4.5E: Exercises
- Page ID
- 31106
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)4.5: Logarithmic Properties
Verbal
1) How does the power rule for logarithms help when solving logarithms with the form \(\log _b(\sqrt[n]{x})\)
- Answer
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Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, \(\log _b \left ( x^{\frac{1}{n}} \right ) = \dfrac{1}{n}\log_{b}(x)\).
2) What does the change-of-base formula do? Why is it useful when using a calculator?
Algebraic
For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.
3) \(\log _b (7x\cdot 2y)\)
- Answer
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\(\log _b (2)+\log _b (7)+\log _b (x)+\log _b (y)\)
4) \(\ln (3ab\cdot 5c)\)
5) \(\log_b \left ( \dfrac{13}{17} \right )\)
- Answer
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\(\log _b (13)-\log _b (17)\)
6) \(\log_4 \left ( \dfrac{\frac{x}{z}}{w} \right )\)
7) \(\ln \left ( \dfrac{1}{4^k} \right )\)
- Answer
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\(-k\ln(4)\)
8) \(\log _2 (y^x)\)
For the following exercises, condense to a single logarithm if possible.
9) \(\ln (7)+\ln (x)+\ln (y)\)
- Answer
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\(\ln(7xy)\)
10) \(\log_3(2)+\log_3(a)+\log_3(11)+\log_3(b)\)
11) \(\log_b(28)-\log_b(7)\)
- Answer
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\(\log_b(4)\)
12) \(\ln (a)-\ln (d)-\ln (c)\)
13) \(-\log_b\left ( \dfrac{1}{7} \right )\)
- Answer
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\(\log_b(7)\)
14) \(\dfrac{1}{3}\ln(8)\)
For the following exercises, use the properties of logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.
15) \(\log \left ( \dfrac{x^{15}y^{13}}{z^{19}} \right )\)
- Answer
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\(15\log (x)+13\log (y)-19\log (z)\)
16) \(\ln \left ( \frac{a^{-2}}{b^{-4}c^{5}} \right )\)
17) \(\log \left ( \sqrt{x^3y^{-4}} \right )\)
- Answer
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\(\frac{3}{2}\log (x)-2\log (y)\)
18) \(\ln \left ( y\sqrt{\frac{y}{1-y}} \right )\)
19) \(\log \left ( x^2y^3 \sqrt[3]{x^2y^5} \right )\)
- Answer
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\(\dfrac{8}{3}\log (x)+\dfrac{14}{3}\log (y)\)
For the following exercises, condense each expression to a single logarithm using the properties of logarithms.
20) \(\log \left ( 2x^4 \right )+\log \left (3x^5 \right )\)
21) \(\ln \left ( 6x^9 \right )-\ln \left (3x^2 \right )\)
- Answer
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\(\ln \left ( 2x^7 \right )\)
22) \(2\log (x)+3\log (x+1)\)
23) \(\log (x)-\dfrac{1}{2}\log (y)+3\log (z)\)
- Answer
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\(\log \left ( \dfrac{xz^3}{\sqrt{y}} \right )\)
24) \(4\log _7(c)+\dfrac{\log _7(a)}{3}+\dfrac{\log _7(b)}{3}\)
For the following exercises, rewrite each expression as an equivalent ratio of logs using the indicated base.
25) \(\log _7(15)\) to base \(e\)
- Answer
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\(\log _7(15)=\dfrac{\ln (15)}{\ln (7)}\)
26) \(\log _{14}(55.875)\) to base \(10\)
For the following exercises, suppose \(\log _5(6)=a\) and \(\log _5(11)=b\)
Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of \(a\) and \(b\) Show the steps for solving.27) \(\log _{11} (5)\)
- Answer
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\(\log _{11} (5)=\dfrac{\log_5 (5)}{\log_5 (11)}=\dfrac{1}{b}\)
28) \(\log _{6} (55)\)
29) \(\log _{11}\left (\dfrac{6}{11} \right )\)
- Answer
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\(\log _{11}\left (\dfrac{6}{11} \right )=\dfrac{\log _{11}\left (\frac{6}{11} \right )}{\log _{5}(11)}=\dfrac{\log _{5}(6)-\log _{5}(11)}{\log _{5}(11)}=\dfrac{a-b}{b}=\dfrac{a}{b}-1\)
Numeric
For the following exercises, use properties of logarithms to evaluate without using a calculator.
30) \(\log _3 \left ( \dfrac{1}{9} \right )-3\log _3 (3)\)
31) \(6\log _8 (2)+\dfrac{\log _8 (64)}{3\log _8 (4)}\)
- Answer
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\(3\)
32) \(2\log _9 (3)-4\log _9 (3)+\log _9 \left (\dfrac{1}{729} \right )\)
For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places.
33) \(\log _3 (22)\)
- Answer
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\(2.81359\)
34) \(\log _8 (65)\)
35) \(\log _6 (5.38)\)
- Answer
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\(0.93913\)
36) \(\log _4 \left (\dfrac{15}{2} \right )\)
37) \(\log _{\frac{1}{2}} (4.7)\)
- Answer
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\(-2.23266\)
Extensions
38) Use the product rule for logarithms to find all \(x\) values such that \(\log _{12} (2x+6)+\log _{12} (x+2)=2\)Show the steps for solving.
39) Use the quotient rule for logarithms to find all \(x\) values such that \(\log _{6} (x+2)-\log _{6} (x-3)=1\)Show the steps for solving.
- Answer
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Rewriting as an exponential equation and solving for \(x\):
\(\begin{align*}
6^1 &= \frac{x+2}{x-3}\\
0 &= \frac{x+2}{x-3}-6\\
0 &= \frac{x+2}{x-3}-\frac{6(x-3)}{(x-3)}\\
0 &= \frac{x+2-6x+18}{x-3}\\
0 &= \frac{x-4}{x-3}\\
x &= 4
\end{align*}\)Checking, we find that \(\log _6(4+2)-\log _6(4-3)=\log _6(6)-\log _6(1)\) is defined, so \(x=4\)
40) Can the power property of logarithms be derived from the power property of exponents using the equation \(b^x=m\)If not, explain why. If so, show the derivation.
41) Prove that \(\log_b(n)=\frac{1}{\log_b(n)}\) for any positive integers \(b>1\) and \(n>1\)
- Answer
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Let \(b\) and \(n\) be positive integers greater than \(1\)Then, by the change-of-base formula, \(\log_b(n)=\frac{\log_n(n)}{\log_n(b)}=\frac{1}{\log_n(b)}\)
42) Does \(\log_{81}(2401)=\log_3(7)\)Verify the claim algebraically.