Cleanup
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% use [nosolutions] flag to hide solutions.
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% use [solutions] flag to show solutions.
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\documentclass[
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solutions
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nosolutions
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]{../../resources/ormc_handout}
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\usepackage{pdfpages}
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@ -32,8 +32,7 @@
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Prepared by Mark on \today
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}
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\vspace{1cm}
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\begin{center}
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\begin{minipage}{6cm}
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Dad says that anyone who can't use
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a slide rule is a cultural illiterate
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@ -41,733 +40,24 @@
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\vspace{1ex}
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\textit{Have Space Suit -- Will Travel, 1958}
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\textit{Have Space Suit --- Will Travel, 1958}
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\end{minipage}
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\end{center}
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\hfill
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\section{Logarithms}
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\definition{}<logdef>
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The \textit{logarithm} is the inverse of the exponent. That is, if $b^p = c$, then $\log_b{c} = p$. \\
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In other words, $\log_b{c}$ asks the question ``what power do I need to raise $b$ to to get $c$?'' \\
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\medskip
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In both $b^p$ and $\log_b{c}$, the number $b$ is called the \textit{base}.
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\problem{}
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Evaluate the following by hand:
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\begin{enumerate}
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\item $\log_{10}{(1000)}$
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\vfill
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\item $\log_2{(64)}$
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\vfill
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\item $\log_2{(\frac{1}{4})}$
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\vfill
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\item $\log_x{(x)}$ for any $x$
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\vfill
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\item $log_x{(1)}$ for any $x$
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\vfill
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\end{enumerate}
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\pagebreak
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\definition{}
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There are a few ways to write logarithms:
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\begin{itemize}
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\item[] $\log{x} = \log_{10}{x}$
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\item[] $\lg{x} = \log_{10}{x}$
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\item[] $\ln{x} = \log_e{x}$
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\end{itemize}
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\definition{}
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The \textit{domain} of a function is the set of values it can take as inputs. \\
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The \textit{range} of a function is the set of values it can produce.
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\medskip
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For example, the domain and range of $f(x) = x$ is $\mathbb{R}$, all real numbers. \\
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The domain of $f(x) = |x|$ is $\mathbb{R}$, and its range is $\mathbb{R}^+ \cup \{0\}$, all positive real numbers and 0. \\
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\medskip
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Note that the domain and range of a function are not always equal.
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\problem{}<expdomain>
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What is the domain of $f(x) = 5^x$? \\
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What is the range of $f(x) = 5^x$?
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\vfill
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\problem{}<logdomain>
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What is the domain of $f(x) = \log{x}$? \\
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What is the range of $f(x) = \log{x}$?
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\vfill
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\pagebreak
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\problem{}<logids>
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Prove the following identities: \\
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\begin{enumerate}[itemsep=2mm]
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\item $\log_b{(b^x)} = x$
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\item $b^{\log_b{x}} = x$
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\item $\log_b{(xy)} = \log_b{(x)} + \log_b{(y)}$
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\item $\log_b{(\frac{x}{y})} = \log_b{(x)} - \log_b{(y)}$
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\item $\log_b{(x^y)} = y \log_b{(x)}$
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\end{enumerate}
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\vfill
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\begin{instructornote}
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A good intro to the following sections is the linear slide rule:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\linearscale{2}{1}{}
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\linearscale{0}{0}{}
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\slideruleind
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{5}
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{1}
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{2 + 3 = 5}
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\end{tikzpicture}
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\end{center}
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Take two linear rulers, offset one, and you add. \\
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If you do the same with a log scale, you multiply! \\
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\vspace{1ex}
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Note that the slide rules above start at 0.
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\linehack{}
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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!
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\end{instructornote}
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\pagebreak
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\section{Introduction}
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Mathematicians, physicists, and engineers needed to quickly solve complex equations even before computers were invented.
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\medskip
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The \textit{slide rule} is an instrument that uses the logarithm to solve this problem. Before you continue, cut out and assemble your slide rule.
