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Intermediate/Slide Rules/parts/6 log.tex

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+\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{}<logcob>
+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