6.E: Exponential and Logarithmic Functions (Exercises)
- Page ID
- 4760
6.1: Introduction to Exponential and Logarithmic Functions
\subsection{Exercises}
In Exercises \ref{expeqnfirst} - \ref{expeqnlast}, solve the equation analytically.
\begin{multicols}{3}
\begin{enumerate}
\item $2^{4x} = 8$ \label{expeqnfirst}
\item $3^{(x - 1)} = 27$
\item $5^{2x-1} = 125$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $4^{2x} = \frac{1}{2}$
\item $8^{x} = \frac{1}{128}$
\item $2^{(x^{3} - x)} = 1$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $3^{7x} = 81^{4-2x}$
\item $9 \cdot 3^{7x} = \left(\frac{1}{9}\right)^{2x}$
\item $3^{2x} = 5$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $5^{-x} = 2$
\item $5^{x} = -2$
\item $3^{(x - 1)} = 29$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $(1.005)^{12x} = 3$
\item $e^{-5730k} = \frac{1}{2}$
\item $2000e^{0.1t} = 4000$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $500\left(1-e^{2x}\right) = 250$
\item $70 + 90e^{-0.1t} = 75$
\item $30-6e^{-0.1x}=20$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $\dfrac{100e^{x}}{e^{x}+2}=50$
\item $\dfrac{5000}{1+2e^{-3t}}=2500$
\item $\dfrac{150}{1 + 29e^{-0.8t}} = 75$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $25\left(\frac{4}{5}\right)^{x} = 10$
\item $e^{2x} = 2e^{x}$
\item $7e^{2x} = 28e^{-6x}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $3^{(x - 1)} = 2^{x}$
\item $3^{(x - 1)} = \left(\frac{1}{2}\right)^{(x + 5)}$
\item $7^{3+7x} = 3^{4-2x}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $e^{2x} - 3e^{x}-10=0$ %Ans $x=\ln(5)$
\item $e^{2x} = e^{x}+6$ %Ans $x=\ln(2)$
\item $4^{x} + 2^{x} = 12$ %Ans $x=\frac{\ln(3)}{\ln(2)}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $e^{x}-3e^{-x}=2$ %Ans $x=\ln(3)$
\item $e^{x}+15e^{-x}=8$ %Ans $x=\ln(2)$, $\ln(5)$
\item $3^{x}+25\cdot3^{-x}=10$ %Ans $x=\frac{\ln(5)}{\ln(3)}$
\label{expeqnlast}
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\end{enumerate}
\end{multicols}
In Exercises \ref{expineqfirst} - \ref{expineqlast}, solve the inequality analytically.
\begin{multicols}{2}
\begin{enumerate}
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\item $e^{x} > 53$ \label{expineqfirst}
\item $1000\left(1.005\right)^{12t} \geq 3000$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $2^{(x^{3} - x)} < 1$
\item $25\left(\frac{4}{5}\right)^{x} \geq 10$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $\dfrac{150}{1 + 29e^{-0.8t}} \leq 130$
\item $\vphantom{\dfrac{150}{1 + 29e^{-0.8t}}} 70 + 90e^{-0.1t} \leq 75$ \label{expineqlast}
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\end{enumerate}
\end{multicols}
In Exercises \ref{calcexpineqfirst} - \ref{calcexpineqlast}, use your calculator to help you solve the equation or inequality.
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\begin{enumerate}
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\item $2^{x} = x^2$ \label{calcexpineqfirst}
\item $e^{x} = \ln(x) + 5$
\item $e^{\sqrt{x}} = x + 1$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $e^{-x} - xe^{-x} \geq 0$
\item $3^{(x - 1)} < 2^{x}$
\item $e^{x} < x^{3} - x$ \label{calcexpineqlast}
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\end{enumerate}
\end{multicols}
\begin{enumerate}
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\item \label{onetoonelogexercise} Since $f(x) = \ln(x)$ is a strictly increasing function, if $0 < a < b$ then $\ln(a) < \ln(b)$. Use this fact to solve the inequality $e^{(3x - 1)} > 6$ without a sign diagram. Use this technique to solve the inequalities in Exercises \ref{expineqfirst} - \ref{expineqlast}. (NOTE: Isolate the exponential function first!)
\item \label{hyperbolicsine} Compute the inverse of $f(x) = \dfrac{e^{x} - e^{-x}}{2}$. State the domain and range of both $f$ and $f^{-1}$.
\item In Example \ref{expfracinverse}, we found that the inverse of $f(x) = \dfrac{5e^{x}}{e^{x}+1}$ was $f^{-1}(x) = \ln\left(\dfrac{x}{5-x}\right)$ but we left a few loose ends for you to tie up.
\begin{enumerate}
\item Show that $\left(f^{-1} \circ f\right)(x) = x$ for all $x$ in the domain of $f$ and that $\left(f \circ f^{-1}\right)(x) = x$ for all $x$ in the domain of $f^{-1}$.
\item Find the range of $f$ by finding the domain of $f^{-1}$.
\item Let $g(x) = \dfrac{5x}{x+1}$ and $h(x) = e^{x}$. Show that $f = g \circ h$ and that $(g \circ h)^{-1} = h^{-1} \circ g^{-1}$.
(We know this is true in general by Exercise \ref{fcircginverse} in Section \ref{InverseFunctions}, but it's nice to see a specific example of the property.)
\end{enumerate}
\item With the help of your classmates, solve the inequality $e^{x} > x^{n}$ for a variety of natural numbers $n$. What might you conjecture about the ``speed'' at which $f(x) = e^{x}$ grows versus any polynomial?
\end{enumerate}
\newpage
\subsection{Answers}
\begin{multicols}{3}
\begin{enumerate}
\item $x = \frac{3}{4}$
\item $x = 4$
\item $x=2$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $x = -\frac{1}{4}$
\item $x = -\frac{7}{3}$
\item $x = -1, \, 0, \, 1$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $x = \frac{16}{15}$
\item $x=-\frac{2}{11}$
\item $x = \frac{\ln(5)}{2\ln(3)}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $x = -\frac{\ln(2)}{\ln(5)}$
\item No solution.
\item $x = \frac{\ln(29) + \ln(3)}{\ln(3)}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $x = \frac{\ln(3)}{12\ln(1.005)}$
\item $k = \frac{\ln\left(\frac{1}{2}\right)}{-5730} = \frac{1}{5730} \ln(2)$
\item $t=\frac{\ln(2)}{0.1} = 10\ln(2)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $x=\frac{1}{2}\ln\left(\frac{1}{2}\right) = -\frac{1}{2}\ln(2)$
\item $t = \frac{\ln\left(\frac{1}{18}\right)}{-0.1} =10 \ln(18)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $x=-10\ln\left(\frac{5}{3}\right) = 10\ln\left(\frac{3}{5}\right)$
\item$x=\ln(2)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $t=\frac{1}{3}\ln(2)$
\item $t = \frac{\ln\left(\frac{1}{29}\right)}{-0.8} = \frac{5}{4}\ln(29)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $x = \frac{\ln\left(\frac{2}{5}\right)}{\ln\left(\frac{4}{5}\right)} = \frac{\ln(2)-\ln(5)}{\ln(4) - \ln(5)}$
\item $x = \ln(2)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $x = -\frac{1}{8} \ln\left(\frac{1}{4} \right) = \frac{1}{4}\ln(2)$
\item $x = \frac{\ln(3)}{\ln(3) - \ln(2)}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $x = \frac{\ln(3) + 5\ln\left(\frac{1}{2}\right)}{\ln(3) - \ln\left(\frac{1}{2}\right)} = \frac{\ln(3)-5\ln(2)}{\ln(3)+\ln(2)}$
\item $x = \frac{4 \ln(3) - 3 \ln(7)}{7 \ln(7) + 2 \ln(3)}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $x=\ln(5)$
\item $x=\ln(3)$
\item $x=\frac{\ln(3)}{\ln(2)}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $x=\ln(3)$
\item $x=\ln(3)$, $\ln(5)$
\item $x=\frac{\ln(5)}{\ln(3)}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $(\ln(53), \infty)$
\item $\left[\frac{\ln(3)}{12\ln(1.005)}, \infty\right)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $(-\infty, -1) \cup (0, 1)$
\item $\left(-\infty, \frac{\ln\left(\frac{2}{5}\right)}{\ln\left(\frac{4}{5}\right)} \right] = \left(-\infty, \frac{\ln(2)-\ln(5)}{\ln(4)-\ln(5)} \right]$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $\left(-\infty, \frac{\ln\left(\frac{2}{377}\right)}{-0.8} \right] = \left(-\infty, \frac{5}{4}\ln\left(\frac{377}{2}\right) \right]$
\item $\left[\frac{\ln\left(\frac{1}{18}\right)}{-0.1}, \infty\right) = [10\ln(18), \infty)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $x \approx -0.76666, \, x = 2, \, x = 4$
\item $x \approx 0.01866, \, x \approx 1.7115$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $x = 0$
\item $(-\infty, 1]$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $\approx (-\infty, 2.7095)$
\item $\approx (2.3217, 4.3717)$
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\end{enumerate}
\end{multicols}
\begin{enumerate}
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\item $x > \frac{1}{3}(\ln(6) + 1)$
\item $f^{-1} = \ln\left(x + \sqrt{x^{2} + 1}\right)$. Both $f$ and $f^{-1}$ have domain $(-\infty, \infty)$ and range $(-\infty, \infty)$.
