12.6E: Exercises
- Page ID
- 50020
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Practice Makes Perfect
In the following exercises, solve for \(x\).
- \(\log _{4} 64=2 \log _{4} x\)
- \(\log 49=2 \log x\)
- \(3 \log _{3} x=\log _{3} 27\)
- \(3 \log _{6} x=\log _{6} 64\)
- \(\log _{5}(4 x-2)=\log _{5} 10\)
- \(\log _{3}\left(x^{2}+3\right)=\log _{3} 4 x\)
- \(\log _{3} x+\log _{3} x=2\)
- \(\log _{4} x+\log _{4} x=3\)
- \(\log _{2} x+\log _{2}(x-3)=2\)
- \(\log _{3} x+\log _{3}(x+6)=3\)
- \(\log x+\log (x+3)=1\)
- \(\log x+\log (x-15)=2\)
- \(\log (x+4)-\log (5 x+12)=-\log x\)
- \(\log (x-1)-\log (x+3)=\log \frac{1}{x}\)
- \(\log _{5}(x+3)+\log _{5}(x-6)=\log _{5} 10\)
- \(\log _{5}(x+1)+\log _{5}(x-5)=\log _{5} 7\)
- \(\log _{3}(2 x-1)=\log _{3}(x+3)+\log _{3} 3\)
- \(\log (5 x+1)=\log (x+3)+\log 2\)
- Answer
-
2. \(x=7\)
4. \(x=4\)
6. \(x=1, x=3\)
8. \(x=8\)
10. \(x=3\)
12. \(x=20\)
14. \(x=3\)
16. \(x=6\)
18. \(x=\frac{5}{3}\)
In the following exercises, solve each exponential equation. Find the exact answer and then approximate it to three decimal places.
- \(3^{x}=89\)
- \(2^{x}=74\)
- \(5^{x}=110\)
- \(4^{x}=112\)
- \(e^{x}=16\)
- \(e^{x}=8\)
- \(\left(\frac{1}{2}\right)^{x}=6\)
- \(\left(\frac{1}{3}\right)^{x}=8\)
- \(4 e^{x+1}=16\)
- \(3 e^{x+2}=9\)
- \(6 e^{2 x}=24\)
- \(2 e^{3 x}=32\)
- \(\frac{1}{4} e^{x}=3\)
- \(\frac{1}{3} e^{x}=2\)
- \(e^{x+1}+2=16\)
- \(e^{x-1}+4=12\)
- Answer
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2. \(x=\frac{\log 74}{\log 2} \approx 6.209\)
4. \(x=\frac{\log 112}{\log 4} \approx 3.404\)
6. \(x=\ln 8 \approx 2.079\)
8. \(x=\frac{\log 8}{\log \frac{1}{3}} \approx-1.893\)
10. \(x=\ln 3-2 \approx-0.901\)
12. \(x=\frac{\ln 16}{3} \approx 0.924\)
14. \(x=\ln 6 \approx 1.792\)
16. \(x=\ln 8+1 \approx 3.079\)
In the following exercises, solve each equation.
- \(3^{3 x+1}=81\)
- \(6^{4 x-17}=216\)
- \(\frac{e^{x^{2}}}{e^{14}}=e^{5 x}\)
- \(\frac{e^{x^{2}}}{e^{x}}=e^{20}\)
- \(\log _{a} 64=2\)
- \(\log _{a} 81=4\)
- \(\ln x=-8\)
- \(\ln x=9\)
- \(\log _{5}(3 x-8)=2\)
- \(\log _{4}(7 x+15)=3\)
- \(\ln e^{5 x}=30\)
- \(\ln e^{6 x}=18\)
- \(3 \log x=\log 125\)
- \(7 \log _{3} x=\log _{3} 128\)
- \(\log _{6} x+\log _{6}(x-5)=\log _{6} 24\)
- \(\log _{9} x+\log _{9}(x-4)=\log _{9} 12\)
- \(\log _{2}(x+2)-\log _{2}(2 x+9)=-\log _{2} x\)
- \(\log _{6}(x+1)-\log _{6}(4 x+10)=\log _{6} \frac{1}{x}\)
- Answer
-
2. \(x=5\)
4. \(x=-4, x=5\)
6. \(a=3\)
8. \(x=e^{9}\)
10. \(x=7\)
12. \(x=3\)
14. \(x=2\)
16. \(x=6\)
18. \(x=5\)
In the following exercises, solve for \(x\), giving an exact answer as well as an approximation to three decimal places.
- \(6^{x}=91\)
- \(\left(\frac{1}{2}\right)^{x}=10\)
- \(7 e^{x-3}=35\)
- \(8 e^{x+5}=56\)
- Answer
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2. \(x=\frac{\log 10}{\log \frac{1}{2}} \approx-3.322\)
4. \(x=\ln 7-5 \approx-3.054\)
In the following exercises, solve.
- Sung Lee invests $\(5,000\) at age \(18\). He hopes the investments will be worth $\(10,000\) when he turns \(25\). If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal? Is that a reasonable expectation?
- Alice invests $\(15,000\) at age \(30\) from the signing bonus of her new job. She hopes the investments will be worth $\(30,000\) when she turns \(40\). If the interest compounds continuously, approximately what rate of growth will she need to achieve her goal?
- Coralee invests $\(5,000\) in an account that compounds interest monthly and earns \(7\)%. How long will it take for her money to double?
- Simone invests $\(8,000\) in an account that compounds interest quarterly and earns \(5\)%. How long will it take for his money to double?
- Researchers recorded that a certain bacteria population declined from \(100,000\) to \(100\) in \(24\) hours. At this rate of decay, how many bacteria will there be in \(16\) hours?
- Researchers recorded that a certain bacteria population declined from \(800,000\) to \(500,000\) in \(6\) hours after the administration of medication. At this rate of decay, how many bacteria will there be in \(24\) hours?
- A virus takes \(6\) days to double its original population \(\left(A=2 A_{0}\right)\). How long will it take to triple its population?
- A bacteria doubles its original population in \(24\) hours \(\left(A=2 A_{0}\right)\). How big will its population be in \(72\) hours?
- Carbon-14 is used for archeological carbon dating. Its half-life is \(5,730\) years. How much of a \(100\)-gram sample of Carbon-14 will be left in \(1000\) years?
- Radioactive technetium-99m is often used in diagnostic medicine as it has a relatively short half-life but lasts long enough to get the needed testing done on the patient. If its half-life is \(6\) hours, how much of the radioactive material form a \(0.5\) ml injection will be in the body in \(24\) hours?
- Answer
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2. \(6.9\)%
4. \(13.9\) years
6. \(122,070\) bacteria
8. \(8\) times as large as the original population
10. \(0.03\) mL
- Explain the method you would use to solve these equations: \(3^{x+1}=81\), \(3^{x+1}=75\). Does your method require logarithms for both equations? Why or why not?
- What is the difference between the equation for exponential growth versus the equation for exponential decay?
- Answer
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2. Answers will vary.
Self Check
a. After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
b. After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?