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\medskip
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There are four scales on your slide rule, each labeled with a letter on the left side:
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\def\sliderulewidth{13}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\tscale{0}{9}{T}
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\kscale{0}{8}{K}
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\abscale{0}{7}{A}
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\abscale{0}{5.5}{B}
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\ciscale{0}{4.5}{CI}
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\cdscale{0}{3.5}{C}
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\cdscale{0}{2}{D}
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\lscale{0}{1}{L}
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\sscale{0}{0}{S}
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\end{tikzpicture}
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\end{center}
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Each scale's ``generating function'' is on the right:
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\begin{itemize}
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\item T: $\tan$
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\item K: $x^3$
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\item A,B: $x^2$
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\item CI: $\frac{1}{x}$
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\item C, D: $x$
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\item L: $\log_{10}(x)$
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\item S: $\sin$
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\end{itemize}
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Once you understand the layout of your slide rule, move on to the next page.
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\pagebreak
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\section{Multiplication}
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We'll use the C and D scales of your slide rule to multiply. \\
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Say we want to multiply $2 \times 3$. First, move the \textit{left-hand index} of the C scale over the smaller number, $2$:
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\def\sliderulewidth{10}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(2)}{1}{C}
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\cdscale{0}{0}{D}
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\end{tikzpicture}
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\end{center}
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Then we'll find the second number, $3$ on the C scale, and read the D scale under it:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(2)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(6)}
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{1}
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{6}
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\end{tikzpicture}
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\end{center}
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Of course, our answer is 6.
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\problem{}
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What is $1.15 \times 2.1$? \\
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Use your slide rule.
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\begin{solution}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(1.15)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(1.15)}
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{1}
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{1.15}
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\slideruleind
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{\cdscalefn(1.15) + \cdscalefn(2.1)}
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{1}
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{2.415}
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\end{tikzpicture}
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\end{center}
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\end{solution}
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\vfill
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Note that your answer isn't exact. $1.15 \times 2.1 = 2.415$, but an answer accurate within two decimal places is close enough for most practical applications. \\
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\pagebreak
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Look at your C and D scales again. They contain every number between 1 and 10, but no more than that.
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What should we do if we want to calculate $32 \times 210$? \\
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\problem{}
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Using your slide rule, calculate $32 \times 210$. \\
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%\hint{$32 = 3.2 \times 10^1$}
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\begin{solution}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(2.1)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(2.1)}
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{1}
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{2.1}
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\slideruleind
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{\cdscalefn(2.1) + \cdscalefn(3.2)}
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{1}
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{6.72}
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\end{tikzpicture}
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\end{center}
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Placing the decimal point correctly is your job. \\
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$10^1 \times 10^2 = 10^3$, so our final answer is $6.72 \times 10^3 = 672$.
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\end{solution}
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\vfill
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%This method of writing numbers is called \textit{scientific notation}. In the form $a \times 10^b$, $a$ is called the \textit{mantissa}, and $b$, the \textit{exponent}. \\
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%You may also see expressions like $4.3\text{e}2$. This is equivalent to $4.3 \times 10^2$, but is more compact.
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\problem{}
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Compute the following:
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\begin{enumerate}
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\item $1.44 \times 52$
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\item $0.38 \times 1.24$
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\item $\pi \times 2.35$
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\end{enumerate}
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\begin{solution}
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\begin{enumerate}
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\item $1.44 \times 52 = 74.88$
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\item $0.38 \times 1.24 = 0.4712$
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\item $\pi \times 2.35 = 7.382$
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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{}<provemult>
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Note that the numbers on your C and D scales are logarithmically spaced.
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\def\sliderulewidth{13}
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{0}{1}{C}
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\cdscale{0}{0}{D}
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\end{tikzpicture}
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\end{center}
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Why does our multiplication procedure work? \\
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%\hint{See \ref{logids}}
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\vfill
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\pagebreak
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Now we want to compute $7.2 \times 5.5$:
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\def\sliderulewidth{10}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\cdscale{\cdscalefn(5.5)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(5.5)}
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{1}
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{5.5}
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\slideruleind
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{\cdscalefn(5.5) + \cdscalefn(7.2)}
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{1}
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{???}
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\end{tikzpicture}
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\end{center}
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No matter what order we go in, the answer ends up off the scale. There must be another way. \\
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\medskip
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Look at the far right of your C scale. There's an arrow pointing to the $10$ tick, labeled \textit{right-hand index}. Move it over the \textit{larger} number, $7.2$:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(7.2)}
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{1}
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{7.2}
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\end{tikzpicture}
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\end{center}
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Now find the smaller number, $5.5$, on the C scale, and read the D scale under it:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(7.2)}
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{1}
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{7.2}
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\slideruleind
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{\cdscalefn(3.96)}
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{1}
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{3.96}
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\end{tikzpicture}
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\end{center}
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Our answer should be about $7 \times 5 = 35$, so let's move the decimal point: $5.5 \times 7.2 = 39.6$. We can do this by hand to verify our answer. \\
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\medskip
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\iftrue
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\problem{}
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Why does this work?