\end{enumerate}
\closegraphsfile
6.2: Properties of Logarithms
\subsection{Exercises}
In Exercises \ref{expandlogfirst} - \ref{expandloglast}, expand the given logarithm and simplify. Assume when necessary that all quantities represent positive real numbers.
\begin{multicols}{3}
\begin{enumerate}
\item $\ln(x^{3}y^{2})$ \vphantom{$\log_{2}\left(\dfrac{128}{x^{2} + 4}\right)$} \label{expandlogfirst}
\item $\log_{2}\left(\dfrac{128}{x^{2} + 4}\right)$
\item $\log_{5}\left(\dfrac{z}{25}\right)^{3}$ \vphantom{$\log_{2}\left(\dfrac{128}{x^{2} + 4}\right)$}
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $\log(1.23 \times 10^{37})$ \vphantom{$\ln\left(\dfrac{\sqrt{z}}{xy}\right)$}
\item $\ln\left(\dfrac{\sqrt{z}}{xy}\right)$
\item $\log_{5} \left(x^2 - 25 \right)$ \vphantom{$\ln\left(\dfrac{\sqrt{z}}{xy}\right)$}
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $\log_{\sqrt{2}} \left(4x^3\right)$
\item $\log_{\frac{1}{3}}(9x(y^{3} - 8))$
\item $\log\left(1000x^3y^5\right)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $\log_{3} \left(\dfrac{x^2}{81y^4}\right)$
\item $\ln\left(\sqrt[4]{\dfrac{xy}{ez}}\right)$
\item $\log_{6} \left(\dfrac{216}{x^3y}\right)^4$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $\log\left(\dfrac{100x\sqrt{y}}{\sqrt[3]{10}}\right)$ \vphantom{$\log_{\frac{1}{2}}\left(\dfrac{4\sqrt[3]{x^2}}{y\sqrt{z}}\right)$}
\item $\log_{\frac{1}{2}}\left(\dfrac{4\sqrt[3]{x^2}}{y\sqrt{z}}\right)$
\item $\ln \left(\dfrac{\sqrt[3]{x}}{10 \sqrt{yz}}\right)$ \vphantom{$\log_{\frac{1}{2}}\left(\dfrac{4\sqrt[3]{x^2}}{y\sqrt{z}}\right)$} \label{expandloglast}
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\end{enumerate}
\end{multicols}
In Exercises \ref{combinelogfirst} - \ref{combineloglast}, use the properties of logarithms to write the expression as a single logarithm.
\begin{multicols}{2}
\begin{enumerate}
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\item $4\ln(x) + 2\ln(y)$ \label{combinelogfirst}
\item $\log_{2}(x) + \log_{2}(y) - \log_{2}(z)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $\log_{3}(x) - 2 \log_{3}(y)$
\item $\frac{1}{2}\log_{3}(x) - 2\log_{3}(y) - \log_{3}(z)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $2 \ln(x) -3 \ln(y) - 4\ln(z)$
\item $\log(x) - \frac{1}{3} \log(z) + \frac{1}{2} \log(y)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $-\frac{1}{3} \ln(x) - \frac{1}{3}\ln(y) + \frac{1}{3} \ln(z)$
\item $\log_{5}(x) - 3$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $3 - \log(x)$
\item $\log_{7}(x) + \log_{7}(x - 3) - 2$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $\ln(x) + \frac{1}{2}$
\item $\log_{2}(x) + \log_{4}(x)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $\log_{2}(x) + \log_{4}(x-1)$
\item $\log_{2}(x) + \log_{\frac{1}{2}}(x - 1)$ \label{combineloglast}
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\end{enumerate}
\end{multicols}
\pagebreak
In Exercises \ref{changeofbasefirst} - \ref{changeofbaselast}, use the appropriate change of base formula to convert the given expression to an expression with the indicated base.
\begin{multicols}{2}
\begin{enumerate}
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\item $7^{x - 1}$ to base $e$ \label{changeofbasefirst}
\item $\log_{3}(x + 2)$ to base 10
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $\left(\dfrac{2}{3}\right)^{x}$ to base $e$
\item $\log(x^{2} + 1)$ to base $e$ \vphantom{$\left(\dfrac{2}{3}\right)^{x}$}\label{changeofbaselast}
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\end{enumerate}
\end{multicols}
In Exercises \ref{changeofbaseapproxfirst} - \ref{changeofbaseapproxlast}, use the appropriate change of base formula to approximate the logarithm.
\begin{multicols}{3}
\begin{enumerate}
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\item $\log_{3}(12)$ \label{changeofbaseapproxfirst}
\item $\log_{5}(80)$
\item $\log_{6}(72)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $\log_{4}\left(\dfrac{1}{10}\right)$
\item $\log_{\frac{3}{5}}(1000)$ \vphantom{$\log_{4}\left(\dfrac{1}{10}\right)$}
\item $\log_{\frac{2}{3}}(50)$ \vphantom{$\log_{4}\left(\dfrac{1}{10}\right)$} \label{changeofbaseapproxlast}
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\end{enumerate}
\end{multicols}
\begin{enumerate}
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\item Compare and contrast the graphs of $y = \ln(x^{2})$ and $y = 2\ln(x)$.
\item Prove the Quotient Rule and Power Rule for Logarithms.
\item Give numerical examples to show that, in general,
\begin{enumerate}
\item $\log_{b}(x + y) \neq \log_{b}(x) + \log_{b}(y)$
\item $\log_{b}(x - y) \neq \log_{b}(x) - \log_{b}(y)$
\item $\log_{b}\left(\dfrac{x}{y}\right) \neq \dfrac{\log_{b}(x)}{\log_{b}(y)}$
\end{enumerate}
\item \label{HendersonHasselbalch} \index{Henderson-Hasselbalch Equation} The Henderson-Hasselbalch Equation: Suppose $HA$ represents a weak acid. Then we have a reversible chemical reaction
\[HA \rightleftharpoons H^{+} + A^{-}.\]
The acid disassociation constant, $K_{a}$, is given by
\[K_{\alpha} = \frac{[H^{+}][A^{-}]}{[HA]} = [H^{+}]\frac{[A^{-}]}{[HA]},\]
where the square brackets denote the concentrations just as they did in Exercise \ref{pHexercise} in Section \ref{IntroExpLogs}. The symbol p$K_{a}$ is defined similarly to pH in that p$K_{a} = -\log(K_{a})$. Using the definition of pH from Exercise \ref{pHexercise} and the properties of logarithms, derive the Henderson-Hasselbalch Equation which states
\[\mbox{pH} = \mbox{p}K_{a} + \log\dfrac{[A^{-}]}{[HA]}\]
\item Research the history of logarithms including the origin of the word `logarithm' itself. Why is the abbreviation of natural log `ln' and not `nl'?
\item There is a scene in the movie `Apollo 13' in which several people at Mission Control use slide rules to verify a computation. Was that scene accurate? Look for other pop culture references to logarithms and slide rules.
