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\problemgrouptitle{Logarithmic Functions}

\begin{problemblock}[3]{Use the properties of logarithms to expand the given expression.}

\item $\displaystyle \log _5  \left( \frac{x+1}{2x+3 } \right)^4$    
\answer{
$4 \left[ \log_5 (x+1)-\log_5(2x+3) \right]$
}

\item $\displaystyle \ln  \frac {(x+1)^2\sqrt{x+4}}{\sqrt[3]{x+2}}$
\answer{
$\displaystyle 2 \ln (x+1)+\frac 12 \ln (x+4)-\frac 13 \ln (x+2)$
}

\item $\displaystyle \ln  \frac {e^{2x}(2x+1)^3}{\sqrt{e^x+1}}$    
\answer{
$\displaystyle 2x+3 \ln (2x+1)-\frac 12 \ln (e^x+1)$
}

\end{problemblock}

\begin{problemblock}{Express the given quantity as a single logarithm.}

\item $\displaystyle \ln (2x)-3\ln(e^x+2)-\frac 12 \ln(x+4)$    
\answer{
$\displaystyle \ln \frac {2x}{(e^x+2)^3\sqrt{x+4}}$
}

\item $\displaystyle 3\log_{10}\left(x^2+1\right)+\frac 13 \log_{10}(x+4)-\frac 14\log_{10}\left(3x^2+5\right))$
\answer{
$\displaystyle\log_{10} \frac{\left(x^2+1\right)^3 \sqrt[3]{x+4}}{\sqrt[4]{3x^2+5}}$
}

\item $\displaystyle 3\ln x+2\log_3 \left(x^3+2\right)-\frac 12 \ln(3x+1)$
\answer{
$\displaystyle\ln \frac{x^3 \left(x^3+2\right)^{2/\ln 3}}{\sqrt {3x+1}}$
}

\end{problemblock}

\begin{problemblock}[2]{Solve for $x$.}

\item $\ln \left(x^2+3\right)=4$   
\answer{
$x=\pm \sqrt{e^4-3}$
}

\item $e^{2x}-5e^x+6=0$   
\answer{
$x=\ln 3,\   x=\ln 2$
}

\item $\ln \left(2x+1\right)+\ln x=\ln \left(x^2+2\right)$   
\answer{
$x=1$
}

\item $\ln \left(e^x-2\right)+\ln e^x=\ln 8$   
\answer{
$x=\ln 4$
}

\item $e^x2^x=5$   
\answer{
$\displaystyle x=\frac{\ln 5}{1+\ln 2}$
}

(Hint: Convert $2^x=\left(e^{\ln 2}\right)^x=e^{x\ln2}$)

\end{problemblock}

\textbook{\clearpage}
\begin{problemblock}[2]{For the following functions find
\begin{enumerate}
\item The domain and the range of $f$, and 
\item $f^{-1}(x)$ and its domain.
\end{enumerate}}

\item $f(x)=\sqrt{2+3e^x}$
\answer{
  \begin{enumerate}
  \item Domain=$\left(-\infty, \infty \right)$,  Range=$\left(\sqrt {2}, \infty\right)$
  \item $f^{-1}(x)=\displaystyle \ln \frac{x^2-2}{3}$, Domain=$\left(\sqrt{2}, \infty \right)$
  \end{enumerate}
}

\item $f(x)=\ln (3x+2)$
\answer{
  \begin{enumerate}
  \item Domain=$\left(- \frac 23, \infty \right)$,  Range=$\left(-\infty, \infty\right)$
  \item $\displaystyle f^{-1}(x)= \frac{e^x-2}{3}$, Domain=$\left(-\infty, \infty \right)$
  \end{enumerate}
}

\item $f(x)=\displaystyle \frac {e^x-2}{e^x+3}$
\answer{
  \begin{enumerate}
  \item Domain=$\left(- \infty, \infty \right)$,  Range=$\left(-\frac 23, 1\right)$
  \item $\displaystyle f^{-1}(x)= \ln \left(\frac{3x+2}{1-x}\right)$, Domain=$\left(-\frac 23, 1\right)$
  \end{enumerate}
}

\item $f(x)=\displaystyle \frac {\ln x+1}{\ln x+2}$
\answer{
  \begin{enumerate}    
  \item Domain=$\mathset{R}-\{e^{-2}\}$,  Range=$\mathset{R}-\{1\}$
  \item $\displaystyle f^{-1}(x)= e^{\frac{1-2x}{x-1}}$, Domain=$\mathset{R}-\{1\}$
  \end{enumerate}
}

\end{problemblock}

\begin{problemblock}[2]{Differentiate the given function.}

\item $f(x)=\ln \left(x^2+e^x \right)$    
\answer{
$\displaystyle f'(x)=\frac {2x+e^x}{x^2+e^x}$
}

\item $y=\log_4  \left(\sin x+3x \right)$    
\answer{
$\displaystyle y'=\frac {\cos x+3}{\left(\ln 4 \right) \left(\sin x+3x\right)}$
}

\item $g(t)=10^{t+2} \ln \left(\ln t+5 \right)$
\answer{
$\displaystyle g'(t)=10^{t+2} \left(\ln 10 \right) \ln \left(\ln t+5 \right)+ \frac {10^{t+2}}{t\left(\ln t+5 \right)}$
}