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\else
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Why does this work? \\
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\medskip
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Consider the following picture, where I've put two D scales next to each other:
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\begin{center}
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\begin{tikzpicture}[scale=0.7]
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\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
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\cdscale{0}{0}{}
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\cdscale{-10}{0}{}
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\draw[
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draw=black,
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]
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(0, 0)
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--
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(0, -0.3)
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node [below] {D};
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\draw[
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draw=black,
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]
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(-10, 0)
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--
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(-10, -0.3)
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node [below] {D};
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\slideruleind
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{-10 + \cdscalefn(7.2)}
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{1}
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{7.2}
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\slideruleind
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{\cdscalefn(7.2)}
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{1}
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{7.2}
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\slideruleind
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{\cdscalefn(3.96)}
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{1}
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{3.96}
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\end{tikzpicture}
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\end{center}
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\medskip
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The second D scale has been moved to the right by $(\log{10})$, so every value on it is $(\log{10})$ smaller than it should be.
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\medskip
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\medskip
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In other words, the answer we get from reverse multiplication is the following: $\log{a} + \log{b} - \log{10}$. \\
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This reduces to $\log{(\frac{a \times b}{10})}$, which explains the misplaced decimal point in $7.2 \times 5.5$.
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\fi
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\vfill
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\pagebreak
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\problem{}
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Compute the following using your slide rule:
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\begin{enumerate}
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\item $9 \times 8$
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\item $15 \times 35$
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\item $42.1 \times 7.65$
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\item $6.5^2$
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\end{enumerate}
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\begin{solution}
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\begin{enumerate}
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\item $9 \times 8 = 72$
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\item $15 \times 35 = 525$
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\item $42.1 \times 7.65 = 322.065$
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\item $6.5^2 = 42.25$
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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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\section{Division}
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Now that you can multiply, division should be easy. All you need to do is work backwards. \\
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Let's look at our first example again: $3 \times 2 = 6$.
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\medskip
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We can easily see that $6 \div 3 = 2$
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(2)}{1}{C}
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\cdscale{0}{0}{D}
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\slideruleind
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{\cdscalefn(6)}
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{1}
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{Align here}
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\slideruleind
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{\cdscalefn(2)}
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{1}
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{2}
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\end{tikzpicture}
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\end{center}
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and that $6 \div 2 = 3$:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\cdscale{\cdscalefn(3)}{-3}{C}
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\cdscale{0}{-4}{D}
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\slideruleind
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{\cdscalefn(6)}
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{-3}
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{Align here}
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\slideruleind
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{\cdscalefn(3)}
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{-3}
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||||
{3}
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||||
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||||
\end{tikzpicture}
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\end{center}
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If your left-hand index is off the scale, read the right-hand one. \\
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Consider $42.25 \div 6.5 = 6.5$:
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\begin{center}
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||||
\begin{tikzpicture}[scale=1]
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||||
\cdscale{\cdscalefn(6.5) - \cdscalefn(10)}{1}{C}
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\cdscale{0}{0}{D}
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||||
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\slideruleind
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||||
{\cdscalefn(4.225)}
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||||
{1}
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||||
{Align here}
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||||
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\slideruleind
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||||
{\cdscalefn(6.5)}
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{1}
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||||
{6.5}
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||||
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\end{tikzpicture}
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||||
\end{center}
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||||
Place your decimal points carefully.