\end{enumerate}
\newpage
\subsection{Answers}
\begin{multicols}{2}
\begin{enumerate}
\item $3\ln(x) + 2\ln(y)$
\item $7 - \log_{2}(x^{2} + 4)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $3\log_{5}(z) - 6$
\item $\log(1.23) + 37$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $\frac{1}{2}\ln(z) - \ln(x) - \ln(y)$
\item $\log_{5}(x-5) + \log_{5}(x+5)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $3\log_{\sqrt{2}}(x) + 4$
\item \small$-2 + \log_{\frac{1}{3}}(x) + \log_{\frac{1}{3}}(y - 2) + \log_{\frac{1}{3}}(y^{2} + 2y + 4)$\normalsize
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $3 + 3\log(x) + 5 \log(y)$
\item $2\log_{3}(x) - 4 - 4\log_{3}(y)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\frac{1}{4} \ln(x) + \frac{1}{4} \ln(y) - \frac{1}{4} - \frac{1}{4} \ln(z)$
\item $12-12\log_{6}(x) - 4\log_{6}(y)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\frac{5}{3}+\log(x)+\frac{1}{2}\log(y)$
\item $-2+\frac{2}{3}\log_{\frac{1}{2}}(x)-\log_{\frac{1}{2}}(y)-\frac{1}{2}\log_{\frac{1}{2}}(z)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\frac{1}{3} \ln(x) - \ln(10) - \frac{1}{2}\ln(y)-\frac{1}{2}\ln(z)$
\item $\ln(x^{4}y^{2})$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{2}\left(\frac{xy}{z}\right)$
\item $\log_{3} \left( \frac{x}{y^2} \right)$
\item $\log_{3}\left(\frac{\sqrt{x}}{y^{2}z}\right)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\ln\left( \frac{x^2}{y^3z^4} \right)$
\item $\log\left(\frac{x \sqrt{y}}{\sqrt[3]{z}} \right)$
\item $\ln\left(\sqrt[3]{\frac{z}{xy}} \right)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{5}\left(\frac{x}{125}\right)$
\item $\log\left(\frac{1000}{x}\right)$
\item $\log_{7}\left(\frac{x(x - 3)}{49}\right)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\ln \left(x \sqrt{e} \right)$
\item $\log_{2}\left(x^{3/2}\right)$
\item $\log_{2}\left(x \sqrt{x-1}\right)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\vphantom{\frac{\log(x + 2)}{\log(3)}}\log_{2}\left(\frac{x}{x - 1}\right)$
\item $\vphantom{\frac{\log(x + 2)}{\log(3)}}7^{x - 1} = e^{(x - 1)\ln(7)}$
\item $\log_{3}(x + 2) = \frac{\log(x + 2)}{\log(3)}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\left(\frac{2}{3}\right)^{x} = e^{x\ln(\frac{2}{3})}$
\item $\log(x^{2} + 1) = \frac{\ln(x^{2} + 1)}{\ln(10)}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{3}(12) \approx 2.26186$
\item $\log_{5}(80) \approx 2.72271$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{6}(72) \approx 2.38685$
\item $\log_{4}\left(\frac{1}{10}\right) \approx -1.66096$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{\frac{3}{5}}(1000) \approx -13.52273$
\item $\log_{\frac{2}{3}}(50) \approx -9.64824$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\closegraphsfile
6.3: Exponential Equations and Inequalities
\subsection{Exercises}
In Exercises \ref{solvelogeqexfirst} - \ref{solvelogeqexlast}, solve the equation analytically.
\begin{multicols}{2}
\begin{enumerate}
\item $\log(3x-1) = \log(4-x)$ \label{solvelogeqexfirst}
\item $\log_{2}\left(x^{3}\right) = \log_{2}(x)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\ln\left(8-x^2\right)=\ln(2-x)$
\item $\log_{5}\left(18-x^2\right) = \log_{5}(6-x)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{3}(7-2x) = 2$
\item $\log_{\frac{1}{2}} (2x-1) = -3$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\ln\left(x^2-99\right) = 0$
\item $\log(x^2-3x) = 1$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{125} \left(\dfrac{3x-2}{2x+3}\right)=\dfrac{1}{3}$
\item $\log\left(\dfrac{x}{10^{-3}}\right) = 4.7$ \vphantom{$\log_{125} \left(\dfrac{3x-2}{2x+3}\right)$} \label{sixfourRichterequ}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $-\log(x) = 5.4$ \vphantom{$10\log\left(\dfrac{x}{10^{-12}}\right)$} \label{sixfourpHequ}
\item $10\log\left(\dfrac{x}{10^{-12}}\right) = 150$ \label{sixfourdecibelequ}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $6-3\log_{5}(2x)=0$
\item $3\ln(x)-2=1-\ln(x)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{3}(x - 4) + \log_{3}(x + 4) = 2$
\item $\log_{5}(2x + 1) + \log_{5}(x + 2) = 1$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{169}(3x + 7) - \log_{169}(5x - 9) = \dfrac{1}{2}$
\item $\ln(x+1) - \ln(x) = 3$ \vphantom{$\log_{169}(3x + 7)$}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $2\log_{7}(x) = \log_{7}(2) + \log_{7}(x+12)$
\item $\log(x) - \log(2) = \log(x+8) - \log(x+2)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$
\item $\ln(\ln(x)) = 3$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\left(\log(x)\right)^2=2\log(x)+15$
\item $\ln(x^{2}) = (\ln(x))^{2}$ \label{solvelogeqexlast}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
In Exercises \ref{solvelogineqexfirst} - \ref{solvelogineqexlast}, solve the inequality analytically.
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\dfrac{1 - \ln(x)}{x^{2}} < 0$ \label{solvelogineqexfirst}
\item $x\ln(x) - x > 0$ \phantom{$\dfrac{1 - \ln(x)}{x^{2}} < 0$}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $10\log\left(\dfrac{x}{10^{-12}}\right) \geq 90$ \label{sixfourdecibelineq}
\item $5.6 \leq \log\left(\dfrac{x}{10^{-3}}\right) \leq 7.1$ \label{sixfourRichterineq}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $2.3 < -\log(x) < 5.4$ \label{sixfourpHineq}
\item $\ln(x^{2}) \leq (\ln(x))^{2}$ \label{solvelogineqexlast}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
In Exercises \ref{logeqcalcexfirst} - \ref{logeqcalcexlast}, use your calculator to help you solve the equation or inequality.
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\ln(x) = e^{-x}$ \label{logeqcalcexfirst}
\item $\ln(x) = \sqrt[4]{x}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\ln(x^{2} + 1) \geq 5$
\item $\ln(-2x^{3} - x^{2} + 13x - 6) < 0$ \label{logeqcalcexlast}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item \label{onetooneexpexercise} Since $f(x) = e^{x}$ is a strictly increasing function, if $a < b$ then $e^{a} < e^{b}$. Use this fact to solve the inequality $\ln(2x + 1) < 3$ without a sign diagram. Use this technique to solve the inequalities in Exercises \ref{sixfourdecibelineq} - \ref{sixfourpHineq}. (Compare this to Exercise \ref{onetoonelogexercise} in Section \ref{ExpEquations}.)
\item Solve $\ln(3 - y) - \ln(y) = 2x + \ln(5)$ for $y$.
\item In Example \ref{logfracinverse} we found the inverse of $f(x) = \dfrac{\log(x)}{1-\log(x)}$ to be $f^{-1}(x) = 10^{\frac{x}{x+1}}$.
\begin{enumerate}
\item Show that $\left(f^{-1} \circ f\right)(x) = x$ for all $x$ in the domain of $f$ and that $\left(f \circ f^{-1}\right)(x) = x$ for all $x$ in the domain of $f^{-1}$.
\item Find the range of $f$ by finding the domain of $f^{-1}$.
\item Let $g(x) = \dfrac{x}{1 - x}$ and $h(x) = \log(x)$. Show that $f = g \circ h$ and $(g \circ h)^{-1} = h^{-1} \circ g^{-1}$.\\
(We know this is true in general by Exercise \ref{fcircginverse} in Section \ref{InverseFunctions}, but it's nice to see a specific example of the property.)
\end{enumerate}
\item \label{inversehyptangent} Let $f(x) = \dfrac{1}{2}\ln\left(\dfrac{1 + x}{1 - x}\right)$. Compute $f^{-1}(x)$ and find its domain and range.
\item Explain the equation in Exercise \ref{sixfourRichterequ} and the inequality in Exercise \ref{sixfourRichterineq} above in terms of the Richter scale for earthquake magnitude. (See Exercise \ref{Richterexercise} in Section \ref{IntroExpLogs}.)
\item Explain the equation in Exercise \ref{sixfourdecibelequ} and the inequality in Exercise \ref{sixfourdecibelineq} above in terms of sound intensity level as measured in decibels. (See Exercise \ref{decibelexercise} in Section \ref{IntroExpLogs}.)
\item Explain the equation in Exercise \ref{sixfourpHequ} and the inequality in Exercise \ref{sixfourpHineq} above in terms of the pH of a solution. (See Exercise \ref{pHexercise} in Section \ref{IntroExpLogs}.)
\item With the help of your classmates, solve the inequality $\sqrt[n]{x} > \ln(x)$ for a variety of natural numbers $n$. What might you conjecture about the ``speed'' at which $f(x) = \ln(x)$ grows versus any principal $n^{\textrm{th}}$ root function?
\end{enumerate}
\newpage
\subsection{Answers}
\begin{multicols}{3}
\begin{enumerate}
\item $x = \frac{5}{4}$
\item $x = 1$
\item $x=-2$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x=-3,\, 4$
\item $x=-1$
\item $x=\frac{9}{2}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x=\pm 10$
\item $x=-2,\, 5$
\item $x = -\frac{17}{7}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x = 10^{1.7}$
\item $x = 10^{-5.4}$
\item $x = 10^{3}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x=\frac{25}{2}$
\item $x=e^{3/4}$
\item $x = 5$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x = \frac{1}{2}$
\item $x = 2$
\item $x = \frac{1}{e^3-1}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x=6$
\item $x=4$
\item $x = 81$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x = e^{e^3}$
\item $x=10^{-3}, \, 10^{5}$
\item $x = 1, \, x = e^{2}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $(e, \infty)$
\item $(e, \infty)$
\item $\left[10^{-3}, \infty \right)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\left[10^{2.6}, 10^{4.1}\right]$
\item $\left(10^{-5.4}, 10^{-2.3}\right)$
\item $(0, 1] \cup [e^{2}, \infty)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $x \approx 1.3098$
\item $x \approx 4.177, \, x \approx 5503.665$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\approx (-\infty, -12.1414) \cup (12.1414, \infty)$
\item $\approx (-3.0281, -3) \cup (0.5, 0.5991) \cup (1.9299, 2)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $-\dfrac{1}{2} < x < \dfrac{e^{3} - 1}{2}$
\item $y = \dfrac{3}{5e^{2x} + 1}$ \vphantom{$\dfrac{e^{3} - 1}{2}$}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\addtocounter{enumi}{1}
\item $f^{-1}(x) = \dfrac{e^{2x} - 1}{e^{2x} + 1} = \dfrac{e^{x} - e^{-x}}{e^{x} + e^{-x}}$. (To see why we rewrite this in this form, see Exercise \ref{andtheresthyperbolic} in Section \ref{Parametric}.) The domain of $f^{-1}$ is $(-\infty, \infty)$ and its range is the same as the domain of $f$, namely $(-1, 1)$.
