105 lines
2.6 KiB
TeX
105 lines
2.6 KiB
TeX
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\section{Logarithms Base 10}
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When we take a logarithm, the resulting number has two parts: the \textit{characteristic} and the \textit{mantissa}. \\
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The characteristic is the integral (whole-numbered) part of the answer, and the mantissa is the fractional part (what comes after the decimal). \\
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\medskip
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For example, $\log_{10}{18} = 1.255$, so in this case the characteristic is $1$ and the mantissa is $0.255$.
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\problem{}
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Approximate the following logs without a slide rule. Find the exact characteristic, and approximate the mantissa.
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\begin{enumerate}
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\item $\log_{10}{20}$
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\item $\log_{2}{18}$
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\end{enumerate}
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\begin{solution}
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\begin{enumerate}
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\item $\log_{10}{20} = 1.30$
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\item $\log_{2}{18} = 4.17$
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\end{enumerate}
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\end{solution}
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\vfill
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Now, find the L scale on your slide rule. As you can see on the right, its generating function is $\log_{10}{x}$.
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\problem{}
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Compute the following logarithms using your slide rule. \\
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You'll have to find the characteristic yourself, but your L scale will give you the mantissa. \\
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Don't forget your log identities!
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\begin{enumerate}
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\item $\log_{10}{20}$
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\item $\log_{10}{15}$
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\item $\log_{10}{150}$
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\item $\log_{10}{0.024}$
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\end{enumerate}
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\begin{solution}
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Careful with number 4.
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\begin{enumerate}
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\item $\log_{10}{20} = 1.30$
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\item $\log_{10}{15} = 1.176$
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\item $\log_{10}{150} = 2.176$
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\item $\log_{10}{0.024} = -1.6197$
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\end{enumerate}
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\end{solution}
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\vfill
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\pagebreak
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%\problem{}
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%Find the following.
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%\begin{enumerate}[itemsep=2mm]
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% \item $\frac{118 \times 0.51}{6.6}$
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% \item $\sqrt{33.8} \times \sqrt[3]{226}$
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% \item $\frac{\sqrt{152}}{\sqrt[3]{4.95}}$
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% \item $\frac{\sqrt{96 \times 250}}{\sqrt{7 \times 0.88}}$
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% \item The area of a circle with radius $1.47$
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% \item The circumference of a circle with radius $31.4$
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% \item The radius of a circle with area $6\pi$
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% \item $\log_{10}{17.38}$
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%\end{enumerate}
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%\vfill
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%\pagebreak
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\section{Logarithms in Any Base}
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Our slide rule easily computes logarithms in base 10, but we can also use it to find logarithms in \textit{any} base.
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\proposition{}<logcob>
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This is usually called the \textit{change-of-base} formula:
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\[
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\log_{b}{a} = \frac{\log_c{a}}{\log_c{b}}
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\]
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\problem{}
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Using log identities, prove \ref{logcob}.
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\vfill
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\problem{}
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Approximate the following:
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\begin{enumerate}
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\item $\log_{2}{56}$
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\item $\log_{5.2}{26}$
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\item $\log_{12}{500}$
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\item $\log_{43}{134}$
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\end{enumerate}
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\begin{solution}
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\begin{enumerate}
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\item $\log_{2}{56} = 5.81$
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\item $\log_{5.2}{26} = 1.97$
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\item $\log_{12}{500} = 2.50$
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\item $\log_{43}{134} = 1.30$
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\end{enumerate}
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\end{solution}
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\vfill
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\pagebreak
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