## Exercises - Triangle of Power Notation and Logarithms

1. 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$}\kern.2ex}}}$ $\newcommand{\dtp}[3]{\vphantom{\dotriangle\LARGE}\Rule{-0.1em}{0pt}{2.5ex}_{\scriptstyle #1} {\overset{\scriptstyle #2} {\dotriangle\LARGE}}\Rule{0em}{0pt}{2.5ex}_{\scriptstyle #3}\Rule{0.1em}{0pt}{0ex}}$ $\newcommand{\itp}[3]{\vphantom{\dotriangle\normalsize}\Rule{-0.7ex}{0pt}{1.8ex}_{#1} \overset{#2}{\dotriangle\normalsize}\Rule{0pt}{0pt}{1.8ex}_{#3}\Rule{0ex}{-0.2em}{0pt}}$ $\newcommand{\stp}[3]{\vphantom{\dotriangle\normalsize}\Rule{-0.7ex}{0pt}{1.8ex}_{\scriptstyle #1} {\overset{\scriptstyle #2}{\dotriangle\normalsize}}\Rule{0pt}{0pt}{1.8ex}_{\scriptstyle #3}\Rule{0ex}{-0.2em}{0pt}}$ $\newcommand{\sstp}[3]{\vphantom{\dotriangle\normalsize}\Rule{-0.7ex}{0pt}{1.8ex}_{\scriptstyle #1} {\overset{\scriptstyle #2}{\dotriangle\normalsize}}\Rule{0pt}{0pt}{1.8ex}_{\scriptstyle #3}\Rule{0ex}{-0.2em}{0pt}}$ $\newcommand{\tripow}[3]{\mathop{\mathchoice{\dtp{#1}{#2}{#3}}{\itp{#1}{#2}{#3}}{\stp{#1}{#2}{#3}}{\sstp{#1}{#2}{#3}}}}$ $\newcommand{\vtp}[3]{\vcenter{\tripow{#1}{#2}{#3}}}$

1.   $\tripow{3}{}{9}$

2.   $\tripow{}{2}{25}$

3.   $\tripow{4}{2}{}$

4.   $\tripow{\frac{1}{9}}{}{3}$

5.   $\tripow{}{4}{\frac{1}{16}}$

6.   $\tripow{5}{0}{}$

1.   $2, \quad \log_3 9$

2.   $5, \quad \sqrt{25}$

3.   $16, \quad 4^2$

4.   $-\frac{1}{2}, \quad \log_{\frac{1}{9}} 3$

5.   $\frac{1}{2}, \quad \sqrt[4]{\frac{1}{16}}$

6.   $1, \quad 5^0$

2. Evaluate the following and then rewrite the original expression in triangle-of-power notation.

1.   $5^3$

2.   $\log_7 343$

3.   $\sqrt[4]{81}$

4.   $25^{1/2}$

5.   $\sqrt{\frac{1}{4}}$

6.   $\log_5 \frac{1}{125}$

7.   $\log_b b^3$ for some $b \gt 0$

8.   $(\sqrt[n]{5})^n$

9.   $b^{\log_b 7}$

1.   $125, \quad \vtp{5}{3}{}$

2.   $3, \quad \vtp{7}{}{343}$

3.   $3, \quad \vtp{}{4}{81}$

4.   $5, \quad \vtp{25}{1/2}{}$

5.   $\frac{1}{2}, \quad \vtp{}{2}{\frac{1}{4}}$

6.   $-3, \quad \vtp{5}{}{\frac{1}{125}}$

7.   $3, \quad \displaystyle{\vtp{b}{}{\vtp{b}{3}{}}}$

8.   $5, \quad \displaystyle{\vtp{\vtp{}{n}{5}}{n}{}}$

9.   $7, \quad \displaystyle{\vtp{b}{\vtp{b}{}{7}}{}}$

3. For each logarithmic equation given, rewrite the same relationship in triangle-of-power notation first, and then as an exponential equation.

1.  $\log_3 9 = 2$

2.  $\log_{10} 0.00001 = -5$

3.  $\log_8 4 = \frac{2}{3}$

4.  $\log_5 1 = 0$

5.  $\log_3 c = n$

6.  $\log_{1/2} 4 = -2$

7.  $\log_b \sqrt{b} = \frac{1}{2}$

8.  $\log 256 = x$
(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.)

