<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\section{Logarithms Base 10}

When we take a logarithm, the resulting number has two parts: the \textit{characteristic} and the \textit{mantissa}. \\
The characteristic is the integral (whole-numbered) part of the answer, and the mantissa is the fractional part (what comes after the decimal). \\

\medskip

For example, $\log_{10}{18} = 1.255$, so in this case the characteristic is $1$ and the mantissa is $0.255$.

\problem{}
Approximate the following logs without a slide rule. Find the exact characteristic, and approximate the mantissa.
\begin{enumerate}
	\item $\log_{10}{20}$
	\item $\log_{2}{18}$
\end{enumerate}

\begin{solution}
	\begin{enumerate}
		\item $\log_{10}{20} = 1.30$
		\item $\log_{2}{18} = 4.17$
	\end{enumerate}
\end{solution}

\vfill

Now, find the L scale on your slide rule. As you can see on the right, its generating function is $\log_{10}{x}$.

\problem{}
Compute the following logarithms using your slide rule. \\
You'll have to find the characteristic yourself, but your L scale will give you the mantissa. \\
Don't forget your log identities!

\begin{enumerate}
	\item $\log_{10}{20}$
	\item $\log_{10}{15}$
	\item $\log_{10}{150}$
	\item $\log_{10}{0.024}$
\end{enumerate}

\begin{solution}
	Careful with number 4.

	\begin{enumerate}
		\item $\log_{10}{20} = 1.30$
		\item $\log_{10}{15} = 1.176$
		\item $\log_{10}{150} = 2.176$
		\item $\log_{10}{0.024} = -1.6197$
	\end{enumerate}
\end{solution}

\vfill
\pagebreak

%\problem{}
%Find the following.
%\begin{enumerate}[itemsep=2mm]
%	\item $\frac{118 \times 0.51}{6.6}$
%	\item $\sqrt{33.8} \times \sqrt[3]{226}$
%	\item $\frac{\sqrt{152}}{\sqrt[3]{4.95}}$
%	\item $\frac{\sqrt{96 \times 250}}{\sqrt{7 \times 0.88}}$
%	\item The area of a circle with radius $1.47$
%	\item The circumference of a circle with radius $31.4$
%	\item The radius of a circle with area $6\pi$
%	\item $\log_{10}{17.38}$
%\end{enumerate}
%\vfill
%\pagebreak

\section{Logarithms in Any Base}

Our slide rule easily computes logarithms in base 10, but we can also use it to find logarithms in \textit{any} base.

\proposition{}&lt;logcob&gt;
This is usually called the \textit{change-of-base} formula:

\[
	\log_{b}{a} = \frac{\log_c{a}}{\log_c{b}}
\]

\problem{}
Using log identities, prove \ref{logcob}.

\vfill

\problem{}
Approximate the following:
\begin{enumerate}
	\item $\log_{2}{56}$
	\item $\log_{5.2}{26}$
	\item $\log_{12}{500}$
	\item $\log_{43}{134}$
\end{enumerate}

\begin{solution}
	\begin{enumerate}
		\item $\log_{2}{56} = 5.81$
		\item $\log_{5.2}{26} = 1.97$
		\item $\log_{12}{500} = 2.50$
		\item $\log_{43}{134} = 1.30$
	\end{enumerate}
\end{solution}


\vfill
\pagebreak</pre></body></html>