\begin{frame}
  \frametitle{Review - Midterm Exam 3}
  
  \begin{exampleblock}{}
    A bacteria culture is growing under ideal conditions and doubling every hour.
    If the initial population is $100$ bacteria, 
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    \emph{How many bacteria will there be after half an hour?}
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    The formula for the population is of the form $P(t) = \pause 100\cdot e^{kt}$.
    \pause\smallskip
    
    Let's determine $k$. \pause After $1$h we have 200 bacteria\pause, thus\vspace{-.5ex}
    \begin{talign}
      200 = 100\cdot e^{k\cdot 1} 
      \mpause[1]{\;\implies\; 2 =  e^{k}}
      \mpause{\;\implies\; k = \ln 2}
    \end{talign}\vspace{-2.5ex}
    \pause\pause\pause
    
    Thus $P(t) = 100\cdot e^{t\ln 2} \pause = 100\cdot 2^{t}$.
    \pause
    
    After half an hour we have $\alert{100\cdot 2^{\frac{1}{2}} \approx 141}$ bacteria.
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    \emph{At what rate will the population be increasing at that point?}\\
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    The rate of growth is $P'(t) = 100\cdot \ln 2 \cdot 2^t$.
    \\\pause
    After half an hour the rate of growth is \alert{$100 \cdot \ln 2 \cdot 2^{\frac{1}{2}}$}.
    \pause\medskip
    
    \emph{When will the bacteria population reach 1000?}\vspace{-.5ex}
    \pause
    \begin{talign}
      1000 = 100\cdot e^{t\ln 2} 
      \mpause[1]{ \iff 10 = e^{t\ln 2} }
      \mpause{ \iff \ln 10 = t\ln 2 }
    \end{talign}\vspace{-3ex}
    \pause\pause\pause
    
    Thus after \alert{$t = \ln 10 / \ln 2 \approx 3.3$} hours.
  \end{exampleblock}
\end{frame}