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\vfill
|
||||
\pagebreak
|
||||
|
||||
\problem{}
|
||||
Compute the following using your slide rule. \\
|
||||
|
||||
\begin{enumerate}
|
||||
\item $135 \div 15$
|
||||
\item $68.2 \div 0.575$
|
||||
\item $(118 \times 0.51) \div 6.6$
|
||||
\end{enumerate}
|
||||
|
||||
\begin{solution}
|
||||
\begin{enumerate}
|
||||
\item $135 \div 15 = 9$
|
||||
\item $68.2 \div 0.575 = 118.609$
|
||||
\item $(118 \times 0.51) \div 6.6 = 9.118$
|
||||
\end{enumerate}
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
|
||||
|
||||
\section{Squares, Cubes, and Roots}
|
||||
|
||||
Now, take a look at scales A and B, and note the label on the right: $x^2$. If C, D are $x$, A and B are $x^2$, and K is $x^3$.
|
||||
|
||||
\medskip
|
||||
|
||||
Finding squares of numbers up to ten is straightforward: just read the scale. \\
|
||||
Square roots are also easy: find your number on B and read its pair on C. \\
|
||||
|
||||
|
||||
\def\sliderulewidth{13}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\abscale{0}{1}{B}
|
||||
\cdscale{0}{0}{C}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
\problem{}
|
||||
Compute the following.
|
||||
\begin{enumerate}
|
||||
\item $1.5^2$
|
||||
\item $3.1^2$
|
||||
\item $7^3$
|
||||
\item $\sqrt{14}$
|
||||
\item $\sqrt[3]{150}$
|
||||
\end{enumerate}
|
||||
|
||||
\begin{solution}
|
||||
\begin{enumerate}
|
||||
\item $1.5^2 = 2.25$
|
||||
\item $3.1^2 = 9.61$
|
||||
\item $7^3 = 343$
|
||||
\item $\sqrt{14} = 3.74$
|
||||
\item $\sqrt[3]{150} = 5.313$
|
||||
\end{enumerate}
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
\problem{}
|
||||
Compute the following.
|
||||
\begin{enumerate}
|
||||
\item $42^2$
|
||||
\item $\sqrt{200}$
|
||||
\item $\sqrt{2000}$
|
||||
\item $\sqrt{0.9}$
|
||||
\item $\sqrt[3]{0.12}$
|
||||
\end{enumerate}
|
||||
|
||||
\begin{solution}
|
||||
\begin{enumerate}
|
||||
\item $42^2 = 1,764$
|
||||
\item $\sqrt{200} = 14.14$
|
||||
\item $\sqrt{2000} = 44.72$
|
||||
\item $\sqrt{0.9} = 0.948$
|
||||
\item $\sqrt[3]{0.12} = 0.493$
|
||||
\end{enumerate}
|
||||
\end{solution}
|
||||
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\section{Inverses}
|
||||
|
||||
Try finding $1 \div 32$ using your slide rule. \\
|
||||
The procedure we learned before doesn't work!
|
||||
|
||||
\medskip
|
||||
|
||||
This is why we have the CI scale, or the ``C Inverse'' scale.
|
||||
|
||||
\problem{}
|
||||
Figure out how the CI scale works and compute the following:
|
||||
\begin{enumerate}[itemsep=1mm]
|
||||
\item $\frac{1}{7}$
|
||||
\item $\frac{1}{120}$
|
||||
\item $\frac{1}{\pi}$
|
||||
\end{enumerate}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\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}
|
||||
|
||||
|
||||
|
||||
\input{parts/0 logarithms.tex}
|
||||
\input{parts/1 intro.tex}
|
||||
\input{parts/2 multiplication.tex}
|
||||
\input{parts/3 division.tex}
|
||||
\input{parts/4 squares.tex}
|
||||
\input{parts/5 inverses.tex}
|
||||
\input{parts/6 log.tex}
|
||||
|
||||
% Make sure the slide rule is on an odd page,
|
||||
% so that double-sided printing won't require
|
||||
% students to tear off problems.