\item $f(x)=\ln \left[ x^2 \sqrt{x^2+3} \left( e^{3x}+1\right) \right]$
\answer{
$\displaystyle f'(x)= \frac 2x+\frac x{x^2+3}+\frac {3 e^{3x}}{e^{3x}+1}$
}

\item $\displaystyle y=\frac {e^{2x}\sqrt{x^2+5}}{\sqrt[3]{x+1}}$
\answer{
$\displaystyle y'=\frac {e^{2x}\sqrt{x^2+5}}{\sqrt[3]{x+1}} \left[ 2+\frac x{x^2+5}-\frac 1{3(x+1)} \right]$
}

\item $\displaystyle f(x)=\left(x^2+e^x \right)^{\ln x}$
\answer{
$\displaystyle f'(x)=\left(x^2+e^x \right)^{\ln x} \left[ \frac{\ln \left( x^2+e^x \right)}{x}+\frac {\left( 2x+e^x \right) \ln x}{x^2+e^x}\right]$
}

\item $\displaystyle y=x^{\sin x}$
\answer{
$\displaystyle y'=x^{\sin x} \left[\cos x \ln x+\frac {\sin x}{x} \right]$
}

\end{problemblock}

\begin{problemblock}[2]{Evaluate the given integral.}

\item $\displaystyle\int \frac {e^{-2x}}{\sqrt{e^{-2x}+3}} \,dx$    
\answer{
$\displaystyle -\sqrt{e^{-2x}+3}+C$
}

\item $\displaystyle\int \frac {e^x}{e^x+5} \,dx$    
\answer{
$\displaystyle  \ln |e^x+5|+C$
}

\item $\displaystyle\int_0^2 \frac {1}{t+9} \,dt$    
\answer{
$\displaystyle \ln \frac {11}{9}$
}

\item $\displaystyle\int \frac {2}{x \left[2+\ln x \right]^3} \,dx$    
\answer{
$\displaystyle -\frac 1{\left( 2+\ln x \right)^2}+C$
}

\item $\displaystyle\int (2x+1) \hspace {.1in}5^{x^2+x}  \,dx$    
\answer{
$\displaystyle \frac {5^{x^2+x}}{\ln 5}+C$
}

\item $\displaystyle\int_1^4 \frac {10^{\sqrt{x}}} {\sqrt{x}} \,dx$    
\answer{
$\displaystyle \frac {180}{\ln 10}$
}

\item $\displaystyle\int \frac {\sin 2x} {3+\cos 2x}$    
\answer{
$\displaystyle -\frac 12 \ln |3+\cos 2x |+C$
}

\item $\displaystyle\int \frac {3} {x \left ( \ln x \right)^4} \,dx$    
\answer{
$\displaystyle -\frac {1}{\left( \ln x \right)^3}+C$
}

\item $\displaystyle\int \frac {\log_4 x} {x} \,dx$    
\answer{
$\displaystyle \frac {\left( \ln x \right)^2}{2\ln 4}+C$
}

\end{problemblock}

\item
  \begin{enumerate}
  \item Show the (remarkable) result that $a^{\ln x}=x^{\ln a}$.
  \item Using the identity from (a) find the integral $\displaystyle\int 5^{\ln x}\,dx$.
  \item Integrate $\displaystyle \int 5^{\ln x}\,dx$ using the substitution $u=\ln x$.
  \item Is $f(x)=5^{\ln x}$ a power function, an exponential function, or neither?
  \end{enumerate}
  \answer{\begin{enumerate}
    \item $\displaystyle a^{\ln x}=\left[e^{\ln a}\right]^{\ln x}=e^{(\ln a)(\ln x)}=\left[e^{\ln x}\right]^{\ln a}=x^{\ln a}$
    \item $\displaystyle \int 5^{\ln x}\,dx=\int x^{\ln 5}\,dx=\frac{1}{1+\ln 5}x^{1+\ln 5}+C$ (Integration Power Rule with $n=\ln 5$)
    \item $u=\ln x\implies du=\frac{dx}{x}$. But solving gives $x=e^u$ so $dx=e^u\,du$.\\[1em]
      $\displaystyle \int 5^{\ln x}\,dx=\int 5^ue^u\,du=\int [e^{\ln 5}]^ue^u\,du=\int e^{u\ln 5}e^u\,du=\int e^{(1+\ln 5)u}du$\\[1em]
      $\displaystyle =\int e^w\frac{dw}{1+\ln 5}=\frac{e^w}{1+\ln 5}+C=\frac{e^{(1+\ln5)u}}{1+\ln 5}+C=\frac{e^{(1+\ln5)\ln x}}{1+\ln 5}+C=\frac{\left(e^{\ln x}\right)^{1+\ln 5}}{1+\ln 5}+C$\\[1em]
      $\displaystyle=\frac{1}{1+\ln 5}x^{1+\ln 5}+C$
    \item Since $5^{\ln x}=x^{\ln 5}$ it is a power function because the base is the variable $x$ and the constant $n=\ln 5$ is in the exponent.
    \end{enumerate}
    }
  
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