\end{enumerate}
\closegraphsfile
6.4: Logarithmic Equations and Inequalities
\subsection{Exercises}
In Exercises \ref{solvelogeqexfirst} - \ref{solvelogeqexlast}, solve the equation analytically.
\begin{multicols}{2}
\begin{enumerate}
\item $\log(3x-1) = \log(4-x)$ \label{solvelogeqexfirst}
\item $\log_{2}\left(x^{3}\right) = \log_{2}(x)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\ln\left(8-x^2\right)=\ln(2-x)$
\item $\log_{5}\left(18-x^2\right) = \log_{5}(6-x)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{3}(7-2x) = 2$
\item $\log_{\frac{1}{2}} (2x-1) = -3$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\ln\left(x^2-99\right) = 0$
\item $\log(x^2-3x) = 1$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{125} \left(\dfrac{3x-2}{2x+3}\right)=\dfrac{1}{3}$
\item $\log\left(\dfrac{x}{10^{-3}}\right) = 4.7$ \vphantom{$\log_{125} \left(\dfrac{3x-2}{2x+3}\right)$} \label{sixfourRichterequ}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $-\log(x) = 5.4$ \vphantom{$10\log\left(\dfrac{x}{10^{-12}}\right)$} \label{sixfourpHequ}
\item $10\log\left(\dfrac{x}{10^{-12}}\right) = 150$ \label{sixfourdecibelequ}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $6-3\log_{5}(2x)=0$
\item $3\ln(x)-2=1-\ln(x)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{3}(x - 4) + \log_{3}(x + 4) = 2$
\item $\log_{5}(2x + 1) + \log_{5}(x + 2) = 1$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{169}(3x + 7) - \log_{169}(5x - 9) = \dfrac{1}{2}$
\item $\ln(x+1) - \ln(x) = 3$ \vphantom{$\log_{169}(3x + 7)$}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $2\log_{7}(x) = \log_{7}(2) + \log_{7}(x+12)$
\item $\log(x) - \log(2) = \log(x+8) - \log(x+2)$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$
\item $\ln(\ln(x)) = 3$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\left(\log(x)\right)^2=2\log(x)+15$
\item $\ln(x^{2}) = (\ln(x))^{2}$ \label{solvelogeqexlast}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
In Exercises \ref{solvelogineqexfirst} - \ref{solvelogineqexlast}, solve the inequality analytically.
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\dfrac{1 - \ln(x)}{x^{2}} < 0$ \label{solvelogineqexfirst}
\item $x\ln(x) - x > 0$ \phantom{$\dfrac{1 - \ln(x)}{x^{2}} < 0$}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $10\log\left(\dfrac{x}{10^{-12}}\right) \geq 90$ \label{sixfourdecibelineq}
\item $5.6 \leq \log\left(\dfrac{x}{10^{-3}}\right) \leq 7.1$ \label{sixfourRichterineq}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $2.3 < -\log(x) < 5.4$ \label{sixfourpHineq}
\item $\ln(x^{2}) \leq (\ln(x))^{2}$ \label{solvelogineqexlast}
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
In Exercises \ref{logeqcalcexfirst} - \ref{logeqcalcexlast}, use your calculator to help you solve the equation or inequality.
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\ln(x) = e^{-x}$ \label{logeqcalcexfirst}
\item $\ln(x) = \sqrt[4]{x}$
\setcounter{HW}{\value{enumi}}
\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
\setcounter{enumi}{\value{HW}}
\item $\ln(x^{2} + 1) \geq 5$
\item $\ln(-2x^{3} - x^{2} + 13x - 6) < 0$ \label{logeqcalcexlast}
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\end{enumerate}
\end{multicols}
\begin{enumerate}
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\item \label{onetooneexpexercise} Since $f(x) = e^{x}$ is a strictly increasing function, if $a < b$ then $e^{a} < e^{b}$. Use this fact to solve the inequality $\ln(2x + 1) < 3$ without a sign diagram. Use this technique to solve the inequalities in Exercises \ref{sixfourdecibelineq} - \ref{sixfourpHineq}. (Compare this to Exercise \ref{onetoonelogexercise} in Section \ref{ExpEquations}.)
\item Solve $\ln(3 - y) - \ln(y) = 2x + \ln(5)$ for $y$.
\item In Example \ref{logfracinverse} we found the inverse of $f(x) = \dfrac{\log(x)}{1-\log(x)}$ to be $f^{-1}(x) = 10^{\frac{x}{x+1}}$.
\begin{enumerate}
\item Show that $\left(f^{-1} \circ f\right)(x) = x$ for all $x$ in the domain of $f$ and that $\left(f \circ f^{-1}\right)(x) = x$ for all $x$ in the domain of $f^{-1}$.
\item Find the range of $f$ by finding the domain of $f^{-1}$.
\item Let $g(x) = \dfrac{x}{1 - x}$ and $h(x) = \log(x)$. Show that $f = g \circ h$ and $(g \circ h)^{-1} = h^{-1} \circ g^{-1}$.\\
(We know this is true in general by Exercise \ref{fcircginverse} in Section \ref{InverseFunctions}, but it's nice to see a specific example of the property.)
\end{enumerate}
\item \label{inversehyptangent} Let $f(x) = \dfrac{1}{2}\ln\left(\dfrac{1 + x}{1 - x}\right)$. Compute $f^{-1}(x)$ and find its domain and range.
\item Explain the equation in Exercise \ref{sixfourRichterequ} and the inequality in Exercise \ref{sixfourRichterineq} above in terms of the Richter scale for earthquake magnitude. (See Exercise \ref{Richterexercise} in Section \ref{IntroExpLogs}.)
\item Explain the equation in Exercise \ref{sixfourdecibelequ} and the inequality in Exercise \ref{sixfourdecibelineq} above in terms of sound intensity level as measured in decibels. (See Exercise \ref{decibelexercise} in Section \ref{IntroExpLogs}.)
\item Explain the equation in Exercise \ref{sixfourpHequ} and the inequality in Exercise \ref{sixfourpHineq} above in terms of the pH of a solution. (See Exercise \ref{pHexercise} in Section \ref{IntroExpLogs}.)
\item With the help of your classmates, solve the inequality $\sqrt[n]{x} > \ln(x)$ for a variety of natural numbers $n$. What might you conjecture about the ``speed'' at which $f(x) = \ln(x)$ grows versus any principal $n^{\textrm{th}}$ root function?
\end{enumerate}
\newpage
\subsection{Answers}
\begin{multicols}{3}
\begin{enumerate}
\item $x = \frac{5}{4}$
\item $x = 1$
\item $x=-2$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $x=-3,\, 4$
\item $x=-1$
\item $x=\frac{9}{2}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $x=\pm 10$
\item $x=-2,\, 5$
\item $x = -\frac{17}{7}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $x = 10^{1.7}$
\item $x = 10^{-5.4}$
\item $x = 10^{3}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $x=\frac{25}{2}$
\item $x=e^{3/4}$
\item $x = 5$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $x = \frac{1}{2}$
\item $x = 2$
\item $x = \frac{1}{e^3-1}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $x=6$
\item $x=4$
\item $x = 81$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $x = e^{e^3}$
\item $x=10^{-3}, \, 10^{5}$
\item $x = 1, \, x = e^{2}$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $(e, \infty)$
\item $(e, \infty)$
\item $\left[10^{-3}, \infty \right)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{3}
\begin{enumerate}
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\item $\left[10^{2.6}, 10^{4.1}\right]$
\item $\left(10^{-5.4}, 10^{-2.3}\right)$
\item $(0, 1] \cup [e^{2}, \infty)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $x \approx 1.3098$
\item $x \approx 4.177, \, x \approx 5503.665$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $\approx (-\infty, -12.1414) \cup (12.1414, \infty)$
\item $\approx (-3.0281, -3) \cup (0.5, 0.5991) \cup (1.9299, 2)$
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item $-\dfrac{1}{2} < x < \dfrac{e^{3} - 1}{2}$
\item $y = \dfrac{3}{5e^{2x} + 1}$ \vphantom{$\dfrac{e^{3} - 1}{2}$}
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\end{enumerate}
\end{multicols}
\begin{enumerate}
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\addtocounter{enumi}{1}
\item $f^{-1}(x) = \dfrac{e^{2x} - 1}{e^{2x} + 1} = \dfrac{e^{x} - e^{-x}}{e^{x} + e^{-x}}$. (To see why we rewrite this in this form, see Exercise \ref{andtheresthyperbolic} in Section \ref{Parametric}.) The domain of $f^{-1}$ is $(-\infty, \infty)$ and its range is the same as the domain of $f$, namely $(-1, 1)$.