|
||||
\checkoddpage
|
||||
\ifoddpage
|
||||
\vfill
|
||||
\pagebreak
|
||||
\ifoddpage\else
|
||||
\vspace*{\fill}
|
||||
\begin{center}
|
||||
{
|
||||
@ -777,13 +67,8 @@
|
||||
\end{center}
|
||||
\vspace{\fill}
|
||||
\pagebreak
|
||||
\else
|
||||
\vfill
|
||||
\pagebreak
|
||||
\fi
|
||||
|
||||
% Slide rule files
|
||||
|
||||
\includepdf[
|
||||
pages=1,
|
||||
fitpaper=true
|
||||
|
||||
103
Intermediate/Slide Rules/parts/0 logarithms.tex
Normal file
103
Intermediate/Slide Rules/parts/0 logarithms.tex
Normal file
@ -0,0 +1,103 @@
|
||||
\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 $c$?'' \\
|
||||
|
||||
\medskip
|
||||
|
||||
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
|
||||
|
||||
|
||||
\definition{}
|
||||
There are a few ways to write logarithms:
|
||||
\begin{itemize}
|
||||
\item[] $\log{x} = \log_{10}{x}$
|
||||
\item[] $\lg{x} = \log_{10}{x}$
|
||||
\item[] $\ln{x} = \log_e{x}$
|
||||
\end{itemize}
|
||||
|
||||
\definition{}
|
||||
The \textit{domain} of a function is the set of values it can take as inputs. \\
|
||||
The \textit{range} of a function is the set of values it can produce.
|
||||
|
||||
\medskip
|
||||
|
||||
For example, the domain and range of $f(x) = x$ is $\mathbb{R}$, all real numbers. \\
|
||||
The domain of $f(x) = |x|$ is $\mathbb{R}$, and its range is $\mathbb{R}^+ \cup \{0\}$, all positive real numbers and 0. \\
|
||||
|
||||
\medskip
|
||||
|
||||
Note that the domain and range of a function are not always equal.
|
||||
|
||||
\problem{}<expdomain>
|
||||
What is the domain of $f(x) = 5^x$? \\
|
||||
What is the range of $f(x) = 5^x$?
|
||||
\vfill
|
||||
|
||||
\problem{}<logdomain>
|
||||
What is the domain of $f(x) = \log{x}$? \\
|
||||
What is the range of $f(x) = \log{x}$?
|
||||
\vfill
|
||||
|
||||
\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. \\
|
||||
If you do the same with a log scale, you multiply! \\
|
||||
\vspace{1ex}
|
||||
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
|
||||
43
Intermediate/Slide Rules/parts/1 intro.tex
Normal file
43
Intermediate/Slide Rules/parts/1 intro.tex
Normal file
@ -0,0 +1,43 @@
|
||||
\section{Introduction}
|
||||
|
||||
Mathematicians, physicists, and engineers needed to quickly solve complex equations even before computers were invented.
|
||||
|
||||
\medskip
|
||||
|
||||
The \textit{slide rule} is an instrument that uses the logarithm to solve this problem. Before you continue, cut out and assemble your slide rule.
|
||||
|
||||
\medskip
|
||||
|
||||
There are four scales on your slide rule, each labeled with a letter on the left side:
|
||||
|
||||
\def\sliderulewidth{13}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\tscale{0}{9}{T}
|
||||
\kscale{0}{8}{K}
|
||||
\abscale{0}{7}{A}
|
||||
|
||||
\abscale{0}{5.5}{B}
|
||||
\ciscale{0}{4.5}{CI}
|
||||
\cdscale{0}{3.5}{C}
|
||||
|
||||
\cdscale{0}{2}{D}
|
||||
\lscale{0}{1}{L}
|
||||
\sscale{0}{0}{S}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
Each scale's ``generating function'' is on the right:
|
||||
\begin{itemize}
|
||||
\item T: $\tan$
|
||||
\item K: $x^3$
|
||||
\item A,B: $x^2$
|
||||
\item CI: $\frac{1}{x}$
|
||||
\item C, D: $x$
|
||||
\item L: $\log_{10}(x)$
|
||||
\item S: $\sin$
|
||||
\end{itemize}
|
||||
|
||||
Once you understand the layout of your slide rule, move on to the next page.