1.   $\vtp{3}{2}{9}, \quad 3^2 = 9$

2.   $\vtp{10}{-5}{0.00001}, \quad 10^{-5} = 0.00001$

3.   $\vtp{8}{\frac{2}{3}}{4}, \quad 8^{2/3} = 4$

4.   $\vtp{5}{0}{1}, \quad 5^0 = 1$

5.   $\vtp{3}{n}{c}, \quad 3^{n} = c$

6.   $\vtp{\frac{1}{2}}{-2}{4}, \quad \left(\frac{1}{2}\right)^{-2} = 4$

7.   $\vtp{b}{\frac{1}{2}}{\sqrt{b}}, \quad b^{\frac{1}{2}} = \sqrt{b}$

8.   $\vtp{2}{x}{256}, \quad 2^x = 256$

4. For each, write three different equations (one using exponents, one using radicals, and one using logarithms) that all convey the same relationship shown.

1.   $\tripow{4}{y}{64}$

2.   $\tripow{x}{3}{5}$

3.   $\tripow{2}{5}{w}$

1.   $\displaystyle{ \begin{array}{rcl} 64 &=& 4^y\\ y &=& \log_4 64\\ 4 &=& \sqrt[y]{64} \end{array}}$

2.   $\displaystyle{ \begin{array}{rcl} 5 &=& x^3\\ 3 &=& \log_x 5\\ x &=& \sqrt[3]{5} \end{array}}$

3.   $\displaystyle{ \begin{array}{rcl} w &=& 2^5\\ 5 &=& \log_2 w\\ 2 &=& \sqrt[5]{w} \end{array}}$

5. 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.

1. $4^{-1/2} = \frac{1}{2}$

2. $9^0 = 1$

3. $10^y=x$

4. $(\frac{1}{64})^{-1/2} = 8$

5. $36^{-3/2} = \frac{1}{216}$

6. $\log_3 81 = 4$

7. $\log 10 = 1$

8. $\log_x 5 = 2$

9. $\log_2 x = y$

10. $\log_5 \frac{1}{25} = -2$

11. $\log_{16} 2 = \frac{1}{4}$

1. $\log_4 \frac{1}{2} = -\frac{1}{2}$

2. $\log_9 1 = 0$

3. $\log_{10} x = y$

4. $\log_{1/64} 8 = -\frac{1}{2}$

5. $\log_{36} \frac{1}{216} = -\frac{3}{2}$

6. $3^4 = 81$

7. $10^1 = 10$

8. $x^2 = 5$

9. $2^y = x$

10. $5^{-2} = \frac{1}{25}$

11. $16^{1/4} = 2$

6. Find the value of each:

1. $\log_{10} 0.000001$

2. $\log_2 (2^2 + 2^2)$

3. $\log_{64} \frac{1}{32}$

4. $\log_{\frac{1}{2}} 16$

5. $\log_{\frac{5}{2}} \frac{8}{125}$

6. $\log_4 \frac{1}{64}$

7. $\log_7 \sqrt[3]{49}$

8. $\log_{\sqrt{3}} 9$

9. $\log_8 \frac{1}{4}$

10. $\log_6 216$

1. $-6$

2. $3$

3. $-\frac{5}{6}$

4. $-4$

5. $-3$

6. $-3$

7. $\frac{2}{3}$

8. $4$

9. $-\frac{2}{3}$

10. $3$

7. 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 $10$.

1. $\log_4 8$

2. $\log_{1/9} \sqrt{27}$

3. $2^{\log_2 5x}$

4. $\log \cfrac{10^5}{10^3}$

5. $\log_5 (3-2)$

6. $\log_4 3 + 2 \log_4 5$     (write using a single logarithm)

7. $\displaystyle{\frac{1}{2} \log_5 49 - \frac{1}{3} \log_5 8 + 13 \log_5 1}$

8. $10^{2\log 3 - 3\log 2}$

9. $(\log_2 3)(\log_3 4)$

10. $\log 2 + \log 5$

11. $3\log 5 - \frac{1}{2}\log 4 + \log 8$

12. $\log_2 5 + \log_2 5^2 + \log_2 5^3 - \log_2 5^6$

1.     $\log_4 8 = \vtp{4}{}{8} = \vtp{4}{}{2^3} = 3 \left( \vtp{4}{}{2} \right) = 3 \cdot \frac{1}{2} = \frac{3}{2}$

or equivalently: $\log_4 8 = \log_{4} 2^3 = 3\log_{4} 2 = 3 \cdot \frac{1}{2} = \frac{3}{2}$