\end{enumerate}
\closegraphsfile
6.5: Applications of Exponential and Logarithmic Functions
\subsection{Exercises}
For each of the scenarios given in Exercises \ref{basicinterestexfirst} - \ref{basicinterestexlast},
\begin{itemize}
\item Find the amount $A$ in the account as a function of the term of the investment $t$ in years.
\item Determine how much is in the account after $5$ years, $10$ years, $30$ years and $35$ years. Round your answers to the nearest cent.
\item Determine how long will it take for the initial investment to double. Round your answer to the nearest year.
\item Find and interpret the average rate of change of the amount in the account from the end of
the fourth year to the end of the fifth year, and from the end of the thirty-fourth year to the
end of the thirty-fifth year. Round your answer to two decimal places.
\end{itemize}
\begin{enumerate}
\item $\$500$ is invested in an account which offers $0.75 \%$, compounded monthly. \label{basicinterestexfirst}
\item $\$500$ is invested in an account which offers $0.75 \%$, compounded continuously.
\item $\$1000$ is invested in an account which offers $1.25 \%$, compounded monthly.
\item $\$1000$ is invested in an account which offers $1.25 \%$, compounded continuously.
\item $\$5000$ is invested in an account which offers $2.125 \%$, compounded monthly.
\item $\$5000$ is invested in an account which offers $2.125 \%$, compounded continuously. \label{basicinterestexlast}
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\end{enumerate}
\begin{enumerate}
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\item Look back at your answers to Exercises \ref{basicinterestexfirst} - \ref{basicinterestexlast}. What can be said about the difference between monthly compounding and continuously compounding the interest in those situations? With the help of your classmates, discuss scenarios where the difference between monthly and continuously compounded interest would be more dramatic. Try varying the interest rate, the term of the investment and the principal. Use computations to support your answer.
\item How much money needs to be invested now to obtain $\$2000$ in 3 years if the interest rate in a savings account is $0.25 \%$, compounded continuously? Round your answer to the nearest cent.
\item How much money needs to be invested now to obtain $\$5000$ in 10 years if the interest rate in a CD is $2.25 \%$, compounded monthly? Round your answer to the nearest cent.
\item On May, 31, 2009, the Annual Percentage Rate listed at Jeff's bank for regular savings accounts was $0.25\%$ compounded monthly. Use Equation \ref{compoundinterest} to answer the following.
\begin{enumerate}
\item If $P = 2000$ what is $A(8)$?
\item Solve the equation $A(t) = 4000$ for $t$.
\item What principal $P$ should be invested so that the account balance is \$2000 is three years?
\end{enumerate}
\item Jeff's bank also offers a 36-month Certificate of Deposit (CD) with an APR of $2.25\%$.
\begin{enumerate}
\item If $P = 2000$ what is $A(8)$?
\item Solve the equation $A(t) = 4000$ for $t$.
\item What principal $P$ should be invested so that the account balance is \$2000 is three years?
\item The Annual Percentage Yield is the \underline{simple} interest rate that returns the same amount of interest after one year as the compound interest does. With the help of your classmates, compute the APY for this investment.
\end{enumerate}
\item A finance company offers a promotion on $\$5000$ loans. The borrower does not have to make any payments for the first three years, however interest will continue to be charged to the loan at $29.9 \%$ compounded continuously. What amount will be due at the end of the three year period, assuming no payments are made? If the promotion is extended an additional three years, and no payments are made, what amount would be due?
\item Use Equation \ref{compoundinterest} to show that the time it takes for an investment to double in value does \underline{not} depend on the principal $P$, but rather, depends only on the APR and the number of compoundings per year. Let $n = 12$ and with the help of your classmates compute the doubling time for a variety of rates $r$. Then look up the Rule of 72 and compare your answers to what that rule says. If you're really interested\footnote{Awesome pun!} in Financial Mathematics, you could also compare and contrast the Rule of 72 with the Rule of 70 and the Rule of 69.
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\end{enumerate}
In Exercises \ref{radioactivefirst} - \ref{radioactivelast}, we list some radioactive isotopes and their associated half-lives. Assume that each decays according to the formula $A(t) = A_{\text{\tiny $0$}}e^{kt}$ where $A_{\text{\tiny $0$}}$ is the initial amount of the material and $k$ is the decay constant. For each isotope:
\begin{itemize}
\item Find the decay constant $k$. Round your answer to four decimal places.
\item Find a function which gives the amount of isotope $A$ which remains after time $t$. (Keep the units of $A$ and $t$ the same as the given data.)
\item Determine how long it takes for $90 \%$ of the material to decay. Round your answer to two decimal places. (HINT: If $90 \%$ of the material decays, how much is left?)
\end{itemize}
\begin{enumerate}
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\item Cobalt 60, used in food irradiation, initial amount 50 grams, half-life of $5.27$ years. \label{radioactivefirst}
\item Phosphorus 32, used in agriculture, initial amount 2 milligrams, half-life $14$ days.
\item Chromium 51, used to track red blood cells, initial amount 75 milligrams, half-life $27.7$ days.
\item Americium 241, used in smoke detectors, initial amount 0.29 micrograms, half-life $432.7$ years.
\item Uranium 235, used for nuclear power, initial amount $1$ kg grams, half-life $704$ million years. \label{radioactivelast}
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\end{enumerate}
\begin{enumerate}
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\item With the help of your classmates, show that the time it takes for $90 \%$ of each isotope listed in Exercises \ref{radioactivefirst} - \ref{radioactivelast} to decay does not depend on the initial amount of the substance, but rather, on only the decay constant $k$. Find a formula, in terms of $k$ only, to determine how long it takes for $90 \%$ of a radioactive isotope to decay.
\item In Example \ref{cardepreciationex} in Section \ref{IntroExpLogs}, the exponential function $V(x) = 25 \left(\frac{4}{5}\right)^{x}$ was used to model the value of a car over time. Use the properties of logs and/or exponents to rewrite the model in the form $V(t) = 25e^{kt}$.
\item The Gross Domestic Product (GDP) of the US (in billions of dollars) $t$ years after the year 2000 can be modeled by: \[ G(t) = 9743.77 e^{0.0514t}\]
\begin{enumerate}
\item Find and interpret $G(0)$.
\item According to the model, what should have been the GDP in 2007? In 2010? (According to the \href{1.usa.gov/iimT40}{\underline{US Department of Commerce}}, the 2007 GDP was $\$14,369.1$ billion and the 2010 GDP was $\$14,657.8$ billion.)
\end{enumerate}
\item The diameter $D$ of a tumor, in millimeters, $t$ days after it is detected is given by: \[D(t) = 15e^{0.0277t} \]
\begin{enumerate}
\item What was the diameter of the tumor when it was originally detected?
\item How long until the diameter of the tumor doubles?
\end{enumerate}
\item Under optimal conditions, the growth of a certain strain of \textit{E. Coli} is modeled by the Law of Uninhibited Growth $N(t) = N_{\text{\tiny $0$}} e^{kt}$ where $N_{\text{\tiny $0$}}$ is the initial number of bacteria and $t$ is the elapsed time, measured in minutes. From numerous experiments, it has been determined that the doubling time of this organism is 20 minutes. Suppose 1000 bacteria are present initially.
\begin{enumerate}
\item Find the growth constant $k$. Round your answer to four decimal places.
\item Find a function which gives the number of bacteria $N(t)$ after $t$ minutes.
\item How long until there are 9000 bacteria? Round your answer to the nearest minute.
\end{enumerate}
\item Yeast is often used in biological experiments. A research technician estimates that a sample of yeast suspension contains 2.5 million organisms per cubic centimeter (cc). Two hours later, she estimates the population density to be 6 million organisms per cc. Let $t$ be the time elapsed since the first observation, measured in hours. Assume that the yeast growth follows the Law of Uninhibited Growth $N(t) = N_{\text{\tiny $0$}} e^{kt}$.
\begin{enumerate}
\item Find the growth constant $k$. Round your answer to four decimal places.
\item Find a function which gives the number of yeast (in millions) per cc $N(t)$ after $t$ hours.
\item What is the doubling time for this strain of yeast?