|
||||
|
||||
\pagebreak
|
||||
284
Intermediate/Slide Rules/parts/2 multiplication.tex
Normal file
284
Intermediate/Slide Rules/parts/2 multiplication.tex
Normal file
@ -0,0 +1,284 @@
|
||||
\section{Multiplication}
|
||||
|
||||
We'll use the C and D scales of your slide rule to multiply. \\
|
||||
|
||||
Say we want to multiply $2 \times 3$. First, move the \textit{left-hand index} of the C scale over the smaller number, $2$:
|
||||
|
||||
\def\sliderulewidth{10}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\cdscale{\cdscalefn(2)}{1}{C}
|
||||
\cdscale{0}{0}{D}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
Then we'll find the second number, $3$ on the C scale, and read the D scale under it:
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\cdscale{\cdscalefn(2)}{1}{C}
|
||||
\cdscale{0}{0}{D}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(6)}
|
||||
{1}
|
||||
{6}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
Of course, our answer is 6.
|
||||
|
||||
\problem{}
|
||||
What is $1.15 \times 2.1$? \\
|
||||
Use your slide rule.
|
||||
|
||||
\begin{solution}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\cdscale{\cdscalefn(1.15)}{1}{C}
|
||||
\cdscale{0}{0}{D}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(1.15)}
|
||||
{1}
|
||||
{1.15}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(1.15) + \cdscalefn(2.1)}
|
||||
{1}
|
||||
{2.415}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
|
||||
Note that your answer isn't exact. $1.15 \times 2.1 = 2.415$, but an answer accurate within two decimal places is close enough for most practical applications. \\
|
||||
|
||||
\pagebreak
|
||||
|
||||
Look at your C and D scales again. They contain every number between 1 and 10, but no more than that.
|
||||
What should we do if we want to calculate $32 \times 210$? \\
|
||||
|
||||
\problem{}
|
||||
Using your slide rule, calculate $32 \times 210$. \\
|
||||
%\hint{$32 = 3.2 \times 10^1$}
|
||||
|
||||
\begin{solution}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\cdscale{\cdscalefn(2.1)}{1}{C}
|
||||
\cdscale{0}{0}{D}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(2.1)}
|
||||
{1}
|
||||
{2.1}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(2.1) + \cdscalefn(3.2)}
|
||||
{1}
|
||||
{6.72}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
Placing the decimal point correctly is your job. \\
|
||||
$10^1 \times 10^2 = 10^3$, so our final answer is $6.72 \times 10^3 = 672$.
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
|
||||
%This method of writing numbers is called \textit{scientific notation}. In the form $a \times 10^b$, $a$ is called the \textit{mantissa}, and $b$, the \textit{exponent}. \\
|
||||
|
||||
%You may also see expressions like $4.3\text{e}2$. This is equivalent to $4.3 \times 10^2$, but is more compact.
|
||||
|
||||
|
||||
\problem{}
|
||||
Compute the following:
|
||||
\begin{enumerate}
|
||||
\item $1.44 \times 52$
|
||||
\item $0.38 \times 1.24$
|
||||
\item $\pi \times 2.35$
|
||||
\end{enumerate}
|
||||
|
||||
\begin{solution}
|
||||
\begin{enumerate}
|
||||
\item $1.44 \times 52 = 74.88$
|
||||
\item $0.38 \times 1.24 = 0.4712$
|
||||
\item $\pi \times 2.35 = 7.382$
|
||||
\end{enumerate}
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\problem{}<provemult>
|
||||
Note that the numbers on your C and D scales are logarithmically spaced.
|
||||
|
||||
\def\sliderulewidth{13}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\cdscale{0}{1}{C}
|
||||
\cdscale{0}{0}{D}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
Why does our multiplication procedure work? \\
|
||||
%\hint{See \ref{logids}}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
Now we want to compute $7.2 \times 5.5$:
|
||||
|
||||
\def\sliderulewidth{10}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.8]
|
||||
\cdscale{\cdscalefn(5.5)}{1}{C}
|
||||
\cdscale{0}{0}{D}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(5.5)}
|
||||
{1}
|
||||
{5.5}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(5.5) + \cdscalefn(7.2)}
|
||||
{1}
|
||||
{???}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
No matter what order we go in, the answer ends up off the scale. There must be another way. \\
|
||||
|
||||
\medskip
|
||||
|
||||
Look at the far right of your C scale. There's an arrow pointing to the $10$ tick, labeled \textit{right-hand index}. Move it over the \textit{larger} number, $7.2$:
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
|
||||
\cdscale{0}{0}{D}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(7.2)}
|
||||
{1}
|
||||
{7.2}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
Now find the smaller number, $5.5$, on the C scale, and read the D scale under it:
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
|
||||
\cdscale{0}{0}{D}
|
||||
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(7.2)}
|
||||
{1}
|
||||
{7.2}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(3.96)}
|
||||
{1}
|
||||
{3.96}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
Our answer should be about $7 \times 5 = 35$, so let's move the decimal point: $5.5 \times 7.2 = 39.6$. We can do this by hand to verify our answer. \\
|
||||
|
||||
\medskip
|
||||
|
||||
\iftrue
|
||||
\problem{}
|
||||
Why does this work?
