies higher up 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 $\left(-\infty ,-2\right)\cup \left(4,\infty \right).$

4.5 Section Exercises

1 .

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

3 .

${\log }_b\left(2\right)+{\log }_b\left(7\right)+{\log }_b\left(\mathrm{10}\right)+{\log }_b\left(y\right)$

5 .

${\log }_b\left(13\right)-{\log }_b\left(17\right)$

thirteen .

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

15 .

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

17 .

$\frac{3}{2}\log (\mathrm{10})-2\log (y)$

nineteen .

$\frac{8}{\mathrm{iii}}\log (x)+\frac{14}{\mathrm{iii}}\log (y)$

21 .

$\ln (\mathrm{ii}{\mathrm{10}}^7)$

23 .

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

25 .

${\log }_7\left(15\right)=\frac{\ln \left(15\right)}{\ln \left(7\right)}$

27 .

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

29 .

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

39 .

$\mathrm{ten}=\mathrm{iv};$ By the quotient dominion: ${\log }_6\left(x+\mathrm{ii}\right)-{\log }_{\mathrm{half\; dozen}}\left(x-3\right)={\log }_6\left(\frac{\mathrm{ten}+\mathrm{ii}}{x-\mathrm{three}}\right)=1.$

Rewriting every bit an exponential equation and solving for $x:$

$$\begin{array}{ll}{\mathrm{vi}}^1\hfill & =\frac{x+\mathrm{two}}{x-\mathrm{iii}}\hfill \\ 0\hfill & =\frac{\mathrm{ten}+2}{x-3}-\mathrm{vi}\hfill \\ 0\hfill & =\frac{x+2}{\mathrm{ten}-\mathrm{iii}}-\frac{6\left(x-3\right)}{\left(x-\mathrm{three}\right)}\hfill \\ 0\hfill & =\frac{\mathrm{10}+2-\mathrm{vi}\mathrm{10}+18}{\mathrm{10}-3}\hfill \\ 0\hfill & =\frac{x-4}{x-3}\hfill \\ \text{}x\hfill & =\mathrm{four}\hfill \end{array}$$

Checking, we find that ${\log }_6\left(4+2\right)-{\log }_6\left(4-\mathrm{three}\right)={\log }_6\left(6\right)-{\log }_6\left(\mathrm{ane}\right)$ is defined, so $x=\mathrm{iv.}$

41 .

Let $b$ and $n$ be positive integers greater than $\mathrm{i.}$ Then, past the change-of-base formula, ${\log }_b\left(\mathrm{northward}\right)=\frac{{\log }_{\mathrm{north}}\left(\mathrm{northward}\right)}{{\log }_n\left(b\right)}=\frac{\mathrm{one}}{{\log }_n\left(b\right)}.$

four.vi Section Exercises

1 .

Make up one's mind 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 and then that each side uses the aforementioned base of operations, then apply the logarithm to each side and use properties of logarithms to solve.

3 .

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

xv .

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

17 .

$\mathrm{1000}=-\frac{\ln \left(38\right)}{3}$

nineteen .

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

23 .

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

29 .

${\mathrm{x}}^{-\mathrm{two}}=\frac{1}{100}$

51 .

$x=9$

53 .

$\mathrm{ten}=\frac{{\mathrm{due\; east}}^{\mathrm{two}}}{3}\approx \mathrm{2.five}$

55 .

$\mathrm{ten}=-\mathrm{v}$

57 .

$\mathrm{ten}=\frac{e+\mathrm{x}}{4}\approx \mathrm{iii.ii}$

59 .

No solution

61 .

$\mathrm{10}=\frac{11}{5}\approx \mathrm{ii.two}$

63 .

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

65 .

about $\$27,710.24$

67 .

about 5 years

69 .

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

71 .

$x=\frac{\log \left(38\right)+5\log \left(3\right)}{\mathrm{iv}\log \left(3\right)}\approx \mathrm{two.078}$

75 .

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

79 .