\end{enumerate}
\item The Law of Uninhibited Growth also applies to situations where an animal is re-introduced into a suitable environment. Such a case is the reintroduction of wolves to Yellowstone National Park. According to the \href{www.nps.gov/yell/naturescienc...rline{National Park Service}}, the wolf population in Yellowstone National Park was 52 in 1996 and 118 in 1999. Using these data, find a function of the form $N(t) = N_{\text{\tiny $0$}} e^{kt}$ which models the number of wolves $t$ years after 1996. (Use $t = 0$ to represent the year 1996. Also, round your value of $k$ to four decimal places.) According to the model, how many wolves were in Yellowstone in 2002? (The recorded number is 272.)
\item \label{PainesvillePopulationTwoPoint} During the early years of a community, it is not uncommon for the population to grow according to the Law of Uninhibited Growth. According to the Painesville Wikipedia entry, in 1860, the Village of Painesville had a population of 2649. In 1920, the population was 7272. Use these two data points to fit a model of the form $N(t) = N_{\text{\tiny $0$}} e^{kt}$ were $N(t)$ is the number of Painesville Residents $t$ years after 1860. (Use $t = 0$ to represent the year 1860. Also, round the value of $k$ to four decimal places.) According to this model, what was the population of Painesville in 2010? (The 2010 census gave the population as 19,563) What could be some causes for such a vast discrepancy? For more on this, see Exercise \ref{PainesvillePopulationManyPoints}.
\item The population of Sasquatch in Bigfoot county is modeled by \[P(t) = \dfrac{120}{1 + 3.167e^{-0.05t}}\] where $P(t)$ is the population of Sasquatch $t$ years after $2010$.
\begin{enumerate}
\item Find and interpret $P(0)$.
\item Find the population of Sasquatch in Bigfoot county in 2013. Round your answer to the nearest Sasquatch.
\item When will the population of Sasquatch in Bigfoot county reach 60? Round your answer to the nearest year.
\item Find and interpret the end behavior of the graph of $y = P(t)$. Check your answer using a graphing utility.
\end{enumerate}
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\end{enumerate}
\begin{enumerate}
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\item The half-life of the radioactive isotope Carbon-14 is about 5730 years.
\begin{enumerate}
\item Use Equation \ref{radioactivedecay} to express the amount of Carbon-14 left from an initial $N$ milligrams as a function of time $t$ in years.
\item What percentage of the original amount of Carbon-14 is left after 20,000 years?
\item If an old wooden tool is found in a cave and the amount of Carbon-14 present in it is estimated to be only 42\% of the original amount, approximately how old is the tool?
\item Radiocarbon dating is not as easy as these exercises might lead you to believe. With the help of your classmates, research radiocarbon dating and discuss why our model is somewhat over-simplified.
\end{enumerate}
\item Carbon-14 cannot be used to date inorganic material such as rocks, but there are many other methods of radiometric dating which estimate the age of rocks. One of them, Rubidium-Strontium dating, uses Rubidium-87 which decays to Strontium-87 with a half-life of 50 billion years. Use Equation \ref{radioactivedecay} to express the amount of Rubidium-87 left from an initial 2.3 micrograms as a function of time $t$ in \emph{billions} of years. Research this and other radiometric techniques and discuss the margins of error for various methods with your classmates.
\item Use Equation \ref{radioactivedecay} to show that $k = -\dfrac{\ln(2)}{h}$ where $h$ is the half-life of the radioactive isotope.
\item A pork roast\footnote{This roast was enjoyed by Jeff and his family on June 10, 2009. This is real data, folks!} was taken out of a hardwood smoker when its internal temperature had reached $180^{\circ}$F and it was allowed to rest in a $75^{\circ}$F house for 20 minutes after which its internal temperature had dropped to $170^{\circ}$F. Assuming that the temperature of the roast follows Newton's Law of Cooling (Equation \ref{newtonslawofcooling}),
\begin{enumerate}
\item Express the temperature $T$ (in $^{\circ}$F) as a function of time $t$ (in minutes).
\item Find the time at which the roast would have dropped to $140^{\circ}$F had it not been carved and eaten.
\end{enumerate}
\item \label{pursuitlog} In reference to Exercise \ref{pursuitfurther} in Section \ref{AlgebraicFunctions}, if Fritzy the Fox's speed is the same as Chewbacca the Bunny's speed, Fritzy's pursuit curve is given by
\[y(x) = \frac{1}{4} x^2-\frac{1}{4} \ln(x)-\frac{1}{4}\]
Use your calculator to graph this path for $x > 0$. Describe the behavior of $y$ as $x \rightarrow 0^{+}$ and interpret this physically.
\item \label{explogsappcircuitone} The current $i$ measured in amps in a certain electronic circuit with a constant impressed voltage of 120 volts is given by $i(t) = 2 - 2e^{-10t}$ where $t \geq 0$ is the number of seconds after the circuit is switched on. Determine the value of $i$ as $t \rightarrow \infty$. (This is called the \textbf{steady state} current.)
\item If the voltage in the circuit in Exercise \ref{explogsappcircuitone} above is switched off after 30 seconds, the current is given by the piecewise-defined function
\[i(t) = \left\{ \begin{array}{rcl} 2 - 2e^{-10t} & \mbox{if} & 0 \leq t < 30 \\ [6pt]
\left(2 - 2e^{-300}\right) e^{-10t+300} & \mbox{if} & t \geq 30 \end{array} \right.\]
With the help of your calculator, graph $y = i(t)$ and discuss with your classmates the physical significance of the two parts of the graph $0 \leq t < 30$ and $t \geq 30$.
\item \label{catenary} In Exercise \ref{parabolicbridgecable} in Section \ref{QuadraticFunctions}, we stated that the cable of a suspension bridge formed a parabola but that a free hanging cable did not. A free hanging cable forms a \underline{catenary} and its basic shape is given by $y = \frac{1}{2}\left(e^{x} + e^{-x}\right)$. Use your calculator to graph this function. What are its domain and range? What is its end behavior? Is it invertible? How do you think it is related to the function given in Exercise \ref{hyperbolicsine} in Section \ref{ExpEquations} and the one given in the answer to Exercise \ref{inversehyptangent} in Section \ref{LogEquations}? When flipped upside down, the catenary makes an arch. The Gateway Arch in St. Louis, Missouri has the shape \[y = 757.7 - \frac{127.7}{2}\left(e^{\frac{x}{127.7}} + e^{-\frac{x}{127.7}}\right)\] where $x$ and $y$ are measured in feet and $-315 \leq x \leq 315$. Find the highest point on the arch.
\item In Exercise \ref{APLcats} in Section \ref{Regression}, we examined the data set given below which showed how two cats and their surviving offspring can produce over 80 million cats in just ten years. It is virtually impossible to see this data plotted on your calculator, so plot $x$ versus $\ln(x)$ as was done on page \pageref{swineflulinearized}. Find a linear model for this new data and comment on its goodness of fit. Find an exponential model for the original data and comment on its goodness of fit.
\medskip
\small
\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|} \hline
Year $x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
Number of & & & & & & & & & & \\
Cats $N(x)$ & 12 & 66 & 382 & 2201 & 12680 & 73041 & 420715 & 2423316 & 13968290 & 80399780 \\ \hline
\end{tabular}
\normalsize
\item \label{PainesvillePopulationManyPoints} This exercise is a follow-up to Exercise \ref{PainesvillePopulationTwoPoint} which more thoroughly explores the population growth of Painesville, Ohio. According to \href{en.Wikipedia.org/wiki/Painesv...ne{Wikipedia}}, the population of Painesville, Ohio is given by
\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|} \hline
Year $t$ & 1860 & 1870 & 1880 & 1890 & 1900 & 1910 & 1920 & 1930 & 1940 & 1950 \\ \hline
Population& 2649 & 3728 & 3841 & 4755 & 5024 & 5501 & 7272 & 10944 & 12235 & 14432 \\ \hline
\end{tabular}
\noindent \begin{tabular}{|l|r|r|r|r|r|} \hline
Year $t$ & 1960 & 1970 & 1980 & 1990 & 2000 \\ \hline
Population& 16116 & 16536 & 16351 & 15699 & 17503 \\ \hline
\end{tabular}
\begin{enumerate}
\item Use a graphing utility to perform an exponential regression on the data from 1860 through 1920 only, letting $t = 0$ represent the year 1860 as before. How does this calculator model compare with the model you found in Exercise \ref{PainesvillePopulationTwoPoint}? Use the calculator's exponential model to predict the population in 2010. (The 2010 census gave the population as 19,563)
\item The logistic model fit to \emph{all} of the given data points for the population of Painesville $t$ years after 1860 (again, using $t = 0$ as 1860) is \[ P(t) = \dfrac{18691}{1+9.8505e^{-0.03617t}} \] According to this model, what should the population of Painesville have been in 2010? (The 2010 census gave the population as 19,563.) What is the population limit of Painesville?
