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Evaluate each of the following and then rewrite the original expression in traditional notation.$\newcommand{\dotriangle}[1]{\raise{-0.7ex}{\vcenter{#1 \kern .2ex\hbox{$\triangle$}\kern.2ex}}}$ $\newcommand{\dtp}[3]{\vphantom{\dotriangle\LARGE}\Rule{-0.1em}{0pt}{2.5ex}_{\scriptstyle #1} {\overset{\scriptstyle #2} {\dotriangle\LARGE}}\Rule{0em}{0pt}{2.5ex}_{\scriptstyle #3}\Rule{0.1em}{0pt}{0ex}}$ $\newcommand{\itp}[3]{\vphantom{\dotriangle\normalsize}\Rule{-0.7ex}{0pt}{1.8ex}_{#1} \overset{#2}{\dotriangle\normalsize}\Rule{0pt}{0pt}{1.8ex}_{#3}\Rule{0ex}{-0.2em}{0pt}}$ $\newcommand{\stp}[3]{\vphantom{\dotriangle\normalsize}\Rule{-0.7ex}{0pt}{1.8ex}_{\scriptstyle #1} {\overset{\scriptstyle #2}{\dotriangle\normalsize}}\Rule{0pt}{0pt}{1.8ex}_{\scriptstyle #3}\Rule{0ex}{-0.2em}{0pt}}$ $\newcommand{\sstp}[3]{\vphantom{\dotriangle\normalsize}\Rule{-0.7ex}{0pt}{1.8ex}_{\scriptstyle #1} {\overset{\scriptstyle #2}{\dotriangle\normalsize}}\Rule{0pt}{0pt}{1.8ex}_{\scriptstyle #3}\Rule{0ex}{-0.2em}{0pt}}$ $\newcommand{\tripow}[3]{\mathop{\mathchoice{\dtp{#1}{#2}{#3}}{\itp{#1}{#2}{#3}}{\stp{#1}{#2}{#3}}{\sstp{#1}{#2}{#3}}}}$ $\newcommand{\vtp}[3]{\vcenter{\tripow{#1}{#2}{#3}}}$
$\tripow{3}{}{9}$
$\tripow{}{2}{25}$
$\tripow{4}{2}{}$
$\tripow{\frac{1}{9}}{}{3}$
$\tripow{}{4}{\frac{1}{16}}$
$\tripow{5}{0}{}$
$2, \quad \log_3 9$
$5, \quad \sqrt{25}$
$16, \quad 4^2$
$-\frac{1}{2}, \quad \log_{\frac{1}{9}} 3$
$\frac{1}{2}, \quad \sqrt[4]{\frac{1}{16}}$
$1, \quad 5^0$
Evaluate the following and then rewrite the original expression in triangle-of-power notation.
$5^3$
$\log_7 343$
$\sqrt[4]{81}$
$25^{1/2}$
$\sqrt{\frac{1}{4}}$
$\log_5 \frac{1}{125}$
$\log_b b^3$ for some $b \gt 0$
$(\sqrt[n]{5})^n$
$b^{\log_b 7}$
$125, \quad \vtp{5}{3}{}$
$3, \quad \vtp{7}{}{343}$
$3, \quad \vtp{}{4}{81}$
$5, \quad \vtp{25}{1/2}{}$
$\frac{1}{2}, \quad \vtp{}{2}{\frac{1}{4}}$
$-3, \quad \vtp{5}{}{\frac{1}{125}}$
$3, \quad \displaystyle{\vtp{b}{}{\vtp{b}{3}{}}}$
$5, \quad \displaystyle{\vtp{\vtp{}{n}{5}}{n}{}}$
$7, \quad \displaystyle{\vtp{b}{\vtp{b}{}{7}}{}}$
For each logarithmic equation given, rewrite the same relationship in triangle-of-power notation first, and then as an exponential equation.
$\log_3 9 = 2$
$\log_{10} 0.00001 = -5$
$\log_8 4 = \frac{2}{3}$
$\log_5 1 = 0$
$\log_3 c = n$
$\log_{1/2} 4 = -2$
$\log_b \sqrt{b} = \frac{1}{2}$
$\log 256 = x$
(Notice in this last equation, we have used a "common log", where the base is meant to be understood. So that you use the correct base, suppose the context for this equation is related to computer science.)
