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When does an extraneous solution occur? How can an extraneous solution be recognized?
When can the one-to-one property of logarithms be used to solve an equation? When can it not be used?
The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.
For the following exercises, use like bases to solve the exponential equation.
${4}^{-3v-2}={4}^{-v}$
$64\cdot {4}^{3x}=16$
$x=-\frac{1}{3}$
${3}^{2x+1}\cdot {3}^{x}=243$
${2}^{-3n}\cdot \frac{1}{4}={2}^{n+2}$
$n=-1$
$625\cdot {5}^{3x+3}=125$
$\frac{{36}^{3b}}{{36}^{2b}}={216}^{2-b}$
$b=\frac{6}{5}$
${\left(\frac{1}{64}\right)}^{3n}\cdot 8={2}^{6}$
For the following exercises, use logarithms to solve.
${9}^{x-10}=1$
$x=10$
$2{e}^{6x}=13$
${e}^{r+10}-10=\mathrm{-42}$
No solution
$2\cdot {10}^{9a}=29$
$-8\cdot {10}^{p+7}-7=\mathrm{-24}$
$p=\mathrm{log}\left(\frac{17}{8}\right)-7$
$7{e}^{3n-5}+5=\mathrm{-89}$
${e}^{-3k}+6=44$
$k=-\frac{\mathrm{ln}\left(38\right)}{3}$
$-5{e}^{9x-8}-8=\mathrm{-62}$
$-6{e}^{9x+8}+2=\mathrm{-74}$
$x=\frac{\mathrm{ln}\left(\frac{38}{3}\right)-8}{9}$
${2}^{x+1}={5}^{2x-1}$
${e}^{2x}-{e}^{x}-132=0$
$x=\mathrm{ln}12\text{}$
$7{e}^{8x+8}-5=\mathrm{-95}$
$10{e}^{8x+3}+2=8$
$x=\frac{\mathrm{ln}\left(\frac{3}{5}\right)-3}{8}$
$4{e}^{3x+3}-7=53$
$8{e}^{-5x-2}-4=\mathrm{-90}$
no solution
${3}^{2x+1}={7}^{x-2}$
${e}^{2x}-{e}^{x}-6=0$
$x=\mathrm{ln}\left(3\right)$
$3{e}^{3-3x}+6=\mathrm{-31}$
For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation.
$\mathrm{log}\left(\frac{1}{100}\right)=\mathrm{-2}$
${10}^{-2}=\frac{1}{100}$
${\mathrm{log}}_{324}\left(18\right)=\frac{1}{2}$
For the following exercises, use the definition of a logarithm to solve the equation.
$5{\mathrm{log}}_{7}n=10$
$n=49$
$-8{\mathrm{log}}_{9}x=16$
$4+{\mathrm{log}}_{2}\left(9k\right)=2$
$k=\frac{1}{36}$
$2\mathrm{log}\left(8n+4\right)+6=10$
$10-4\mathrm{ln}\left(9-8x\right)=6$
$x=\frac{9-e}{8}$
For the following exercises, use the one-to-one property of logarithms to solve.