|
||||
|
||||
\else
|
||||
Why does this work? \\
|
||||
|
||||
\medskip
|
||||
|
||||
Consider the following picture, where I've put two D scales next to each other:
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.7]
|
||||
\cdscale{\cdscalefn(7.2) - \cdscalefn(10)}{1}{C}
|
||||
\cdscale{0}{0}{}
|
||||
\cdscale{-10}{0}{}
|
||||
|
||||
\draw[
|
||||
draw=black,
|
||||
]
|
||||
(0, 0)
|
||||
--
|
||||
(0, -0.3)
|
||||
node [below] {D};
|
||||
|
||||
\draw[
|
||||
draw=black,
|
||||
]
|
||||
(-10, 0)
|
||||
--
|
||||
(-10, -0.3)
|
||||
node [below] {D};
|
||||
|
||||
\slideruleind
|
||||
{-10 + \cdscalefn(7.2)}
|
||||
{1}
|
||||
{7.2}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(7.2)}
|
||||
{1}
|
||||
{7.2}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(3.96)}
|
||||
{1}
|
||||
{3.96}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
\medskip
|
||||
|
||||
The second D scale has been moved to the right by $(\log{10})$, so every value on it is $(\log{10})$ smaller than it should be.
|
||||
|
||||
\medskip
|
||||
|
||||
\medskip
|
||||
In other words, the answer we get from reverse multiplication is the following: $\log{a} + \log{b} - \log{10}$. \\
|
||||
This reduces to $\log{(\frac{a \times b}{10})}$, which explains the misplaced decimal point in $7.2 \times 5.5$.
|
||||
\fi
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\problem{}
|
||||
Compute the following using your slide rule:
|
||||
\begin{enumerate}
|
||||
\item $9 \times 8$
|
||||
\item $15 \times 35$
|
||||
\item $42.1 \times 7.65$
|
||||
\item $6.5^2$
|
||||
\end{enumerate}
|
||||
|
||||
\begin{solution}
|
||||
\begin{enumerate}
|
||||
\item $9 \times 8 = 72$
|
||||
\item $15 \times 35 = 525$
|
||||
\item $42.1 \times 7.65 = 322.065$
|
||||
\item $6.5^2 = 42.25$
|
||||
\end{enumerate}
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
92
Intermediate/Slide Rules/parts/3 division.tex
Normal file
92
Intermediate/Slide Rules/parts/3 division.tex
Normal file
@ -0,0 +1,92 @@
|
||||
\section{Division}
|
||||
|
||||
Now that you can multiply, division should be easy. All you need to do is work backwards. \\
|
||||
Let's look at our first example again: $3 \times 2 = 6$.
|
||||
|
||||
\medskip
|
||||
|
||||
We can easily see that $6 \div 3 = 2$
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\cdscale{\cdscalefn(2)}{1}{C}
|
||||
\cdscale{0}{0}{D}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(6)}
|
||||
{1}
|
||||
{Align here}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(2)}
|
||||
{1}
|
||||
{2}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
and that $6 \div 2 = 3$:
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\cdscale{\cdscalefn(3)}{-3}{C}
|
||||
\cdscale{0}{-4}{D}
|
||||
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(6)}
|
||||
{-3}
|
||||
{Align here}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(3)}
|
||||
{-3}
|
||||
{3}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
If your left-hand index is off the scale, read the right-hand one. \\
|
||||
Consider $42.25 \div 6.5 = 6.5$:
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\cdscale{\cdscalefn(6.5) - \cdscalefn(10)}{1}{C}
|
||||
\cdscale{0}{0}{D}
|
||||
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(4.225)}
|
||||
{1}
|
||||
{Align here}
|
||||
|
||||
\slideruleind
|
||||
{\cdscalefn(6.5)}
|
||||
{1}
|
||||
{6.5}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
Place your decimal points carefully.
