Chapter 6

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

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

exponential; the population decreases by a proportional rate. .

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

The forest represented by the function $B(t)=82{(1.029)}^t.$

After $t=20$ years, forest A will have $43$ more trees than forest B.

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

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

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

$f(x)=2000{(0.1)}^x$

$f(x)={\left(\frac{1}{6}\right)}^{-\frac{3}{5}}{\left(\frac{1}{6}\right)}^{\frac{x}{5}}\approx 2.93{\left(0.699\right)}^x$

Linear

Neither

Linear

$\$10,250$

$\$13,268.58$

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

$\$\mathrm{4,572.56}$

$4\%$

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

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

$\$669.42$

$f(-1)=-4$

$f(-1)\approx -0.2707$

$f(3)\approx 483.8146$

$y=3\cdot 5^x$

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

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

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

Let $f$ be the exponential decay function $f(x)=a\cdot {\left(\frac{1}{b}\right)}^x$ such that $b>1.$ Then for some number $n>0,$ $f(x)=a\cdot {\left(\frac{1}{b}\right)}^x=a{\left(b^{-1}\right)}^x=a{\left({\left(e^n\right)}^{-1}\right)}^x=a{\left(e^{-n}\right)}^x=a{\left(e\right)}^{-nx}.$

$47,622$ fox

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

$\$35,838.76$

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

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 us the limit of the function’s values as the independent variable gets either extremely large or extremely small.

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

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

$g(x)=2{\left(\frac{1}{4}\right)}^x;$ y-intercept: $\left(0,\text{2}\right);$ Domain: all real numbers; Range: all real numbers greater than $0.$

B

A

E

D

C

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

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

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

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

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

$y=-2^x+3$

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

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

$h(-7)=-58$

$x\approx -2.953$

$x\approx -0.222$

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

The graphs of $g(x)$ and $h(x)$ are the same and are a horizontal shift to the right of the graph of $f(x);$ For any 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 $f(x-n).$

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 result of raising $b$ to the power of $y.$

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 then properties of exponents can be applied to solve for $x.$

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 notation used is $\ln \left(x\right).$

$a^c=b$

$x^y=64$

${15}^b=a$

${13}^a=142$

$e^n=w$

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

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

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

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

$\text{ln}(h)=k$

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

$x=3^3=27$

$x=9^{\frac{1}{2}}=3$

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

$x=e^2$

$32$

$1.06$

$14.125$

$\frac{1}{2}$

$4$

$-\text{3}$

$-12$

$0$

$10$

$\text{2}.\text{7}0\text{8}$

$0.151$

No, the function has no defined value for $x=0.$ To verify, suppose $x=0$ is in the domain of the function $f(x)=\log (x).$ Then there is some number $n$ such that $n=\log (0).$ Rewriting as 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 $f(x)=\log (x).$

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

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

$2$

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

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

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

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

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

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

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

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

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

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

Domain: $\left(1,\infty \right);$ Range: $\left(-\infty ,\infty \right);$ Vertical asymptote: $x=1;$ x-intercept: $\left(\frac{5}{4},0\right);$ y-intercept: DNE

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

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

B

C

B

C

C

$f(x)={\log }_2(-(x-1))$

$f(x)=3{\log }_4(x+2)$

$x=2$

$x\approx \text{2}\text{.303}$

$x\approx -0.472$

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

Recall that the argument of a logarithmic function must be positive, so we determine where $\frac{x+2}{x-4}>0$ . From the graph of the function $f\left(x\right)=\frac{x+2}{x-4},$ 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 $\left(-\infty ,-2\right)\cup \left(4,\infty \right).$