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: $\log_{\frac{1}{9}} \sqrt{27} = \log_{\frac{1}{9}} 3^{\frac{3}{2}} = \frac{3}{2} \dot \log_{\frac{1}{9}} 3 = \frac{3}{2} \cdot \left(-\frac{1}{2}\right) = -\frac{3}{4}$

3.     $2^{\log_2 5x} = 5x$,     or equivalently: $\displaystyle{\vtp{2}{\vtp{2}{}{5x}}{} = 5x}$

4.     $\log \cfrac{10^5}{10^3} = \log 10^2 = \log_{10} 10^2 = 2$

5.     $\log_5 (3-2) = \log_5 1 = 0$

6.     $\vtp{4}{}{3} + 2 \left(\vtp{4}{}{5}\right) = \vtp{4}{}{3} + \vtp{4}{}{25} = \vtp{4}{}{75}$

or equivalently:   $\log_4 3 + 2 \log_4 5 = \log_4 3 + \log_4 25 = \log_4 75$

7.     $\displaystyle{ \begin{array}[t]{rcl} \frac{1}{2} \left(\vtp{5}{}{49}\right) - \frac{1}{3} \left(\vtp{5}{}{8}\right) + 13 \left(\vtp{5}{}{1}\right) &=& \vtp{5}{}{49^{1/2}} - \vtp{5}{}{8^{1/3}} + 13 \cdot 0\\ &=& \vtp{5}{}{7} - \vtp{5}{}{2}\\ &=& \vtp{5}{}{7/2} \end{array}}$

or equivalently:

$\displaystyle{ \begin{array}[t]{rcl} \frac{1}{2}\log_5 49 - \frac{1}{3}\log_5 8 + 13\log_5 1 &=& \log_5 49^{1/2} - \log_5 8^{1/3} + 13 \cdot 0\\ &=& \log_5 7 - \log_5 2\\ &=& \log_5 \frac{7}{2} \end{array}}$

8.     $10^{2\log 3 - 3\log 2} = 10^{\log 9 - \log 8} = 10^{\log \frac{9}{8}} = \frac{9}{8}$

9.     $(\log_2 3)(\log_3 4) = {\vtp{2}{}{3}} \cdot {\vtp{3}{}{4}} = \vtp{2}{}{3} \cdot {\cfrac{\vtp{2}{}{4}}{\vtp{2}{}{3}}} = \vtp{2}{}{4} = 2$

or equivalently:     $(\log_2 3)(\log_3 4) = (\log_2 3) \cdot \cfrac{\log_2 4}{\log_2 3} = \log_2 4 = 2$

10.     $\log 2 + \log 5 = \log 10 = 1$

11.     $3\log 5 - \frac{1}{2}\log 4 + \log 8 = \log 125 - \log 2 + \log 8 = \log \frac{125}{2} + \log 8 = \log 500$

12.     $\log_2 5 + \log_2 5^2 + \log_2 5^3 - \log_2 5^6 = 1 + 2 + 3 - 6 = 0$

8. 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 $10$)

1. $\log 2$

2. $\log 64$

3. $\log \sqrt{40}$

4. $\log \sqrt[3]{5}$

5. $\log 0.8$

1. $\log 2 = \log 4^{1/2} = \frac{1}{2} \log 4 \doteq \frac{1}{2} \cdot (.6021) = 0.30105$

2. $\log 64 = \log 4^3 = 3 \log 4 \doteq 3 \cdot (.6021) = 1.8063$

3. $\displaystyle{\begin{array}[t]{rcl} \log \sqrt{40} &=& \frac{1}{2} \log 40 = \frac{1}{2} \log (5 \cdot 4^{3/2}) = \frac{1}{2} (\log 5 + \frac{3}{2} \log 4)\\ &\doteq& \frac{1}{2} (0.6990 + \frac{3}{2} (.6021)) = 0.801075 \end{array}}$

4. $\log 5^{1/3} = \frac{1}{3} \log 5 \doteq \frac{1}{3} \cdot (.6990) = 0.233$

5. $\log 0.8 = \log \frac{8}{10} = \log \frac{4}{5} = \log 4 - \log 5 \doteq (.6021 - .6990) = -0.0969$