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\problem{}
|
||||
Compute the following using your slide rule. \\
|
||||
|
||||
\begin{enumerate}
|
||||
\item $135 \div 15$
|
||||
\item $68.2 \div 0.575$
|
||||
\item $(118 \times 0.51) \div 6.6$
|
||||
\end{enumerate}
|
||||
|
||||
\begin{solution}
|
||||
\begin{enumerate}
|
||||
\item $135 \div 15 = 9$
|
||||
\item $68.2 \div 0.575 = 118.609$
|
||||
\item $(118 \times 0.51) \div 6.6 = 9.118$
|
||||
\end{enumerate}
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
62
Intermediate/Slide Rules/parts/4 squares.tex
Normal file
62
Intermediate/Slide Rules/parts/4 squares.tex
Normal file
@ -0,0 +1,62 @@
|
||||
\section{Squares, Cubes, and Roots}
|
||||
|
||||
Now, take a look at scales A and B, and note the label on the right: $x^2$. If C, D are $x$, A and B are $x^2$, and K is $x^3$.
|
||||
|
||||
\medskip
|
||||
|
||||
Finding squares of numbers up to ten is straightforward: just read the scale. \\
|
||||
Square roots are also easy: find your number on B and read its pair on C. \\
|
||||
|
||||
|
||||
\def\sliderulewidth{13}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\abscale{0}{1}{B}
|
||||
\cdscale{0}{0}{C}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
\problem{}
|
||||
Compute the following.
|
||||
\begin{enumerate}
|
||||
\item $1.5^2$
|
||||
\item $3.1^2$
|
||||
\item $7^3$
|
||||
\item $\sqrt{14}$
|
||||
\item $\sqrt[3]{150}$
|
||||
\end{enumerate}
|
||||
|
||||
\begin{solution}
|
||||
\begin{enumerate}
|
||||
\item $1.5^2 = 2.25$
|
||||
\item $3.1^2 = 9.61$
|
||||
\item $7^3 = 343$
|
||||
\item $\sqrt{14} = 3.74$
|
||||
\item $\sqrt[3]{150} = 5.313$
|
||||
\end{enumerate}
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
\problem{}
|
||||
Compute the following.
|
||||
\begin{enumerate}
|
||||
\item $42^2$
|
||||
\item $\sqrt{200}$
|
||||
\item $\sqrt{2000}$
|
||||
\item $\sqrt{0.9}$
|
||||
\item $\sqrt[3]{0.12}$
|
||||
\end{enumerate}
|
||||
|
||||
\begin{solution}
|
||||
\begin{enumerate}
|
||||
\item $42^2 = 1,764$
|
||||
\item $\sqrt{200} = 14.14$
|
||||
\item $\sqrt{2000} = 44.72$
|
||||
\item $\sqrt{0.9} = 0.948$
|
||||
\item $\sqrt[3]{0.12} = 0.493$
|
||||
\end{enumerate}
|
||||
\end{solution}
|
||||
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
19
Intermediate/Slide Rules/parts/5 inverses.tex
Normal file
19
Intermediate/Slide Rules/parts/5 inverses.tex
Normal file
@ -0,0 +1,19 @@
|
||||
\section{Inverses}
|
||||
|
||||
Try finding $1 \div 32$ using your slide rule. \\
|
||||
The procedure we learned before doesn't work!
|
||||
|
||||
\medskip
|
||||
|
||||
This is why we have the CI scale, or the ``C Inverse'' scale.
|
||||
|
||||
\problem{}
|
||||
Figure out how the CI scale works and compute the following:
|
||||
\begin{enumerate}[itemsep=1mm]
|
||||
\item $\frac{1}{7}$
|
||||
\item $\frac{1}{120}$
|
||||
\item $\frac{1}{\pi}$
|
||||
\end{enumerate}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
105
Intermediate/Slide Rules/parts/6 log.tex
Normal file
105
Intermediate/Slide Rules/parts/6 log.tex
Normal file
@ -0,0 +1,105 @@
|
||||
\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
|
||||
Loading…
x
Reference in New Issue
Block a user