4.5 Exponential and logarithmic equations  (Page 7/8)

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?

$4^{-3v-2}=4^{-v}$

$64\cdot 4^{3x}=16$

$3^{2x+1}\cdot 3^x=243$

$2^{-3n}\cdot \frac{1}{4}=2^{n+2}$

$625\cdot 5^{3x+3}=125$

$\frac{{36}^{3b}}{{36}^{2b}}={216}^{2-b}$

${\left(\frac{1}{64}\right)}^{3n}\cdot 8=2^6$

$9^{x-10}=1$

$2e^{6x}=13$

$e^{r+10}-10=\mathrm{-42}$

$2\cdot {10}^{9a}=29$

$-8\cdot {10}^{p+7}-7=\mathrm{-24}$

$7e^{3n-5}+5=\mathrm{-89}$

$e^{-3k}+6=44$

$-5e^{9x-8}-8=\mathrm{-62}$

$-6e^{9x+8}+2=\mathrm{-74}$

$2^{x+1}=5^{2x-1}$

$e^{2x}-e^x-132=0$

$7e^{8x+8}-5=\mathrm{-95}$

$10e^{8x+3}+2=8$

$4e^{3x+3}-7=53$

$8e^{-5x-2}-4=\mathrm{-90}$

$3^{2x+1}=7^{x-2}$

$e^{2x}-e^x-6=0$

$3e^{3-3x}+6=\mathrm{-31}$

$\log \left(\frac{1}{100}\right)=\mathrm{-2}$

${\log }_{324}\left(18\right)=\frac{1}{2}$

$5{\log }_{7}n=10$

$-8{\log }_{9}x=16$

$4+{\log }_2\left(9k\right)=2$

$2\log \left(8n+4\right)+6=10$

$10-4\ln \left(9-8x\right)=6$

$\ln \left(10-3x\right)=\ln \left(-4x\right)$

${\log }_{13}\left(5n-2\right)={\log }_{13}\left(8-5n\right)$

$\log \left(x+3\right)-\log \left(x\right)=\log \left(74\right)$

$\ln \left(-3x\right)=\ln \left(x^2-6x\right)$

${\log }_4\left(6-m\right)={\log }_{4}3m$

$\ln \left(x-2\right)-\ln \left(x\right)=\ln \left(54\right)$

${\log }_9\left(2n^2-14n\right)={\log }_9\left(-45+n^2\right)$

$\ln \left(x^2-10\right)+\ln \left(9\right)=\ln \left(10\right)$

$\log (x+12)=\log (x)+\log (12)$

$\ln (x)+\ln (x-3)=\ln (7x)$

${\log }_2(7x+6)=3$

$\ln \left(7\right)+\ln \left(2-4x^2\right)=\ln \left(14\right)$

${\log }_8\left(x+6\right)-{\log }_8\left(x\right)={\log }_8\left(58\right)$

$\ln \left(3\right)-\ln \left(3-3x\right)=\ln \left(4\right)$

${\log }_3\left(3x\right)-{\log }_3\left(6\right)={\log }_3\left(77\right)$

${\log }_9\left(x\right)-5=\mathrm{-4}$

${\log }_3\left(x\right)+3=2$

$\ln \left(3x\right)=2$

$\ln \left(x-5\right)=1$

$\log \left(4\right)+\log \left(-5x\right)=2$

$-7+{\log }_3\left(4-x\right)=\mathrm{-6}$

$\ln \left(4x-10\right)-6=-5$

$\log \left(4-2x\right)=\log \left(-4x\right)$

${\log }_{11}\left(-2x^2-7x\right)={\log }_{11}\left(x-2\right)$

$\ln \left(2x+9\right)=\ln \left(-5x\right)$

${\log }_9\left(3-x\right)={\log }_9\left(4x-8\right)$

$\log \left(x^2+13\right)=\log \left(7x+3\right)$

$\frac{3}{{\log }_2\left(10\right)}-\log \left(x-9\right)=\log \left(44\right)$

$\ln \left(x\right)-\ln \left(x+3\right)=\ln \left(6\right)$

An account with an initial deposit of $\text{\$6,500}$ earns $7.25\%$ annual interest, compounded continuously. How much will the account be worth after 20 years?

The formula for measuring sound intensity in decibels $D$ is defined by the equation $D=10\log \left(\frac{I}{I_0}\right),$ where $I$ is the intensity of the sound in watts per square meter and $I_0={10}^{-12}$ 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 $8.3\cdot {10}^2$ watts per square meter?

The population of a small town is modeled by the equation $P=1650e^{0.5t}$ where $t$ is measured in years. In approximately how many years will the town’s population reach $\text{20,000?}$

$1000{\left(1.03\right)}^t=5000$ using the common log.

$e^{5x}=17$ using the natural log

$3{\left(1.04\right)}^{3t}=8$ using the common log

$3^{4x-5}=38$ using the common log

$50e^{-0.12t}=10$ using the natural log

$7e^{3x-5}+7.9=47$

$\ln \left(3\right)+\ln \left(4.4x+6.8\right)=2$

$\log \left(-0.7x-9\right)=1+5\log \left(5\right)$

Atmospheric pressure $P$ in pounds per square inch is represented by the formula $P=14.7e^{-0.21x},$ 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 $8.369$ pounds per square inch? ( Hint : there are 5280 feet in a mile)

The magnitude M of an earthquake is represented by the equation $M=\frac{2}{3}\log \left(\frac{E}{E_0}\right)$ where $E$ is the amount of energy released by the earthquake in joules and $E_0={10}^{4.4}$ is the assigned minimal measure released by an earthquake. To the nearest hundredth, what would the magnitude be of an earthquake releasing $1.4\cdot {10}^{13}$ joules of energy?

Use the definition of a logarithm along with the one-to-one property of logarithms to prove that $b^{{\log }_{b}x}=x.$

Recall the formula for continually compounding interest, $y=A{e}^{kt}.$ Use the definition of a logarithm along with properties of logarithms to solve the formula for time $t$ such that $t$ is equal to a single logarithm.

Recall the compound interest formula $A=a{\left(1+\frac{r}{k}\right)}^{kt}.$ Use the definition of a logarithm along with properties of logarithms to solve the formula for time $t.$

Newton’s Law of Cooling states that the temperature $T$ of an object at any time t can be described by the equation $T=T_s+\left(T_0-T_s\right)e^{-kt},$ where $T_s$ is the temperature of the surrounding environment, $T_0$ is the initial temperature of the object, and $k$ is the cooling rate. Use the definition of a logarithm along with properties of logarithms to solve the formula for time $t$ such that $t$ is equal to a single logarithm.