$\vtp{3}{2}{9}, \quad 3^2 = 9$
$\vtp{10}{-5}{0.00001}, \quad 10^{-5} = 0.00001$
$\vtp{8}{\frac{2}{3}}{4}, \quad 8^{2/3} = 4$
$\vtp{5}{0}{1}, \quad 5^0 = 1$
$\vtp{3}{n}{c}, \quad 3^{n} = c$
$\vtp{\frac{1}{2}}{-2}{4}, \quad \left(\frac{1}{2}\right)^{-2} = 4$
$\vtp{b}{\frac{1}{2}}{\sqrt{b}}, \quad b^{\frac{1}{2}} = \sqrt{b}$
$\vtp{2}{x}{256}, \quad 2^x = 256$
For each, write three different equations (one using exponents, one using radicals, and one using logarithms) that all convey the same relationship shown.
$\tripow{4}{y}{64}$
$\tripow{x}{3}{5}$
$\tripow{2}{5}{w}$
$\displaystyle{ \begin{array}{rcl} 64 &=& 4^y\\ y &=& \log_4 64\\ 4 &=& \sqrt[y]{64} \end{array}}$
$\displaystyle{ \begin{array}{rcl} 5 &=& x^3\\ 3 &=& \log_x 5\\ x &=& \sqrt[3]{5} \end{array}}$
$\displaystyle{ \begin{array}{rcl} w &=& 2^5\\ 5 &=& \log_2 w\\ 2 &=& \sqrt[5]{w} \end{array}}$
If the statement given is in exponential form, rewrite it in logarithmic form. If instead the statement is in logarithmic form, write it in exponential form.
$4^{-1/2} = \frac{1}{2}$
$9^0 = 1$
$10^y=x$
$(\frac{1}{64})^{-1/2} = 8$
$36^{-3/2} = \frac{1}{216}$
$\log_3 81 = 4$
$\log 10 = 1$
$\log_x 5 = 2$
$\log_2 x = y$
$\log_5 \frac{1}{25} = -2$
$\log_{16} 2 = \frac{1}{4}$
$\log_4 \frac{1}{2} = -\frac{1}{2}$
$\log_9 1 = 0$
$\log_{10} x = y$
$\log_{1/64} 8 = -\frac{1}{2}$
$\log_{36} \frac{1}{216} = -\frac{3}{2}$
$3^4 = 81$
$10^1 = 10$
$x^2 = 5$
$2^y = x$
$5^{-2} = \frac{1}{25}$
$16^{1/4} = 2$
Find the value of each:
$\log_{10} 0.000001$
$\log_2 (2^2 + 2^2)$
$\log_{64} \frac{1}{32}$
$\log_{\frac{1}{2}} 16$
$\log_{\frac{5}{2}} \frac{8}{125}$
$\log_4 \frac{1}{64}$
$\log_7 \sqrt[3]{49}$
$\log_{\sqrt{3}} 9$
$\log_8 \frac{1}{4}$
$\log_6 216$
$-6$
$3$
$-\frac{5}{6}$
$-4$
$-3$
$-3$
$\frac{2}{3}$
$4$
$-\frac{2}{3}$
$3$
Simplify each expression completely. Expressions containing multiple logarithms should be simplified to a single logarithm -- or even more preferably when possible, a simplified expression containing no logarithms. You may use either traditional logarithm notation or triangle-of-power notation, as desired. Assume any common logs encountered involve base $10$.
$\log_4 8$
$\log_{1/9} \sqrt{27}$
$2^{\log_2 5x}$
$\log \cfrac{10^5}{10^3}$
$\log_5 (3-2)$
$\log_4 3 + 2 \log_4 5$ (write using a single logarithm)
$\displaystyle{\frac{1}{2} \log_5 49 - \frac{1}{3} \log_5 8 + 13 \log_5 1}$
$10^{2\log 3 - 3\log 2}$
$(\log_2 3)(\log_3 4)$
$\log 2 + \log 5$
$3\log 5 - \frac{1}{2}\log 4 + \log 8$
$\log_2 5 + \log_2 5^2 + \log_2 5^3 - \log_2 5^6$
$\log_4 8 = \vtp{4}{}{8} = \vtp{4}{}{2^3} = 3 \left( \vtp{4}{}{2} \right) = 3 \cdot \frac{1}{2} = \frac{3}{2}$
or equivalently: $\log_4 8 = \log_{4} 2^3 = 3\log_{4} 2 = 3 \cdot \frac{1}{2} = \frac{3}{2}$
$\log_{\frac{1}{9}} \sqrt{27} = \vtp{\frac{1}{9}}{}{\sqrt{27}} = \vtp{\frac{1}{9}}{}{3^{3/2}} = \frac{3}{2} \left(\vtp{\frac{1}{9}}{}{3} \right) = \frac{3}{2} \cdot (-\frac{1}{2}) = -\frac{3}{4}$
