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\begin{coverpages}
\begin{center}{\huge MATH 1112 - EXAM 4 FALL 2016 \\
SOLUTION}
\end{center}

\begin{flushleft}\vspace{0.5in}
Friday 18 November 2016 \\
Instructor: Tom Cuchta
\end{flushleft}

\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}
\vspace*{10pt}


\end{coverpages}

\begin{questions}
	\question[10] Determine if the function is one-to-one. If it is one-to-one, find its inverse function.
	\begin{parts}
		\part[5] $f(x)=(x+2)^2$ \\
		\textit{Solution:} This function is not one-to-one (it does not pass the horizontal line test).
		\part[5] $f(x)=3x+5$ \\
		\textit{Solution:} This function is one-to-one. To find its inverse consider
		$$y=3x+5$$
		and solve for $x$ by subtracting $5$ and dividing by $3$ to get
		$$x = \dfrac{y-5}{3}.$$
		This shows that $f^{-1}(y)=\dfrac{y-5}{3}$. 
	\end{parts}

	\question[10] Sketch the graph of the given function.
		\begin{parts}
			\part[5] $f(x)=e^{x-1}$ \\
			\textit{Solution:} \\
			\begin{tikzpicture}[scale=2]
      \draw[->] (-2,0) -- (2,0) node[right] {$x$};
      \draw[->] (0,-1) -- (0,3) node[above] {$y$};
      \draw[scale=1,domain=-2:2.3,smooth,variable=\x,black,very thick] plot ({\x},{(2.71)^(\x-1)});
      \draw (0.3,0.2) node {$(0,\frac{1}{e})$};
      \draw[fill=black] (0,0.369) circle (0.25ex);
      
      \draw[fill=black] (1,1) circle (0.25ex);
      \draw (1.3,0.8) node {$(1,1)$};
      \draw[fill=black] (2,2.71) circle (0.25ex);
      \draw (2.3,2.71) node {$(2,3)$};
    \end{tikzpicture}
			\part[5] $f(x)=\ln(x)+1$ \\
						\begin{tikzpicture}[scale=1]
      \draw[->] (0,0) -- (5,0) node[right] {$x$};
      \draw[->] (0,-3) -- (0,3) node[above] {$y$};
      \draw[scale=1,domain=0.015:5,smooth,variable=\x,black,very thick] plot ({\x},{ln(\x)+1)});
      \draw (1.3,0.7) node {$(1,1)$};
      \draw[fill=black] (1,1) circle (0.5ex);
      \draw[fill=black] (2.71,2) circle (0.5ex);
      \draw (3,1.65) node {$(e,2)$};
    \end{tikzpicture}
		\end{parts}
\newpage		
	\question[15] Find an exact value for the logarithm.
	\begin{parts}
		\part[5] $\log_4(16)$ \\
		\textit{Solution:} Calculate
		$$\log_4(16) = \log_4(4^2) = 2.$$
		\part[5] $\log_3(27)$ \\
		\textit{Solution:} Calculate
		$$\log_3(27) = \log_3(3^3) = 3.$$
		\part[5] $\ln(\sqrt{e})$ \\
		\textit{Solution:} Calculate
		$$\ln(\sqrt{e}) = \ln(e^{\frac{1}{2}})=\dfrac{1}{2}.$$
	\end{parts}
	
	\question[15] Write as a sum or difference of logarithms.
	\begin{parts}
		\part[7] $\log_a(10x^2y^3)$ \\
		\textit{Solution:} Calculate
		$$\begin{array}{ll}
		\log_a(10x^2y^3) &= \log_a(10) + \log_a(x^2) + \log_a(y^3) \\
		&= \log_a(10) + 2\log_a(x) + 3\log_a(y).
		\end{array}$$
		\part[8] $\log_a \left( \sqrt{\dfrac{a^4 b^3}{3}} \right)$ \\
		\textit{Solution:} Calculate
		$$\begin{array}{ll}
		\log_a \left( \sqrt{ \dfrac{a^4b^3}{3} } \right) &= \log_a \left( \left( \dfrac{a^4b^3}{3} \right)^{\frac{1}{2}} \right) \\
		&=\dfrac{1}{2} \log_a \left( \dfrac{a^4 b^3}{3} \right) \\
		&= \dfrac{1}{2} \Bigg[ \log_a(a^4) + \log_a(b^3) - \log_a(3) \Bigg] \\
		&=\dfrac{1}{2} \Bigg[ 4 + 3 \log_a(b) - \log_a(3) \Bigg]
		\end{array}$$
	\end{parts}
	\question[15] Express as a single logarithm.
	\begin{parts}
		\part[7] $2\log_a(x)+\log_a(y)$ \\
		\textit{Solution:} Calculate
		$$2\log_a(x)+\log_a(y) = \log_a(x^2)+\log_a(y) = \log_a(x^2y).$$
		\part[8] $\dfrac{1}{2} \log_a(x) + 4\log_a(y) - 3\log_a(x)$ \\
		\textit{Solution:} Calculate 
		$$\begin{array}{ll}
		\dfrac{1}{2}\log_a(x) + 4\log_a(y)-3\log_a(x) &= \log_a(\sqrt{x})+\log_a(y^4) - \log_a(x^3) \\
		&= \log_a \left( \dfrac{\sqrt{x}y^4}{x^3} \right).
		\end{array}$$
	\end{parts}
	