$\mathrm{ln}\left(10-3x\right)=\mathrm{ln}\left(-4x\right)$
${\mathrm{log}}_{13}\left(5n-2\right)={\mathrm{log}}_{13}\left(8-5n\right)$
$n=1$
$\mathrm{log}\left(x+3\right)-\mathrm{log}\left(x\right)=\mathrm{log}\left(74\right)$
$\mathrm{ln}\left(-3x\right)=\mathrm{ln}\left({x}^{2}-6x\right)$
No solution
${\mathrm{log}}_{4}\left(6-m\right)={\mathrm{log}}_{4}3m$
$\mathrm{ln}\left(x-2\right)-\mathrm{ln}\left(x\right)=\mathrm{ln}\left(54\right)$
No solution
${\mathrm{log}}_{9}\left(2{n}^{2}-14n\right)={\mathrm{log}}_{9}\left(-45+{n}^{2}\right)$
$\mathrm{ln}\left({x}^{2}-10\right)+\mathrm{ln}\left(9\right)=\mathrm{ln}\left(10\right)$
$x=\pm \frac{10}{3}$
For the following exercises, solve each equation for $\text{\hspace{0.17em}}x.$
$\mathrm{log}(x+12)=\mathrm{log}(x)+\mathrm{log}(12)$
$\mathrm{ln}(x)+\mathrm{ln}(x-3)=\mathrm{ln}(7x)$
$x=10$
${\mathrm{log}}_{2}(7x+6)=3$
$\mathrm{ln}\left(7\right)+\mathrm{ln}\left(2-4{x}^{2}\right)=\mathrm{ln}\left(14\right)$
$x=0$
${\mathrm{log}}_{8}\left(x+6\right)-{\mathrm{log}}_{8}\left(x\right)={\mathrm{log}}_{8}\left(58\right)$
$\mathrm{ln}\left(3\right)-\mathrm{ln}\left(3-3x\right)=\mathrm{ln}\left(4\right)$
$x=\frac{3}{4}$
${\mathrm{log}}_{3}\left(3x\right)-{\mathrm{log}}_{3}\left(6\right)={\mathrm{log}}_{3}\left(77\right)$
For the following exercises, solve the equation for $\text{\hspace{0.17em}}x,$ if there is a solution . Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.
${\mathrm{log}}_{9}\left(x\right)-5=\mathrm{-4}$
$x=9$
${\mathrm{log}}_{3}\left(x\right)+3=2$
$\mathrm{ln}\left(3x\right)=2$
$x=\frac{{e}^{2}}{3}\approx 2.5$
$\mathrm{ln}\left(x-5\right)=1$
$\mathrm{log}\left(4\right)+\mathrm{log}\left(-5x\right)=2$
$x=-5$
$-7+{\mathrm{log}}_{3}\left(4-x\right)=\mathrm{-6}$
$\mathrm{ln}\left(4x-10\right)-6=-5$
$x=\frac{e+10}{4}\approx 3.2$
$\mathrm{log}\left(4-2x\right)=\mathrm{log}\left(-4x\right)$
${\mathrm{log}}_{11}\left(-2{x}^{2}-7x\right)={\mathrm{log}}_{11}\left(x-2\right)$
No solution
$\mathrm{ln}\left(2x+9\right)=\mathrm{ln}\left(-5x\right)$
${\mathrm{log}}_{9}\left(3-x\right)={\mathrm{log}}_{9}\left(4x-8\right)$
$x=\frac{11}{5}\approx 2.2$
$\mathrm{log}\left({x}^{2}+13\right)=\mathrm{log}\left(7x+3\right)$
$\frac{3}{{\mathrm{log}}_{2}\left(10\right)}-\mathrm{log}\left(x-9\right)=\mathrm{log}\left(44\right)$
$x=\frac{101}{11}\approx 9.2$
$\mathrm{ln}\left(x\right)-\mathrm{ln}\left(x+3\right)=\mathrm{ln}\left(6\right)$
For the following exercises, solve for the indicated value, and graph the situation showing the solution point.
An account with an initial deposit of $\text{\hspace{0.17em}}\text{\$6,500}\text{\hspace{0.17em}}$ earns $\text{\hspace{0.17em}}7.25\%\text{\hspace{0.17em}}$ annual interest, compounded continuously. How much will the account be worth after 20 years?
about $\text{\hspace{0.17em}}\$27,710.24$
The formula for measuring sound intensity in decibels $\text{\hspace{0.17em}}D\text{\hspace{0.17em}}$ is defined by the equation $\text{\hspace{0.17em}}D=10\mathrm{log}\left(\frac{I}{{I}_{0}}\right),\text{}$ where $\text{\hspace{0.17em}}I\text{\hspace{0.17em}}$ is the intensity of the sound in watts per square meter and $\text{\hspace{0.17em}}{I}_{0}={10}^{-12}\text{\hspace{0.17em}}$ is the lowest level of sound that the average person can hear. How many decibels are emitted from a jet plane with a sound intensity of $\text{\hspace{0.17em}}8.3\cdot {10}^{2}\text{\hspace{0.17em}}$ watts per square meter?
