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Intermediate Algebra 2e

10.5 Solve Exponential and Logarithmic Equations

Intermediate Algebra 2e10.5 Solve Exponential and Logarithmic Equations

Learning Objectives

By the end of this section, you will be able to:

  • Solve logarithmic equations using the properties of logarithms
  • Solve exponential equations using logarithms
  • Use exponential models in applications

Be Prepared 10.13

Before you get started, take this readiness quiz.

Solve: x2=16.x2=16.
If you missed this problem, review Example 6.46.

Be Prepared 10.14

Solve: x25x+6=0.x25x+6=0.
If you missed this problem, review Example 6.45.

Be Prepared 10.15

Solve: x(x+6)=2x+5.x(x+6)=2x+5.
If you missed this problem, review Example 6.47.

Solve Logarithmic Equations Using the Properties of Logarithms

In the section on logarithmic functions, we solved some equations by rewriting the equation in exponential form. Now that we have the properties of logarithms, we have additional methods we can use to solve logarithmic equations.

If our equation has two logarithms we can use a property that says that if logaM=logaNlogaM=logaN then it is true that M=N.M=N. This is the One-to-One Property of Logarithmic Equations.

One-to-One Property of Logarithmic Equations

For M>0,N>0,a>0,M>0,N>0,a>0, and a1a1 is any real number:

IflogaM=logaN,thenM=N.IflogaM=logaN,thenM=N.

To use this property, we must be certain that both sides of the equation are written with the same base.

Remember that logarithms are defined only for positive real numbers. Check your results in the original equation. You may have obtained a result that gives a logarithm of zero or a negative number.

Example 10.38

Solve: 2log5x=log581.2log5x=log581.

Try It 10.75

Solve: 2log3x=log3362log3x=log336

Try It 10.76

Solve: 3logx=log643logx=log64

Another strategy to use to solve logarithmic equations is to condense sums or differences into a single logarithm.

Example 10.39

Solve: log3x+log3(x8)=2.log3x+log3(x8)=2.

Try It 10.77

Solve: log2x+log2(x2)=3log2x+log2(x2)=3

Try It 10.78

Solve: log2x+log2(x6)=4log2x+log2(x6)=4

When there are logarithms on both sides, we condense each side into a single logarithm. Remember to use the Power Property as needed.

Example 10.40

Solve: log4(x+6)log4(2x+5)=log4x.log4(x+6)log4(2x+5)=log4x.

Try It 10.79

Solve: log(x+2)log(4x+3)=logx.log(x+2)log(4x+3)=logx.

Try It 10.80

Solve: log(x2)log(4x+16)=log1x.log(x2)log(4x+16)=log1x.

Solve Exponential Equations Using Logarithms

In the section on exponential functions, we solved some equations by writing both sides of the equation with the same base. Next we wrote a new equation by setting the exponents equal.

It is not always possible or convenient to write the expressions with the same base. In that case we often take the common logarithm or natural logarithm of both sides once the exponential is isolated.

Example 10.41

Solve 5x=11.5x=11. Find the exact answer and then approximate it to three decimal places.

Try It 10.81

Solve 7x=43.7x=43. Find the exact answer and then approximate it to three decimal places.

Try It 10.82

Solve 8x=98.8x=98. Find the exact answer and then approximate it to three decimal places.

When we take the logarithm of both sides we will get the same result whether we use the common or the natural logarithm (try using the natural log in the last example. Did you get the same result?) When the exponential has base e, we use the natural logarithm.

Example 10.42

Solve 3ex+2=24.3ex+2=24. Find the exact answer and then approximate it to three decimal places.

Try It 10.83

Solve 2ex2=18.2ex2=18. Find the exact answer and then approximate it to three decimal places.

Try It 10.84

Solve 5e2x=25.5e2x=25. Find the exact answer and then approximate it to three decimal places.

Use Exponential Models in Applications

In previous sections we were able to solve some applications that were modeled with exponential equations. Now that we have so many more options to solve these equations, we are able to solve more applications.