	\question[10] Under certain conditions an initial population of $1000$ bacteria grows exponentially with a growth rate of $5\%$ per hour. 
	\begin{parts}
		\part[5] Find the exponential growth function. \\
		\textit{Solution:} Note that $5\% = \dfrac{5}{100}=0.05$. The exponential growth function is 
		$$P(t) = 1000 e^{0.05t}.$$ 
		\part[5] Find the doubling time. \\
		\textit{Solution:} The doubling time is the solution $t$ to the equation
		$$2(1000)=1000e^{0.05t}.$$
		Dividing by $1000$ yields
		$$2 = e^{0.05t}.$$
		Now take $\ln$ of both sides to get
		$$\ln(2) = 0.05t.$$
		Finally, divide by $0.05$ to get
		$$t = \dfrac{\ln(2)}{0.05}.$$
	\end{parts}
	\question[25] Solve the equation.
	\begin{parts}
		\part[6] $13e^{t}=11$ \\
		\textit{Solution:} First divide by $13$ to get
		$$e^t = \dfrac{11}{13}.$$
		Now take $\ln$ of both sides to get
		$$\ln(e^t) = \ln \left( \dfrac{11}{13} \right).$$
		On the left the $\ln$ and the base $e$ cancel to yield
		$$t = \ln \left( \dfrac{11}{13} \right).$$
		\part[6] $\log_2(x+1)=3$ \\
		\textit{Solution:} Plug both sides into $2^x$ to get
		$$2^{\log_2(x+1)}=2^3.$$
		On the left, the $2$ and the $\log_2$ cancel and the right side reduces to $8$ to yield
		$$x+1 = 8.$$
		Therefore $x=7$. 
		\part[6] $\ln(x)-\ln(x-1)=\ln(2)$ \\
		\textit{Solution:} Use the fact that the sum of logs is the log of the product to combine the logarithms on the left, yielding
		$$\ln(x(x-1))=\ln(2).$$
		Now plug both sides into $e^x$ to get
		$$e^{\ln(x(x-1))} = e^{\ln(2)}.$$
		The $e$ and $\ln$ cancel each other on both sides yielding
		$$x(x-1)=2.$$
		This is the equation
		$$x^2-x=2,$$
		or
		$$x^2-x-2=0.$$
		Solve this quadratic any way you wish (it factors or you could use quadratic formula) to get solutions $x=-1$ and $x=2$. (Note that $x=-1$ cannot be plugged into $\ln(x)$, so we should drop that solution).
		\part[7] $2^{x+3} = 5^{x+1}$ \\
		\textit{Solution:} This equation can be solved by taking \textit{any} logarithm. We will provide a solution taking $\log_2$ of both sides. Do this to get
		$$\log_2(2^{x+3}) = \log_2( 5^{x+1}).$$
		On the left, the $2$ and $\log_2$ cancel and on the right pull the power in the front yielding
		$$x+3 = (x+1) \log_2(5).$$
		Now collect all $x$'s on the left and all non-$x$'s on the right:
		$$x - \log_2(5)x = \log_2(5) - 3.$$
		Factor the $x$ on the left side to get
		$$x \left( 1 - \log_2(5) \right) = \log_2(5)-3.$$
		Now divide both sides by $1-\log_2(5)$ to get
		$$x = \dfrac{\log_2(5)-3}{1-\log_2(5)}.$$
		\textit{Note: if you chose a different logarithm, your answer may look different than this!}
		\vfill
	\end{parts}

\end{questions}
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