\end{enumerate}
\item According to \href{www.ohiobiz.com/census/Lake.p...line{OhioBiz}}, the census data for Lake County, Ohio is as follows:
\small
\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|} \hline
Year $t$ & 1860 & 1870 & 1880 & 1890 & 1900 & 1910 & 1920 & 1930 & 1940 & 1950 \\ \hline
Population& 15576 & 15935 & 16326 & 18235 & 21680 & 22927 & 28667 & 41674 & 50020 & 75979 \\ \hline
\end{tabular}
\noindent \begin{tabular}{|l|r|r|r|r|r|} \hline
Year $t$ & 1960 & 1970 & 1980 & 1990 & 2000 \\ \hline
Population& 148700 & 197200 & 212801 & 215499 & 227511 \\ \hline
\end{tabular}
\normalsize
\begin{enumerate}
\item Use your calculator to fit a logistic model to these data, using $x = 0$ to represent the year 1860.
\item Graph these data and your logistic function on your calculator to judge the reasonableness of the fit.
\item Use this model to estimate the population of Lake County in 2010. (The 2010 census gave the population to be 230,041.)
\item According to your model, what is the population limit of Lake County, Ohio?
\end{enumerate}
\item According to \href{www.facebook.com/press/info.p...ine{facebook}}, the number of active users of facebook has grown significantly since its initial launch from a Harvard dorm room in February 2004. The chart below has the approximate number $U(x)$ of active users, in \underline{millions}, $x$ months after February 2004. For example, the first entry $(10, 1)$ means that there were $1$ million active users in December 2004 and the last entry $(77, 500)$ means that there were $500$ million active users in July 2010.
\medskip
\small
\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline
Month $x$ & 10 & 22 & 34 & 38 & 44 & 54 & 59 & 60 & 62 & 65 & 67 & 70 & 72 & 77 \\ \hline
Active Users in & & & & & & & & & & & & & & \\
Millions $U(x)$ & 1 & 5.5 & 12 & 20 & 50 & 100 & 150 & 175 & 200 & 250 & 300 & 350 & 400 & 500\\ \hline
\end{tabular}
\normalsize
\medskip
With the help of your classmates, find a model for this data.
\item Each Monday during the registration period before the Fall Semester at LCCC, the Enrollment Planning Council gets a report prepared by the data analysts in Institutional Effectiveness and Planning.\footnote{The authors thank Dr. Wendy Marley and her staff for this data and Dr. Marcia Ballinger for the permission to use it in this problem.} While the ongoing enrollment data is analyzed in many different ways, we shall focus only on the overall headcount. Below is a chart of the enrollment data for Fall Semester 2008. It starts 21 weeks before ``Opening Day'' and ends on ``Day 15'' of the semester, but we have relabeled the top row to be $x = 1$ through $x = 24$ so that the math is easier. (Thus, $x = 22$ is Opening Day.)
\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|} \hline
Week $x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline
Total & & & & & & & & \\
Headcount & 1194 & 1564 & 2001 & 2475 & 2802 & 3141 & 3527 & 3790 \\ \hline
\end{tabular}
\medskip
\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|} \hline
Week $x$ & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline
Total & & & & & & & & \\
Headcount & 4065 & 4371 & 4611 & 4945 & 5300 & 5657 & 6056 & 6478 \\ \hline
\end{tabular}
\medskip
\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|} \hline
Week $x$ & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24\\ \hline
Total & & & & & & & & \\
Headcount & 7161 & 7772 & 8505 & 9256 & 10201 & 10743 & 11102 & 11181 \\ \hline
\end{tabular}
\medskip
With the help of your classmates, find a model for this data. Unlike most of the phenomena we have studied in this section, there is no single differential equation which governs the enrollment growth. Thus there is no scientific reason to rely on a logistic function even though the data plot may lead us to that model. What are some factors which influence enrollment at a community college and how can you take those into account mathematically?
\item When we wrote this exercise, the Enrollment Planning Report for Fall Semester 2009 had only 10 data points for the first 10 weeks of the registration period. Those numbers are given below.
\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|} \hline
Week $x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline
Total & & & & & & & & & & \\
Headcount & 1380 & 2000 & 2639 & 3153 & 3499 & 3831 & 4283 & 4742 & 5123 & 5398 \\ \hline
\end{tabular}
With the help of your classmates, find a model for this data and make a prediction for the Opening Day enrollment as well as the Day 15 enrollment. (WARNING: The registration period for 2009 was one week shorter than it was in 2008 so Opening Day would be $x = 21$ and Day 15 is $x = 23$.)
\end{enumerate}
\newpage
\subsection{Answers}
\begin{enumerate}
\item \begin{itemize} \item $A(t) = 500\left(1 + \frac{0.0075}{12}\right)^{12t}$
\item $A(5) \approx \$ 519.10$, $A(10) \approx \$ 538.93$, $A(30) \approx \$ 626.12$, $A(35) \approx \$ 650.03$
\item It will take approximately $92$ years for the investment to double.
\item The average rate of change from the end of the fourth year to the end of the fifth year is approximately $3.88$. This means that the investment is growing at an average rate of $\$3.88$ per year at this point. The average rate of change from the end of the thirty-fourth year to the end of the thirty-fifth year is approximately $4.85$. This means that the investment is growing at an average rate of $\$4.85$ per year at this point.
\end{itemize}
\item \begin{itemize} \item $A(t) = 500e^{0.0075t}$
\item $A(5) \approx \$ 519.11$, $A(10) \approx \$ 538.94$, $A(30) \approx \$ 626.16$, $A(35) \approx \$ 650.09$
\item It will take approximately $92$ years for the investment to double.
\item The average rate of change from the end of the fourth year to the end of the fifth year is approximately $3.88$. This means that the investment is growing at an average rate of $\$3.88$ per year at this point. The average rate of change from the end of the thirty-fourth year to the end of the thirty-fifth year is approximately $4.86$. This means that the investment is growing at an average rate of $\$4.86$ per year at this point.
\end{itemize}
\item \begin{itemize} \item $A(t) = 1000\left(1 + \frac{0.0125}{12}\right)^{12t}$
\item $A(5) \approx \$ 1064.46$, $A(10) \approx \$ 1133.07$, $A(30) \approx \$ 1454.71$, $A(35) \approx \$ 1548.48$
\item It will take approximately $55$ years for the investment to double.
\item The average rate of change from the end of the fourth year to the end of the fifth year is approximately $13.22$. This means that the investment is growing at an average rate of $\$13.22$ per year at this point. The average rate of change from the end of the thirty-fourth year to the end of the thirty-fifth year is approximately $19.23$. This means that the investment is growing at an average rate of $\$19.23$ per year at this point.
\end{itemize}
\item \begin{itemize} \item $A(t) = 1000e^{0.0125t}$
\item $A(5) \approx \$ 1064.49$, $A(10) \approx \$ 1133.15$, $A(30) \approx \$ 1454.99$, $A(35) \approx \$ 1548.83$
\item It will take approximately $55$ years for the investment to double.
\item The average rate of change from the end of the fourth year to the end of the fifth year is approximately $13.22$. This means that the investment is growing at an average rate of $\$13.22$ per year at this point. The average rate of change from the end of the thirty-fourth year to the end of the thirty-fifth year is approximately $19.24$. This means that the investment is growing at an average rate of $\$19.24$ per year at this point.
\end{itemize}
\item \begin{itemize} \item $A(t) = 5000\left(1 + \frac{0.02125}{12}\right)^{12t}$
\item $A(5) \approx \$ 5559.98$, $A(10) \approx \$ 6182.67$, $A(30) \approx \$ 9453.40$, $A(35) \approx \$ 10512.13$
\item It will take approximately $33$ years for the investment to double.
\item The average rate of change from the end of the fourth year to the end of the fifth year is approximately $116.80$. This means that the investment is growing at an average rate of $\$116.80$ per year at this point. The average rate of change from the end of the thirty-fourth year to the end of the thirty-fifth year is approximately $220.83$. This means that the investment is growing at an average rate of $\$220.83$ per year at this point.
\end{itemize}
\item \begin{itemize} \item $A(t) = 5000e^{0.02125t}$
\item $A(5) \approx \$ 5560.50$, $A(10) \approx \$ 6183.83$, $A(30) \approx \$ 9458.73$, $A(35) \approx \$ 10519.05$
\item It will take approximately $33$ years for the investment to double.
\item The average rate of change from the end of the fourth year to the end of the fifth year is approximately $116.91$. This means that the investment is growing at an average rate of $\$116.91$ per year at this point. The average rate of change from the end of the thirty-fourth year to the end of the thirty-fifth year is approximately $221.17$. This means that the investment is growing at an average rate of $\$221.17$ per year at this point.
