Chapter 6 Test Algebra 1

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6.one Exponential Functions

ane.

$g(\mathrm{10})={0.875}^{\mathrm{ten}}$ and $j(\mathrm{ten})={\mathrm{1095.vi}}^{-2\mathrm{ten}}$ represent exponential functions.

3.

About $1.548$ billion people; by the year 2031, India'due south population will exceed Mainland china'due south by about 0.001 billion, or one million people.

four.

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

5.

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

6.

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

7.

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

10.

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

12.

iii.77E-26 (This is calculator notation for the number written as $\mathrm{three.77}\times {\mathrm{ten}}^{-26}$ in scientific note. While the output of an exponential function is never nil, this number is so shut to zero that for all practical purposes we can take nothing as the answer.)

6.2 Graphs of Exponential Functions

1.

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

two.

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 $y=0.$

six.

$f(x)=-\frac{1}{\mathrm{iii}}{e}^x-2;$ the domain is $\left(-\infty ,\infty \right);$ the range is $\left(-\infty ,\mathrm{-ii}\right);$ the horizontal asymptote is $y=\mathrm{-ii.}$

six.3 Logarithmic Functions

one.

1. ⓐ ${\log }_{\mathrm{ten}}\left(\mathrm{ane},000,000\right)=6$ is equivalent to ${10}^{\mathrm{half\; dozen}}=\mathrm{one},000,000$
2. ⓑ ${\log }_{\mathrm{five}}\left(25\right)=\mathrm{ii}$ is equivalent to ${\mathrm{v}}^{\mathrm{ii}}=25$

ii.

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

3.

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

iv.

${\log }_2\left(\frac{1}{32}\right)=-\mathrm{five}$

5.

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

six.

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

seven.

The difference in magnitudes was near $\mathrm{3.929.}$

8.

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

6.iv Graphs of Logarithmic Functions

3.

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

four.

The domain is $\left(-4,\infty \right),$ 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 $x=0.$

six.

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 $\mathrm{ten}=2.$

8.

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

11.

$f(\mathrm{10})=2\ln (x+3)-\mathrm{one}$

6.5 Logarithmic Backdrop

ane.

${\log }_{b}2+{\log }_{b}2+{\log }_{b}2+{\log }_{b}k=3{\log }_{b}2+{\log }_{b}k$

2.

${\log }_3\left(\mathrm{ten}+3\right)-{\log }_3\left(x-1\right)-{\log }_3\left(\mathrm{10}-2\right)$

half-dozen.

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

8.

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

9.

$\log \left(\frac{3\cdot \mathrm{v}}{4\cdot 6}\right);$ can as well exist written $\log \left(\frac{5}{\mathrm{viii}}\right)$ by reducing the fraction to lowest terms.

10.

$\log \left(\frac{\mathrm{v}{\left(\mathrm{10}-\mathrm{ane}\right)}^3\sqrt{x}}{\left(7\mathrm{ten}-\mathrm{one}\right)}\right)$

11.

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

12.

The pH increases past well-nigh 0.301.

14.

$\frac{\ln 100}{\ln \mathrm{v}}\approx \frac{4.6051}{1.6094}=\mathrm{ii.861}$

six.half-dozen Exponential and Logarithmic Equations

4.

The equation has no solution.

5.

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

6.

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

seven.

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

thirteen.

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

6.vii Exponential and Logarithmic Models

1.

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

ii.

less than 230 years, 229.3157 to be exact

3.

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

half-dozen.

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

7.

$y=3e^{\left(\ln \mathrm{0.v}\right)x}$

6.viii Fitting Exponential Models to Information

1.

1. ⓐ The exponential regression model that fits these information is $y=522.88585984{\left(1.19645256\right)}^x.$
2. ⓑ If spending continues at this rate, the graduate'south credit carte debt will be $iv,499.38 afterwards one year.

2.

1. ⓐ The logarithmic regression model that fits these information is $y=141.91242949+10.45366573\ln (x)$
2. ⓑ If sales go along 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 $y=\frac{25.65665979}{1+6.113686306{\mathrm{eastward}}^{-0.3852149008x}}.$
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 carrying capacity is 25,657.

6.one Department Exercises

1.

Linear functions accept a abiding charge per unit of modify. Exponential functions increase based on a percentage of the original.

3.

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

five.

exponential; the population decreases by a proportional rate. .

