72 lines
1.7 KiB
TeX
72 lines
1.7 KiB
TeX
\section{Logarithms}
|
|
|
|
\definition{}<logdef>
|
|
The \textit{logarithm} is the inverse of the exponent. That is, if $b^p = c$, then $\log_b{c} = p$. \par
|
|
In other words, $\log_b{c}$ asks the question ``what power do I need to raise $b$ to to get $c$?''
|
|
|
|
\vspace{2mm}
|
|
|
|
In both $b^p$ and $\log_b{c}$, the number $b$ is called the \textit{base}.
|
|
|
|
|
|
\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}
|
|
|
|
\pagebreak
|
|
|
|
|
|
|
|
\problem{}<logids>
|
|
Prove the following identities:
|
|
|
|
\begin{enumerate}[itemsep=2mm]
|
|
\item $\log_b{(b^x)} = x$
|
|
\item $b^{\log_b{x}} = x$
|
|
\item $\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$
|
|
\item $\log_b{(\frac{x}{y})} = \log_b{(x)} - \log_b{(y)}$
|
|
\item $\log_b{(x^y)} = y \log_b{(x)}$
|
|
\end{enumerate}
|
|
|
|
\vfill
|
|
|
|
\begin{instructornote}
|
|
A good intro to the following sections is the linear slide rule:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=1]
|
|
\linearscale{2}{1}{}
|
|
\linearscale{0}{0}{}
|
|
|
|
\slideruleind
|
|
{5}
|
|
{1}
|
|
{2 + 3 = 5}
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Take two linear rulers, offset one, and you add. \par
|
|
If you do the same with a log scale, you multiply!
|
|
|
|
\vspace{2mm}
|
|
|
|
Note that the slide rules above start at 0.
|
|
|
|
\linehack{}
|
|
|
|
After assembling the paper slide rule, you can make a visor with some transparent tape. Wrap a strip around the slide rule, sticky side out, and stick it to itself to form a ring. Cover the sticky side with another layer of tape, and trim the edges to make them straight. Use the edge of the visor to read your slide rule!
|
|
\end{instructornote}
|
|
|
|
\pagebreak |