\end{itemize}
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\end{enumerate}
\begin{enumerate}
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\addtocounter{enumi}{1}
\item $P = \frac{2000}{e^{0.0025 \cdot 3}} \approx \$ 1985.06$
\item $P = \frac{5000}{\left(1 + \frac{0.0225}{12}\right)^{12 \cdot 10}} \approx \$ 3993.42$
\item \begin{enumerate}
\item $A(8) = 2000\left(1 + \frac{0.0025}{12}\right)^{12 \cdot 8} \approx \$2040.40$
\item $t = \dfrac{\ln(2)}{12 \ln\left(1 + \frac{0.0025}{12}\right)} \approx 277.29$ years
\item $P = \dfrac{2000}{\left(1 + \frac{0.0025}{12}\right)^{36}} \approx \$1985.06$
\end{enumerate}
\item \begin{enumerate}
\item $A(8) = 2000\left(1 + \frac{0.0225}{12}\right)^{12 \cdot 8} \approx \$2394.03$
\item $t = \dfrac{\ln(2)}{12 \ln\left(1 + \frac{0.0225}{12}\right)} \approx 30.83$ years
\item $P = \dfrac{2000}{\left(1 + \frac{0.0225}{12}\right)^{36}} \approx \$1869.57$
\item $\left(1 + \frac{0.0225}{12}\right)^{12} \approx 1.0227$ so the APY is 2.27\%
\end{enumerate}
\item $A(3) = 5000e^{0.299 \cdot 3} \approx \$12,226.18$, $A(6) = 5000e^{0.299 \cdot 6} \approx \$30,067.29$
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\end{enumerate}
\begin{multicols}{2}
\begin{enumerate}
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\addtocounter{enumi}{1}
\item \begin{itemize} \item $k = \frac{\ln(1/2)}{5.27} \approx -0.1315$
\item $A(t) = 50e^{-0.1315t}$
\item $t = \frac{\ln(0.1)}{-0.1315} \approx 17.51$ years.
\end{itemize}
\item \begin{itemize} \item $k = \frac{\ln(1/2)}{14} \approx -0.0495$
\item $A(t) = 2e^{-0.0495t}$
\item $t = \frac{\ln(0.1)}{-0.0495} \approx 46.52$ days.
\end{itemize}
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\end{enumerate}
\end{multicols}
\begin{multicols}{2}
\begin{enumerate}
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\item \begin{itemize} \item $k = \frac{\ln(1/2)}{27.7} \approx -0.0250$
\item $A(t) = 75e^{-0.0250t}$
\item $t = \frac{\ln(0.1)}{-0.025} \approx 92.10$ days.
\end{itemize}
\item \begin{itemize} \item $k = \frac{\ln(1/2)}{432.7} \approx -0.0016$
\item $A(t) = 0.29e^{-0.0016t}$
\item $t = \frac{\ln(0.1)}{-0.0016} \approx 1439.11$ years.
\end{itemize}
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\end{enumerate}
\end{multicols}
\begin{enumerate}
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\item \begin{itemize} \item $k = \frac{\ln(1/2)}{704} \approx -0.0010$
\item $A(t) = e^{-0.0010t}$
\item $t = \frac{\ln(0.1)}{-0.0010} \approx 2302.58$ million years, or $2.30$ billion years.
\end{itemize}
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\end{enumerate}
\begin{multicols}{2}
\begin{enumerate}
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\item $t = \frac{\ln(0.1)}{k} = -\frac{\ln(10)}{k}$
\item $V(t) = 25e^{\ln\left(\frac{4}{5}\right)t} \approx 25e^{-0.22314355t}$
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\end{enumerate}
\end{multicols}
\begin{enumerate}
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\item \begin{enumerate} \item $G(0) = 9743.77$ This means that the GDP of the US in 2007 was $\$9743.77$ billion dollars.
\item $G(7) = 13963.24$ and $G(10) = 16291.25$, so the model predicted a GDP of $\$ 13,963.24$ billion in 2007 and $\$ 16,291.25$ billion in 2010.
\end{enumerate}
\item \begin{enumerate} \item $D(0) = 15$, so the tumor was 15 millimeters in diameter when it was first detected.
\item $t = \frac{\ln(2)}{0.0277} \approx 25$ days.
\end{enumerate}
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\end{enumerate}
\begin{multicols}{2}
\begin{enumerate}
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\item \begin{enumerate} \item $k = \frac{\ln(2)}{20} \approx 0.0346$
\item $N(t) = 1000e^{0.0346 t}$
\item $t = \frac{\ln(9)}{0.0346} \approx 63$ minutes
\end{enumerate}
\item \begin{enumerate} \item $k = \frac{1}{2}\frac{\ln(6)}{2.5} \approx 0.4377$
\item $N(t) = 2.5e^{0.4377 t}$
\item $t = \frac{\ln(2)}{0.4377} \approx 1.58$ hours
\end{enumerate}
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\end{enumerate}
\end{multicols}
\begin{enumerate}
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\item $N_{\text{\tiny $0$}} = 52$, $k = \frac{1}{3} \ln\left( \frac{118}{52}\right) \approx 0.2731$, $N(t) = 52e^{0.2731t}$. $N(6) \approx 268$.
\item $N_{\text{\tiny $0$}} = 2649$, $k = \frac{1}{60} \ln\left( \frac{7272}{2649}\right) \approx 0.0168$, $N(t) = 2649e^{0.0168t}$. $N(150) \approx 32923$, so the population of Painesville in 2010 based on this model would have been 32,923.
\item \begin{enumerate} \item $P(0) = \frac{120}{4.167} \approx 29$. There are 29 Sasquatch in Bigfoot County in 2010.
\item $P(3) = \frac{120}{1+3.167e^{-0.05(3)}} \approx 32$ Sasquatch.
\item $t = 20 \ln(3.167) \approx 23$ years.
\item As $t \rightarrow \infty$, $P(t) \rightarrow 120$. As time goes by, the Sasquatch Population in Bigfoot County will approach 120. Graphically, $y = P(x)$ has a horizontal asymptote $y=120$.
\end{enumerate}
\item \begin{enumerate}
\item $A(t) = Ne^{-\left(\frac{\ln(2)}{5730}\right)t} \approx Ne^{-0.00012097t}$
\item $A(20000) \approx 0.088978 \cdot N$ so about 8.9\% remains
\item $t \approx \dfrac{\ln(.42)}{-0.00012097} \approx 7171$ years old
\end{enumerate}
\item $A(t) = 2.3e^{-0.0138629t}$
\pagebreak
\addtocounter{enumi}{1}
\item \begin{enumerate}
\item $T(t) = 75 + 105e^{-0.005005t}$
\item The roast would have cooled to $140^{\circ}$F in about 95 minutes.
\end{enumerate}
\item From the graph, it appears that as $x \rightarrow 0^{+}$, $y \rightarrow \infty$. This is due to the presence of the $\ln(x)$ term in the function. This means that Fritzy will never catch Chewbacca, which makes sense since Chewbacca has a head start and Fritzy only runs as fast as he does.
\begin{center}
\includegraphics[width=2in]{./ExpLogsGraphics/PURSUIT03.jpg}
\smallskip
$y(x) = \frac{1}{4} x^2-\frac{1}{4} \ln(x)-\frac{1}{4}$
\end{center}
\item The steady state current is 2 amps.
\addtocounter{enumi}{2}
\item The linear regression on the data below is $y = 1.74899x + 0.70739$ with $r^{2} \approx 0.999995$. This is an excellent fit.
\scriptsize
\noindent \begin{tabular}{|l|r|r|r|r|r|r|r|r|r|r|} \hline
$x$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
$\ln(N(x))$ & 2.4849 & 4.1897 & 5.9454 & 7.6967 & 9.4478 & 11.1988 & 12.9497 & 14.7006 & 16.4523 & 18.2025 \\ \hline
\end{tabular}
\normalsize
$N(x) = 2.02869(5.74879)^{x} = 2.02869e^{1.74899x}$ with $r^{2} \approx 0.999995$. This is also an excellent fit and corresponds to our linearized model because $\ln(2.02869) \approx 0.70739$.
\item \begin{enumerate} \item The calculator gives: $y = 2895.06 (1.0147)^{x}$. Graphing this along with our answer from Exercise \ref{PainesvillePopulationTwoPoint} over the interval $[0,60]$ shows that they are pretty close. From this model, $y(150) \approx 25840$ which once again overshoots the actual data value.
\item $P(150) \approx 18717$, so this model predicts 17,914 people in Painesville in 2010, a more conservative number than was recorded in the 2010 census. As $t \rightarrow \infty$, $P(t) \rightarrow 18691$. So the limiting population of Painesville based on this model is 18,691 people.
\enlargethispage{\baselineskip}
\end{enumerate}
\item \begin{enumerate} \item $y = \dfrac{242526}{1+874.62e^{-0.07113x}}$, where $x$ is the number of years since 1860.
\item The plot of the data and the curve is below.
\centerline{\includegraphics[width=1.75in]{./ExpLogsGraphics/LAKECOUNTYLOGISTIC.jpg} }
\item $y(140) \approx 232889$, so this model predicts 232,889 people in Lake County in 2010.
\item As $x \rightarrow \infty$, $y \rightarrow 242526$, so the limiting population of Lake County based on this model is 242,526 people.
\end{enumerate}
\end{enumerate}
\closegraphsfile