\section{Logarithms}

\definition{}<logdef>
The \textit{logarithm} is the inverse of the exponent. That is, if $b^p = c$, then $\log_b{c} = p$. \\
In other words, $\log_b{c}$ asks the question ``what power do I need to raise $b$ to to get $c$?'' \\

\problem{}
Evaluate the following by hand:

\begin{enumerate}
	\item $\log_{10}{(1000)}$
	\vfill
	\item $\log_2{(64)}$
	\vfill
	\item $\log_2{(\frac{1}{4})}$
	\vfill
	\item $\log_x{(x)}$ for any $x$
	\vfill
	\item $log_x{(1)}$ for any $x$
	\vfill
\end{enumerate}

\problem{}<logids>
Prove the following:

\begin{enumerate}[itemsep=2mm]
	\item $\log_b{(b^x)} = x$
	\vfill
	\item $b^{\log_b{x}} = x$
	\vfill
	\item $\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$
	\vfill
	\item $\log_b{(\frac{x}{y})} = \log_b{(x)} - \log_b{(y)}$
	\vfill
	\item $\log_b{(x^y)} = y \log_b{(x)}$
	\vfill
\end{enumerate}


\begin{instructornote}
	A good intro to the following sections is the linear slide rule:
	\note{(note that these rules start at 0)}
	\begin{center}
		\begin{tikzpicture}[scale=0.5]
			\linearscale{2}{1}{}
			\linearscale{0}{0}{}

			\slideruleind
				{5}
				{1}
				{2 + 3 = 5}
	\end{tikzpicture}
	\end{center}

	Take two linear rules, offset one, and you add.
	Do the same with a log scale, and you multiply! \\

	\linehack{}

	After assembling the paper slide rule, you can make a visor with some transparent tape.
\end{instructornote}

\pagebreak