\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}