# 6.6 Exponential and logarithmic equations  (Page 7/8)

 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?

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.

## Algebraic

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.

${9}^{x-10}=1$

$x=10$

$2{e}^{6x}=13$

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

No solution

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

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

$p=\mathrm{log}\left(\frac{17}{8}\right)-7$

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

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

$k=-\frac{\mathrm{ln}\left(38\right)}{3}$

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

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

$x=\frac{\mathrm{ln}\left(\frac{38}{3}\right)-8}{9}$

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

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

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

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

$x=\frac{\mathrm{ln}\left(\frac{3}{5}\right)-3}{8}$

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

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

no solution

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

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

$x=\mathrm{ln}\left(3\right)$

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

For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation.

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

${10}^{-2}=\frac{1}{100}$

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

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

${\mathrm{log}}_{9}\left(2{n}^{2}-14n\right)={\mathrm{log}}_{9}\left(-45+{n}^{2}\right)$

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

$x=±\frac{10}{3}$

For the following exercises, solve each equation for $\text{\hspace{0.17em}}x.$

$\mathrm{log}\left(x+12\right)=\mathrm{log}\left(x\right)+\mathrm{log}\left(12\right)$

$\mathrm{ln}\left(x\right)+\mathrm{ln}\left(x-3\right)=\mathrm{ln}\left(7x\right)$

$x=10$

${\mathrm{log}}_{2}\left(7x+6\right)=3$

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

$x=0$

${\mathrm{log}}_{8}\left(x+6\right)-{\mathrm{log}}_{8}\left(x\right)={\mathrm{log}}_{8}\left(58\right)$

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

$x=\frac{3}{4}$

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

## Graphical

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.

${\mathrm{log}}_{9}\left(x\right)-5=-4$

$x=9$

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

$\mathrm{ln}\left(3x\right)=2$

$x=\frac{{e}^{2}}{3}\approx 2.5$

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

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

$x=-5$

$-7+{\mathrm{log}}_{3}\left(4-x\right)=-6$

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

$x=\frac{e+10}{4}\approx 3.2$

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

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

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 $\text{\hspace{0.17em}}\text{20,000?}$

## Technology

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

$\frac{\mathrm{ln}\left(17\right)}{5}\approx 0.567$

$3{\left(1.04\right)}^{3t}=8\text{\hspace{0.17em}}$ using the common log

${3}^{4x-5}=38\text{\hspace{0.17em}}$ using the common log

$50{e}^{-0.12t}=10\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}5.83$

## Extensions

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.

$t=\mathrm{ln}\left({\left(\frac{T-{T}_{s}}{{T}_{0}-{T}_{s}}\right)}^{-\text{\hspace{0.17em}}\frac{1}{k}}\right)$

Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1