9. 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 $2$)

1. $\log 21$

2. $\log 343$

3. $\log \ (3 \cdot 7^3)$

4. $\log \frac{49}{3}$

5. $\log \frac{7}{9}$

6. $\log \frac{1}{27\sqrt{7}}$

1. $\log 21 = \log (7 \cdot 3) = \log 7 + \log 3 \doteq 2.8074 + 1.5850 = 4.3924$

2. $\log 343 = \log 7^3 = 3 \log 7 \doteq 3 \cdot 2.8074 = 8.4222$

3. $\log \ (3 \cdot 7^3) = \log 3 + \log 7^3 = \log 3 + 3 \log 7 \doteq 1.5850 + 3 (2.8074) = 10.0072$

4. $\displaystyle{ \begin{array}[t]{rcl} \log \frac{49}{3} &=& \log 49 - \log 3\\ &=& \log 7^2 - \log 3\\ &=& 2 \log 7 - \log 3\\ &\doteq& 2(2.8074)-1.5850\\ &=& 4.0298 \end{array}}$

5. $\displaystyle{ \begin{array}[t]{rcl} \log \frac{7}{9} &=& \log 7 - \log 9 = \log 7 - \log 3^2 = \log 7 - 2\log 3\\ &\doteq& 2.8074 - 2 \cdot (1.5850) = -0.3626 \end{array}}$

6. $\displaystyle{ \begin{array}[t]{rcl} \log \frac{1}{27\sqrt{7}} &=& -\log 27\sqrt{7} = -(\log 27 + \log \sqrt{7}) = -(\log 3^3 + \log 7^{1/2})\\ &=& -(3\log 3 + \frac{1}{2} \log 7) \doteq -(3 \cdot 1.5850 + \frac{1}{2} \cdot 2.8074) = -6.1587 \end{array}}$

10. 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.

11. Find the values of the following:

1. $\ln \sqrt{e}$

2. $\ln \sqrt[3]{e^2}$

3. $\ln (e^2 \cdot e^3)$

4. $\ln (e^2)^3$

5. $\ln \frac{1}{\sqrt[3]{e^2}}$

6. $\ln \left(\frac{e^{3/2}}{e^2 \sqrt{e}} \right)$

7. $\ln \frac{\sqrt{e^3}}{e}$

8. $e^{-\ln 3}$

9. $e^{\frac{1}{2}\ln \frac{1}{16} - \frac{2}{3} \ln 27} - \ln e^{\frac{5}{4}}$

1. $\frac{1}{2}$

2. $\frac{2}{3}$

3. $5$

4. $6$

5. $-\frac{2}{3}$

6. $-1$

7. $\frac{1}{2}$

8. $\frac{1}{3}$

9. $\frac{-11}{9}$

12. 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:

1. $e^y = 3$

2. $e^5 = x$

3. $\ln x = 3$

4. $\ln 3x = -2$

5. $\ln e^2 = 2$

1. $\ln 3 = y$

2. $\ln x = 5$

3. $e^3 = x$

4. $e^{-2} = 3x$

5. $e^2 = e^2$

13. Write each expression using only one logarithm:

1. $3\ln 5 - \frac{1}{2}\ln 4 + \ln 8$
2. $\frac{1}{3} \log_4 x^6 + 5 \log_4 xy - 7 \log_2 x$

1. $\ln 500$

2. $\displaystyle{\frac{1}{2} \log \frac{y^5}{x^7}}$

14. 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: $$\ln \frac{a^3 \sqrt{b}}{c^4 e^6}$$

$3 \ln a + \frac{1}{2} \ln b - 4 \ln c + 6$

15. If the following is written in the form $\log_2 z$, for some expression $z$, find $z$. $$\log_2 x + \log_{\sqrt[3]{2}} y - 5$$

$\displaystyle{z = \frac{xy^3}{32}}$

16. Rewrite the expression below so that the only exponent and/or radical that appers is on $e$ (the natural base). $$\frac{\sqrt[4]{x+1}}{(x+2)^6 \cdot \sqrt{x+3}}$$

$\displaystyle{e^{\left( \frac{1}{4} \ln (x+1) - 6 \ln (x+2) - \frac{1}{2} \ln (x+3) \right)}}$