7.

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

9.

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

11.

After $t=\mathrm{twenty}$ years, woods A volition have $43$ more trees than forest B.

13.

Answers will vary. Sample response: For a number of years, the population of forest A volition increasingly exceed wood B, but considering 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 agree. Some factors that might influence the long-term validity of the exponential growth model are drought, an epidemic that culls the population, and other ecology 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.

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

21.

$f(x)={\left(\frac{\mathrm{ane}}{6}\right)}^{-\frac{\mathrm{three}}{5}}{\left(\frac{1}{\mathrm{six}}\right)}^{\frac{\mathrm{ten}}{\mathrm{five}}}\approx 2.93{\left(0.699\right)}^x$

31.

$\$\mathrm{xiii},268.58$

33.

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

39.

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

41.

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

47.

$f(-\mathrm{ane})\approx -0.2707$

49.

$f(\mathrm{three})\approx 483.8146$

53.

$y\approx 18\cdot {1.025}^x$

55.

$y\approx \mathrm{0.ii}\cdot {1.95}^{\mathrm{ten}}$

57.

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

59.

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

63.

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

65.

$\$35,838.76$

67.

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

6.2 Section Exercises

i.

An asymptote is a line that the graph of a function approaches, as $\mathrm{ten}$ either increases or decreases without bound. The horizontal asymptote of an exponential office tells the states the limit of the function'due south values equally the independent variable gets either extremely big or extremely small.

iii.

$\mathrm{thousand}(\mathrm{10})=4{\left(3\right)}^{-\mathrm{ten}};$ y-intercept: $(0,4);$ Domain: all real numbers; Range: all real numbers greater than $0.$

5.

$\mathrm{1000}(x)=-{10}^x+7;$ y-intercept: $\left(0,6\right);$ Domain: all real numbers; Range: all real numbers less than $\mathrm{vii.}$

7.

$g(x)=2{\left(\frac{1}{\mathrm{four}}\right)}^x;$ y-intercept: $\left(0,\text{2}\right);$ Domain: all real numbers; Range: all real numbers greater than $0.$

9.

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

27.

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

29.

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

31.

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

33.

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

35.

$f(x)=4^{x-5}$

37.

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

39.

$y=-{\mathrm{two}}^x+3$

41.

$y=-2{\left(3\right)}^x+7$

43.

$\mathrm{yard}(\mathrm{vi})=800+\frac{1}{3}\approx 800.3333$

51.

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

53.

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

half dozen.3 Section Exercises

1.

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 equally the logarithm, ${\log }_{b}x,$ and the value of $x$ is the result of raising $b$ to the power of $y.$

3.

Since the equation of a logarithm is equivalent to an exponential equation, the logarithm tin be converted to the exponential equation $b^y=x,$ and then properties of exponents tin 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 ${\log }_{\mathrm{due\; east}}\left(x\right),$ the annotation used is $\ln \left(x\right).$

17.

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

19.

${\log }_{19}y=\mathrm{10}$

21.

${\log }_{\mathrm{due\; north}}\left(103\right)=4$

23.

${\log }_y\left(\frac{39}{100}\right)=\mathrm{10}$

27.

$\mathrm{ten}={\mathrm{two}}^{-3}=\frac{1}{8}$

29.

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

31.

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

33.

$\mathrm{ten}=6^{-\mathrm{iii}}=\frac{1}{216}$

59.

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

61.

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

63.

No; $\ln \left(\mathrm{i}\right)=0,$ then $\frac{\ln \left({\mathrm{due\; east}}^{\mathrm{one.725}}\right)}{\ln \left(1\right)}$ is undefined.

six.iv Section 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 function, at that place is a corresponding point $(b,a)$ on the graph of its changed exponential part.

3.

Shifting the part right or left and reflecting the function most the y-axis will bear upon its domain.

5.

No. A horizontal asymptote would advise a limit on the range, and the range of whatever logarithmic function in general class is all real numbers.

7.

Domain: $\left(-\infty ,\frac{1}{2}\right);$ 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: $\mathrm{10}=\mathrm{five}$

13.

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

15.

Domain: $\left(-3,\infty \right);$ Vertical asymptote: $\mathrm{10}=-3$

17.

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

19.

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

21.

