\begin{frame}
  \frametitle{Logarithmic Functions: Laws of Logarithm}
  
  \begin{block}{}
    If $x,y > 0$, then
    \begin{enumerate}
    \pause
      \item $\log_a (xy) = \log_a(x) + \log_a(y)$
    \smallskip\pause
      \item $\log_a (\frac{x}{y}) = \log_a(x) - \log_a(y)$
    \smallskip\pause
      \item $\log_a (x^r) = r \log_a x$
    \end{enumerate}
  \end{block}
  \pause\medskip

  \begin{exampleblock}{}
    \begin{malign}
      \log_2 80 - \log_2 5 = \mpause[1]{ \log_2 (\frac{80}{5}) = }\mpause[2]{ \log_2 16 = }\mpause[3]{ 4}
    \end{malign}
  \end{exampleblock}
  \pause\pause\pause
  \medskip
  
  We can proof the laws from the laws for exponents.
  \pause
  \begin{enumerate}
    \item $\log_a (xy) = z \pause \iff a^z = xy$\\[1ex] \pause 
          and \;\;$a^{\log_a(x) + \log_a(y)} = \pause a^{\log_a(x)} \cdot a^{\log_a(y)} = \pause xy$ 
          \medskip
%     \item $\log_a (\frac{x}{y}) = z \pause \iff a^z = \frac{x}{y}$\\[1ex] \pause 
%           and \;\;$a^{\log_a(x) - \log_a(y)} = \pause \frac{a^{\log_a(x)}}{ a^{\log_a(y)}} = \pause \frac{x}{y}$ 
    \item[3.] $\log_a (x^r) = z \pause \iff a^z = x^r$\\[1ex] \pause 
          and \;\;$a^{r\log_a(x)} = \pause (a^{\log_a(x)})^r = \pause x^r$ 
  \end{enumerate}
  \vspace{10cm}
\end{frame}