The population of a small town is modeled by the equation $\text{\hspace{0.17em}}P=1650{e}^{0.5t}\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is measured in years. In approximately how many years will the town’s population reach $\text{\hspace{0.17em}}\text{20,000?}$
about 5 years
For the following exercises, solve each equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to 3 decimal places .
$1000{\left(1.03\right)}^{t}=5000\text{\hspace{0.17em}}$ using the common log.
${e}^{5x}=17\text{\hspace{0.17em}}$ using the natural log
$\frac{\mathrm{ln}(17)}{5}\approx 0.567$
$3{\left(1.04\right)}^{3t}=8\text{\hspace{0.17em}}$ using the common log
${3}^{4x-5}=38\text{\hspace{0.17em}}$ using the common log
$x=\frac{\mathrm{log}\left(38\right)+5\mathrm{log}\left(3\right)\text{}}{4\mathrm{log}\left(3\right)}\approx 2.078$
$50{e}^{-0.12t}=10\text{\hspace{0.17em}}$ using the natural log
For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.
$7{e}^{3x-5}+7.9=47$
$x\approx 2.2401$
$\mathrm{ln}\left(3\right)+\mathrm{ln}\left(4.4x+6.8\right)=2$
$\mathrm{log}\left(-0.7x-9\right)=1+5\mathrm{log}\left(5\right)$
$x\approx -\text{44655}.\text{7143}$
Atmospheric pressure $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ in pounds per square inch is represented by the formula $\text{\hspace{0.17em}}P=14.7{e}^{-0.21x},$ where $x$ is the number of miles above sea level. To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of $\text{\hspace{0.17em}}8.369\text{\hspace{0.17em}}$ pounds per square inch? ( Hint : there are 5280 feet in a mile)
The magnitude M of an earthquake is represented by the equation $\text{\hspace{0.17em}}M=\frac{2}{3}\mathrm{log}\left(\frac{E}{{E}_{0}}\right)\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}E\text{\hspace{0.17em}}$ is the amount of energy released by the earthquake in joules and $\text{\hspace{0.17em}}{E}_{0}={10}^{4.4}\text{\hspace{0.17em}}$ is the assigned minimal measure released by an earthquake. To the nearest hundredth, what would the magnitude be of an earthquake releasing $\text{\hspace{0.17em}}1.4\cdot {10}^{13}\text{\hspace{0.17em}}$ joules of energy?
about $\text{\hspace{0.17em}}5.83$
Use the definition of a logarithm along with the one-to-one property of logarithms to prove that $\text{\hspace{0.17em}}{b}^{{\mathrm{log}}_{b}x}=x.$
Recall the formula for continually compounding interest, $\text{\hspace{0.17em}}y=A{e}^{kt}.\text{\hspace{0.17em}}$ Use the definition of a logarithm along with properties of logarithms to solve the formula for time $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is equal to a single logarithm.
$t=\mathrm{ln}\left({\left(\frac{y}{A}\right)}^{\frac{1}{k}}\right)$
Recall the compound interest formula $\text{\hspace{0.17em}}A=a{\left(1+\frac{r}{k}\right)}^{kt}.\text{\hspace{0.17em}}$ Use the definition of a logarithm along with properties of logarithms to solve the formula for time $\text{\hspace{0.17em}}t.$
Newton’s Law of Cooling states that the temperature $\text{\hspace{0.17em}}T\text{\hspace{0.17em}}$ of an object at any time t can be described by the equation $\text{\hspace{0.17em}}T={T}_{s}+\left({T}_{0}-{T}_{s}\right){e}^{-kt},$ where $\text{\hspace{0.17em}}{T}_{s}\text{\hspace{0.17em}}$ is the temperature of the surrounding environment, $\text{\hspace{0.17em}}{T}_{0}\text{\hspace{0.17em}}$ is the initial temperature of the object, and $\text{\hspace{0.17em}}k\text{}$ is the cooling rate. Use the definition of a logarithm along with properties of logarithms to solve the formula for time $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is equal to a single logarithm.
$t=\mathrm{ln}\left({\left(\frac{T-{T}_{s}}{{T}_{0}-{T}_{s}}\right)}^{-\text{\hspace{0.17em}}\frac{1}{k}}\right)$
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