Domain: $\left(1,\infty \right);$ Range: $\left(-\infty ,\infty \right);$ Vertical asymptote: $\mathrm{10}=1;$ x-intercept: $\left(\frac{5}{\mathrm{four}},0\right);$ y-intercept: DNE

23.

Domain: $\left(-\infty ,0\right);$ Range: $\left(-\infty ,\infty \right);$ Vertical asymptote: $x=0;$ x-intercept: $\left(-{\mathrm{due\; east}}^2,0\right);$ y-intercept: DNE

25.

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

47.

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

49.

$f(x)=\mathrm{three}{\log }_4(\mathrm{10}+\mathrm{two})$

57.

The graphs of $f(x)={\log }_{\frac{\mathrm{one}}{2}}\left(\mathrm{10}\right)$ and $g(\mathrm{10})=-{\log }_2\left(x\right)$ appear to exist the same; Conjecture: for any positive base $b\ne 1,$ ${\log }_b\left(x\right)=-{\log }_{\frac{\mathrm{ane}}{b}}\left(x\right).$

59.

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

6.v Section Exercises

1.

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

3.

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

5.

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

xiii.

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

15.

$15\log (\mathrm{10})+\mathrm{xiii}\log (y)-\mathrm{nineteen}\log (z)$

17.

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

xix.

$\frac{8}{3}\log (x)+\frac{14}{3}\log (y)$

21.

$\ln (2{\mathrm{ten}}^7)$

23.

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

25.

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

27.

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

29.

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

39.

$x=4;$ By the quotient rule: ${\log }_{\mathrm{half\; dozen}}\left(\mathrm{ten}+\mathrm{two}\right)-{\log }_{\mathrm{vi}}\left(x-3\right)={\log }_6\left(\frac{\mathrm{ten}+\mathrm{ii}}{x-3}\right)=\mathrm{ane.}$

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

$$\begin{array}{ll}{\mathrm{six}}^{\mathrm{i}}\hfill & =\frac{x+2}{\mathrm{10}-\mathrm{iii}}\hfill \\ 0\hfill & =\frac{x+2}{x-3}-6\hfill \\ 0\hfill & =\frac{\mathrm{10}+2}{x-3}-\frac{6\left(x-\mathrm{iii}\right)}{\left(\mathrm{10}-3\right)}\hfill \\ 0\hfill & =\frac{\mathrm{10}+2-6\mathrm{ten}+\mathrm{xviii}}{\mathrm{ten}-3}\hfill \\ 0\hfill & =\frac{x-4}{x-3}\hfill \\ \text{}x\hfill & =\mathrm{iv}\hfill \end{array}$$

Checking, we detect that ${\log }_6\left(4+2\right)-{\log }_6\left(4-3\right)={\log }_6\left(\mathrm{vi}\right)-{\log }_{\mathrm{six}}\left(1\right)$ is defined, so $x=4.$

41.

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

six.vi Section Exercises

i.

Determine first if the equation tin be rewritten so that each side uses the same base of operations. If then, 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 belongings can be used if both sides of the equation can be rewritten as a single logarithm with the aforementioned base. If so, the arguments tin can be prepare equal to each other, and the resulting equation tin be solved algebraically. The one-to-one belongings cannot be used when each side of the equation cannot exist rewritten equally a unmarried logarithm with the aforementioned base.

xv.

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

17.

$k=-\frac{\ln \left(38\right)}{\mathrm{three}}$

19.

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

23.

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

29.

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

51.

$x=\mathrm{ix}$

53.

$x=\frac{e^2}{3}\approx 2.5$

55.

$x=-\mathrm{five}$

57.

$x=\frac{\mathrm{due\; east}+\mathrm{ten}}{4}\approx 3.2$

59.

No solution

61.

$\mathrm{ten}=\frac{\mathrm{xi}}{\mathrm{v}}\approx 2.2$

63.

$x=\frac{101}{\mathrm{xi}}\approx \mathrm{ix.2}$

65.

about $\$27,710.24$

67.

most 5 years

69.

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

71.

$x=\frac{\log \left(38\right)+\mathrm{v}\log \left(3\right)}{4\log \left(3\right)}\approx 2.078$

75.

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

79.

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

81.

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

6.7 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 corporeality of time it takes for one-half of the initial amount of that substance or quantity to decay.

three.