or equivalently: $\log_{\frac{1}{9}} \sqrt{27} = \log_{\frac{1}{9}} 3^{\frac{3}{2}} = \frac{3}{2} \dot \log_{\frac{1}{9}} 3 = \frac{3}{2} \cdot \left(-\frac{1}{2}\right) = -\frac{3}{4}$
$2^{\log_2 5x} = 5x$, or equivalently: $\displaystyle{\vtp{2}{\vtp{2}{}{5x}}{} = 5x}$
$\log \cfrac{10^5}{10^3} = \log 10^2 = \log_{10} 10^2 = 2$
$\log_5 (3-2) = \log_5 1 = 0$
$\vtp{4}{}{3} + 2 \left(\vtp{4}{}{5}\right) = \vtp{4}{}{3} + \vtp{4}{}{25} = \vtp{4}{}{75}$
or equivalently: $\log_4 3 + 2 \log_4 5 = \log_4 3 + \log_4 25 = \log_4 75$
$\displaystyle{ \begin{array}[t]{rcl} \frac{1}{2} \left(\vtp{5}{}{49}\right) - \frac{1}{3} \left(\vtp{5}{}{8}\right) + 13 \left(\vtp{5}{}{1}\right) &=& \vtp{5}{}{49^{1/2}} - \vtp{5}{}{8^{1/3}} + 13 \cdot 0\\ &=& \vtp{5}{}{7} - \vtp{5}{}{2}\\ &=& \vtp{5}{}{7/2} \end{array}}$
or equivalently:
$\displaystyle{ \begin{array}[t]{rcl} \frac{1}{2}\log_5 49 - \frac{1}{3}\log_5 8 + 13\log_5 1 &=& \log_5 49^{1/2} - \log_5 8^{1/3} + 13 \cdot 0\\ &=& \log_5 7 - \log_5 2\\ &=& \log_5 \frac{7}{2} \end{array}}$
$10^{2\log 3 - 3\log 2} = 10^{\log 9 - \log 8} = 10^{\log \frac{9}{8}} = \frac{9}{8}$
$(\log_2 3)(\log_3 4) = {\vtp{2}{}{3}} \cdot {\vtp{3}{}{4}} = \vtp{2}{}{3} \cdot {\cfrac{\vtp{2}{}{4}}{\vtp{2}{}{3}}} = \vtp{2}{}{4} = 2$
or equivalently: $(\log_2 3)(\log_3 4) = (\log_2 3) \cdot \cfrac{\log_2 4}{\log_2 3} = \log_2 4 = 2$
$\log 2 + \log 5 = \log 10 = 1$
$3\log 5 - \frac{1}{2}\log 4 + \log 8 = \log 125 - \log 2 + \log 8 = \log \frac{125}{2} + \log 8 = \log 500$
$\log_2 5 + \log_2 5^2 + \log_2 5^3 - \log_2 5^6 = 1 + 2 + 3 - 6 = 0$
Use only the approximate values $\log 4 \doteq 0.6021$ and $\log 5 \doteq 0.6990$ and the properties of logarithms to approximate the value of each expression below. (You may assume the "common log" here involves base $10$)
$\log 2$
$\log 64$
$\log \sqrt{40}$
$\log \sqrt[3]{5}$
$\log 0.8$
$\log 2 = \log 4^{1/2} = \frac{1}{2} \log 4 \doteq \frac{1}{2} \cdot (.6021) = 0.30105$
$\log 64 = \log 4^3 = 3 \log 4 \doteq 3 \cdot (.6021) = 1.8063$
$\displaystyle{\begin{array}[t]{rcl} \log \sqrt{40} &=& \frac{1}{2} \log 40 = \frac{1}{2} \log (5 \cdot 4^{3/2}) = \frac{1}{2} (\log 5 + \frac{3}{2} \log 4)\\ &\doteq& \frac{1}{2} (0.6990 + \frac{3}{2} (.6021)) = 0.801075 \end{array}}$
$\log 5^{1/3} = \frac{1}{3} \log 5 \doteq \frac{1}{3} \cdot (.6990) = 0.233$
$\log 0.8 = \log \frac{8}{10} = \log \frac{4}{5} = \log 4 - \log 5 \doteq (.6021 - .6990) = -0.0969$
Use only the approximate values $\log 7 \doteq 2.8074$ and $\log 3 \doteq 1.5850$ and the properties of logarithms to approximate the values of the following expressions. (Assume the "common log" here involves base $2$)
$\log 21$
$\log 343$
$\log \ (3 \cdot 7^3)$
$\log \frac{49}{3}$
$\log \frac{7}{9}$
$\log \frac{1}{27\sqrt{7}}$
$\log 21 = \log (7 \cdot 3) = \log 7 + \log 3 \doteq 2.8074 + 1.5850 = 4.3924$
