Skip to main content
Mathematics LibreTexts

1.5.1: Exercises 1.5

  • Page ID
    62723
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    Terms and Concepts

    Exercise \(\PageIndex{1}\)

    Explain the relationship between logarithmic functions and exponential functions.

    Answer

    For the same base, they are inverses of each other.

    Exercise \(\PageIndex{2}\)

    What questions are you answering when you evaluate \(\log_5{(25)}\)?

    Answer

    \(25\) is \(5\) raised to what power?

    Exercise \(\PageIndex{3}\)

    What is the value of the base for \(\ln{(x)}\)?

    Answer

    \(e\)

    Exercise \(\PageIndex{4}\)

    Explain why logarithms help solve exponential statements.

    Answer

    Logarithms help solve exponential statements because logarithms and exponentials are inverse functions.

    Problems

    Evaluate the given statement in exercises \(\PageIndex{5}\) – \(\PageIndex{8}\).

    Exercise \(\PageIndex{5}\)

    \(\displaystyle \log_3{(81)}\)

    Answer

    \(4\)

    Exercise \(\PageIndex{6}\)

    \(\displaystyle \ln{(e^{5.7})}\)

    Answer

    \(5.7\)

    Exercise \(\PageIndex{7}\)

    \(\displaystyle e^{-\ln{(x})}\)

    Answer

    \(\displaystyle \frac{1}{x}\)

    Exercise \(\PageIndex{8}\)

    \(\displaystyle 4^{\log_2{(2^2)}}\)

    Answer

    \(\displaystyle 16\)

    Write the given statement as a single simplified logarithm in exercises \(\PageIndex{9}\) – \(\PageIndex{12}\).

    Exercise \(\PageIndex{9}\)

    \(\displaystyle 4 \log_3{(2x)}-\log_3{(y^2)}\)

    Answer

    \(\displaystyle \log_3{\Big( \frac{16x^4}{y^2}\Big)}\)

    Exercise \(\PageIndex{10}\)

    \(\displaystyle \frac{2}{3} \ln{(x)} + 3\ln{(2y)}\)

    Answer

    \(\displaystyle \ln{(8x^{2/3}y^3)}\)

    Exercise \(\PageIndex{11}\)

    \(\displaystyle (2x)\log_2{(3)}+ \log_2{(5)}\)

    Answer

    \(\displaystyle \log_2{(5 (3^{2x}))}\)

    Exercise \(\PageIndex{12}\)

    \(\displaystyle 3\ln{(xy)}-2\ln{(x^2y)}\)

    Answer

    \(\displaystyle \ln{\Big( \frac{y}{x} \Big)}\)

    In exercises \(\PageIndex{13}\) – \(\PageIndex{17}\), solve the given problem for \(x\), if possible. If a problem cannot be solved, explain why.

    Exercise \(\PageIndex{13}\)

    \(\displaystyle 5^x = 25\)

    Answer

    \(\displaystyle x=2\)

    Exercise \(\PageIndex{14}\)

    \(\displaystyle 5^x = -5\)

    Answer

    Not possible; we cannot raise a positive number to a power and get a negative number.

    Exercise \(\PageIndex{15}\)

    \(\displaystyle 5^x = 0\)

    Answer

    Not possible; we cannot raise a positive number to a power and get a zero.

    Exercise \(\PageIndex{16}\)

    \(\displaystyle 5^x = 0.2\)

    Answer

    \(x=-1\)

    Exercise \(\PageIndex{17}\)

    \(\displaystyle 5^x = 1\)

    Answer

    \(x=0\)

    In exercises \(\PageIndex{18}\) – \(\PageIndex{25}\), solve the given problem for \(x\).

    Exercise \(\PageIndex{18}\)

    \(\displaystyle 3^{x-6} =2\)

    Answer

    \(x=6+\log_3{(2)}\)

    Exercise \(\PageIndex{19}\)

    \(\displaystyle 4^{2x-5} = 3\)

    Answer

    \(\displaystyle x=\frac{5+\log_4{(3)}}{2}\)

    Exercise \(\PageIndex{20}\)

    \(\displaystyle 2^{5x+6} = 4\)

    Answer

    \(\displaystyle x=\frac{-4}{5}\)

    Exercise \(\PageIndex{21}\)

    \(\displaystyle 6^{x+\pi} = 2\)

    Answer

    \(\displaystyle x=\log_6{(2)} - \pi\)

    Exercise \(\PageIndex{22}\)

    \(\displaystyle \bigg( \frac{1}{6}\bigg)^{-3x-2}=36^{x+1}\)

    Answer

    \(\displaystyle x=0\)

    Exercise \(\PageIndex{23}\)

    \(\displaystyle -15=-8\ln{(3x)}+7\)

    Answer

    \(\displaystyle x=\frac{1}{3} e^{11/4}\)

    Exercise \(\PageIndex{24}\)

    \(\displaystyle 2^x=3^{x-1}\)

    Answer

    \(\displaystyle x=-\frac{\ln{(3)}}{\ln{(2)} -\ln{(3)}}\)

    Exercise \(\PageIndex{25}\)

    \(\displaystyle 8= 4\ln{(2x+5)}\)

    Answer

    \(\displaystyle x=\frac{e^2-5}{2}\)


    This page titled 1.5.1: Exercises 1.5 is shared under a CC BY-NC license and was authored, remixed, and/or curated by Amy Givler Chapman, Meagan Herald, Jessica Libertini.

    • Was this article helpful?