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 x 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 betwixt numbers that differ by a great amount. For case, the mass of Saturn is 95 times greater than the mass of Earth. This is the same every bit saying that the mass of Saturn is almost ${10}^{\text{two}}$ times, or two orders of magnitude greater, than the mass of Earth.

7.

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

11.

exponential; $f(x)={1.2}^x$

thirteen.

logarithmic

xv.

logarithmic

23.

$4$ half-lives; $\mathrm{eight.xviii}$ minutes

25.

$\begin{array}{l}\mathrm{Chiliad}=\frac{2}{\mathrm{iii}}\log \left(\frac{\mathrm{Due\; south}}{S_0}\right)\hfill \\ \log \left(\frac{S}{S_0}\right)=\frac{3}{2}K\hfill \\ \frac{\mathrm{Due\; south}}{S_0}={10}^{\frac{3G}{2}}\hfill \\ \mathrm{South}=S_0{10}^{\frac{\mathrm{iii}M}{\mathrm{two}}}\hfill \end{array}$

27.

Let $y=b^{\mathrm{10}}$ for some non-negative real number $b$ such that $b\ne 1.$ And then,

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

29.

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

33.

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

35.

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

37.

$f(t)=1350{\mathrm{eastward}}^{(0.03466t)};$ afterwards 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.

about $\text{113}$ minutes

47.

$\log \left(x\right)=1.5;x\approx 31.623$

49.

MMS magnitude: $\mathrm{v.82}$

6.8 Section Exercises

1.

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

three.

Regression analysis is the process of finding an equation that best fits a given prepare of data points. To perform a regression analysis on a graphing utility, showtime list the given points using the STAT then EDIT menu. Side by side graph the scatter plot using the STAT PLOT feature. The shape of the data points on the scatter graph tin can help make up one's mind which regression feature to utilise. In one case this is determined, select the appropriate regression analysis control from the STAT and then CALC bill of fare.

five.

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

xi.

$P(0)=22$ ; 175

fifteen.

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

19.

about $6.8$ months.

27.

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

33.

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

35.

When $f\left(x\right)=250,\mathrm{10}\approx \mathrm{three.half\; dozen}$

37.

$y=5.063+1.934\text{log}\left(x\right)$

43.

When $f\left(10\right)\approx \mathrm{2.three}$

45.

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

47.

$f(\mathrm{ten})=\frac{25.081}{1+\mathrm{three.182}{e}^{-0.545x}}$

55.

When $f\left(x\right)=68,x\approx \mathrm{iv.nine}$

57.

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

59.

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

Review Exercises

ane.

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

iii.

$y=0.25{\left(3\right)}^{\mathrm{10}}$

7.

continuous decay; the growth rate is negative.

9.

domain: all existent numbers; range: all real numbers strictly greater than nothing; y-intercept: (0, 3.5);

xi.

$g(x)=7{\left(\mathrm{half-dozen.v}\right)}^{-\mathrm{10}};$ y-intercept: $(0,\text{seven});$ Domain: all real numbers; Range: all real numbers greater than $0.$

13.

${17}^x=4913$

15.

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

17.

$x={64}^{\frac{1}{\mathrm{three}}}=4$

19.

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

21.

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

25.

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

27.

${\text{log}}_8\left(65\mathrm{ten}y\right)$

29.

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

31.

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

33.

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

35.

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

37.

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

45.

$x=\ln \left(11\right)$

51.

well-nigh $5.45$ years

53.

$f^{-1}\left(\mathrm{ten}\right)=\sqrt[\mathrm{three}]{{\mathrm{two}}^{4\mathrm{ten}}-\mathrm{i}}$

55.

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

61.

exponential

63.

$y=4{\left(\mathrm{0.ii}\right)}^x;$ $y=4e^{\text{-ane.609438}x}$

67.

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

Practice Test

v.

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

vii.

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

nine.

$x={\left(\frac{1}{7}\right)}^2=\frac{1}{49}$

11.

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

13.

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

fifteen.

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

17.

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

19.

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

21.

$a=\frac{\ln \left(\mathrm{iv}\right)+\mathrm{viii}}{\mathrm{ten}}$

29.

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

31.

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

33.

logarithmic

35.

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

37.

logistic; $y=\frac{18.41659}{\mathrm{one}+7.54644{\mathrm{east}}^{-0.68375x}}$

Chapter 6 Test Algebra 1,

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

Posted by: robinsonwitilly.blogspot.com

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