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

81 .

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

4.7 Section Exercises

1 .

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

three .

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

5 .

An social club of magnitude is the nearest power of ten by which a quantity exponentially grows. It is also an estimate position on a logarithmic calibration; Sample response: Orders of magnitude are useful when making comparisons between numbers that differ past a cracking 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 most ${\mathrm{ten}}^{\text{two}}$ times, or 2 orders of magnitude greater, than the mass of Globe.

seven .

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

11 .

exponential; $f(\mathrm{10})={\mathrm{1.two}}^{\mathrm{ten}}$

13 .

logarithmic

fifteen .

logarithmic

23 .

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

25 .

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

27 .

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

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

29 .

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

33 .

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

35 .

$r\approx -0.0667,$ Then the hourly decay rate is about $\mathrm{half-dozen.67}\%$

37 .

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

39 .

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

43 .

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

45 .

near $\text{113}$ minutes

47 .

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

49 .

MMS magnitude: $5.82$

four.8 Section Exercises

1 .

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

iii .

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

v .

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: $\left(0,15\right)$

19 .

about $\mathrm{6.viii}$ months.

27 .

$f\left(x\right)=776.682{\left(\mathrm{i.426}\right)}^{\mathrm{10}}$

33 .

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

35 .

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

37 .

$y=\mathrm{v.063}+1.934\text{log}\left(\mathrm{10}\right)$

43 .

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

45 .

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

47 .

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

55 .

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

57 .

$f\left(\mathrm{10}\right)=1.034341{\left(1.281204\right)}^x$ ; $\mathrm{1000}\left(\mathrm{10}\right)=\mathrm{four.035510}$ ; the regression curves are symmetrical almost $y=x$ , so information technology appears that they are inverse functions.

59 .

$f^{-1}\left(\mathrm{10}\right)=\frac{\text{ln}(a)-\text{ln}(\frac{c}{x}-\mathrm{one})}{b}$

Review Exercises

one .

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

three .

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

5 .

$\mathrm{\$}42,888.18$

seven .

continuous decay; the growth rate is negative.

nine .

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

xi .

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

13 .

${17}^x=4913$

15 .

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

17 .

$x={64}^{\frac{\mathrm{ane}}{\mathrm{iii}}}=4$

nineteen .

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

21 .

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

25 .

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

27 .

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

29 .

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

31 .

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

33 .

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

35 .

${\log }_7\left(\frac{v^3{w}^6}{\sqrt[3]{u}}\right)$

37 .

$x=\frac{\frac{\log \left(125\right)}{\log \left(5\right)}+17}{12}=\frac{5}{\mathrm{iii}}$

45 .

$\mathrm{10}=\ln \left(11\right)$

51 .

about $5.45$ years

53 .

$f^{-1}\left(\mathrm{ten}\right)=\sqrt[\mathrm{iii}]{{\mathrm{ii}}^{4x}-1}$

55 .

$f(t)=300{\left(0.83\right)}^t;$
$f(24)\approx 3.43g$

61 .

exponential

63 .

$y=4{\left(\mathrm{0.ii}\right)}^x;$ $y=4e^{\text{-1.609438}x}$

67 .

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

Practice Test

5 .

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

seven .

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

9 .

$x={\left(\frac{1}{\mathrm{vii}}\right)}^{\mathrm{two}}=\frac{1}{49}$

11 .

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

thirteen .

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

15 .

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

17 .

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

19 .

$x=\frac{\frac{\ln \left(1000\right)}{\ln \left(16\right)}+5}{3}\approx 2.497$

21 .

$a=\frac{\ln \left(\mathrm{four}\right)+\mathrm{eight}}{10}$

29 .

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

31 .

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

33 .

logarithmic

35 .

exponential; $y=15.10062{\left(1.24621\right)}^x$

37 .

logistic; $y=\frac{18.41659}{1+\mathrm{vii.54644}{\mathrm{eastward}}^{-0.68375x}}$

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

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