$\log 343 = \log 7^3 = 3 \log 7 \doteq 3 \cdot 2.8074 = 8.4222$
$\log \ (3 \cdot 7^3) = \log 3 + \log 7^3 = \log 3 + 3 \log 7 \doteq 1.5850 + 3 (2.8074) = 10.0072$
$\displaystyle{ \begin{array}[t]{rcl} \log \frac{49}{3} &=& \log 49 - \log 3\\ &=& \log 7^2 - \log 3\\ &=& 2 \log 7 - \log 3\\ &\doteq& 2(2.8074)-1.5850\\ &=& 4.0298 \end{array}}$
$\displaystyle{ \begin{array}[t]{rcl} \log \frac{7}{9} &=& \log 7 - \log 9 = \log 7 - \log 3^2 = \log 7 - 2\log 3\\ &\doteq& 2.8074 - 2 \cdot (1.5850) = -0.3626 \end{array}}$
$\displaystyle{ \begin{array}[t]{rcl} \log \frac{1}{27\sqrt{7}} &=& -\log 27\sqrt{7} = -(\log 27 + \log \sqrt{7}) = -(\log 3^3 + \log 7^{1/2})\\ &=& -(3\log 3 + \frac{1}{2} \log 7) \doteq -(3 \cdot 1.5850 + \frac{1}{2} \cdot 2.8074) = -6.1587 \end{array}}$
Assuming unique factorizations of positive integers into primes, prove that $\log_3 5$ is irrational.
Argue indirectly. Suppose not. Then $\log_3 5$ is rational, meaning there exist integers $p$ and $q$ with $\log_3 5 = \frac{p}{q}$. Hence, $q\log_3 5 = p$ which implies $\log_3 5^q = p$. Writing this in exponential form yields $3^p = 5^q$. As both $3$ and $5$ are primes and $p$ and $q$ are integers, we have just written two different prime factorizations for the same value. This is impossible, so our original assumption that $\log_3 5$ was rational must have been incorrect. Hence, $\log_3 5$ is irrational.
Find the values of the following:
$\ln \sqrt{e}$
$\ln \sqrt[3]{e^2}$
$\ln (e^2 \cdot e^3)$
$\ln (e^2)^3$
$\ln \frac{1}{\sqrt[3]{e^2}}$
$\ln \left(\frac{e^{3/2}}{e^2 \sqrt{e}} \right)$
$\ln \frac{\sqrt{e^3}}{e}$
$e^{-\ln 3}$
$e^{\frac{1}{2}\ln \frac{1}{16} - \frac{2}{3} \ln 27} - \ln e^{\frac{5}{4}}$
$\frac{1}{2}$
$\frac{2}{3}$
$5$
$6$
$-\frac{2}{3}$
$-1$
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{-11}{9}$
If the given statement is in exponential form, write it in logarithmic form. If instead it is in logarithmic form, write it in exponential form:
$e^y = 3$
$e^5 = x$
$\ln x = 3$
$\ln 3x = -2$
$\ln e^2 = 2$
$\ln 3 = y$
$\ln x = 5$
$e^3 = x$
$e^{-2} = 3x$
$e^2 = e^2$
Write each expression using only one logarithm:
$\ln 500$
$\displaystyle{\frac{1}{2} \log \frac{y^5}{x^7}}$
Expand the following expression to a sum and/or difference of multiples of logs of single variables or constant values (i.e., where no variable appears), and with no exponents or radicals present: $$\ln \frac{a^3 \sqrt{b}}{c^4 e^6}$$
$3 \ln a + \frac{1}{2} \ln b - 4 \ln c + 6$
If the following is written in the form $\log_2 z$, for some expression $z$, find $z$. $$\log_2 x + \log_{\sqrt[3]{2}} y - 5$$
$\displaystyle{z = \frac{xy^3}{32}}$
Rewrite the expression below so that the only exponent and/or radical that appers is on $e$ (the natural base). $$\frac{\sqrt[4]{x+1}}{(x+2)^6 \cdot \sqrt{x+3}}$$
$\displaystyle{e^{\left( \frac{1}{4} \ln (x+1) - 6 \ln (x+2) - \frac{1}{2} \ln (x+3) \right)}}$