We will again use the Compound Interest Formulas and so we list them here for reference.

Compound Interest

For a principal, P, invested at an interest rate, r, for t years, the new balance, A is:

A=P(1+rn)ntwhen compoundedntimes a year. A=Pertwhen compounded continuously.A=P(1+rn)ntwhen compoundedntimes a year. A=Pertwhen compounded continuously.

Example 10.43

Jermael’s parents put $10,000 in investments for his college expenses on his first birthday. They hope the investments will be worth $50,000 when he turns 18. If the interest compounds continuously, approximately what rate of growth will they need to achieve their goal?

Try It 10.85

Hector invests $10,000$10,000 at age 21. He hopes the investments will be worth $150,000$150,000 when he turns 50. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal?

Try It 10.86

Rachel invests $15,000$15,000 at age 25. She hopes the investments will be worth $90,000$90,000 when she turns 40. If the interest compounds continuously, approximately what rate of growth will she need to achieve her goal?

We have seen that growth and decay are modeled by exponential functions. For growth and decay we use the formula A=A0ekt.A=A0ekt. Exponential growth has a positive rate of growth or growth constant, kk, and exponential decay has a negative rate of growth or decay constant, k.

Exponential Growth and Decay

For an original amount, A0,A0, that grows or decays at a rate, k, for a certain time, t, the final amount, A, is:

A=A0ektA=A0ekt

We can now solve applications that give us enough information to determine the rate of growth. We can then use that rate of growth to predict other situations.

Example 10.44

Researchers recorded that a certain bacteria population grew from 100 to 300 in 3 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment?

Try It 10.87

Researchers recorded that a certain bacteria population grew from 100 to 500 in 6 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment?

Try It 10.88

Researchers recorded that a certain bacteria population declined from 700,000 to 400,000 in 5 hours after the administration of medication. At this rate of decay, how many bacteria will there be 24 hours from the start of the experiment?

Radioactive substances decay or decompose according to the exponential decay formula. The amount of time it takes for the substance to decay to half of its original amount is called the half-life of the substance.

Similar to the previous example, we can use the given information to determine the constant of decay, and then use that constant to answer other questions.

Example 10.45

The half-life of radium-226 is 1,590 years. How much of a 100 mg sample will be left in 500 years?

Try It 10.89

The half-life of magnesium-27 is 9.45 minutes. How much of a 10-mg sample will be left in 6 minutes?

Try It 10.90

The half-life of radioactive iodine is 60 days. How much of a 50-mg sample will be left in 40 days?

Section 10.5 Exercises

Practice Makes Perfect

Solve Logarithmic Equations Using the Properties of Logarithms

In the following exercises, solve for x.

288.

log 4 64 = 2 log 4 x log 4 64 = 2 log 4 x

289.

log 49 = 2 log x log 49 = 2 log x

290.

3 log 3 x = log 3 27 3 log 3 x = log 3 27

291.

3 log 6 x = log 6 64 3 log 6 x = log 6 64

292.

log 5 ( 4 x 2 ) = log 5 10 log 5 ( 4 x 2 ) = log 5 10

293.

log 3 ( x 2 + 3 ) = log 3 4 x log 3 ( x 2 + 3 ) = log 3 4 x

294.

log 3 x + log 3 x = 2 log 3 x + log 3 x = 2

295.

log 4 x + log 4 x = 3 log 4 x + log 4 x = 3

296.

log 2 x + log 2 ( x 3 ) = 2 log 2 x + log 2 ( x 3 ) = 2

297.

log 3 x + log 3 ( x + 6 ) = 3 log 3 x + log 3 ( x + 6 ) = 3

298.

log x + log ( x + 3 ) = 1 log x + log ( x + 3 ) = 1

299.

log x + log ( x 15 ) = 2 log x + log ( x 15 ) = 2

300.

log ( x + 4 ) log ( 5 x + 12 ) = log x log ( x + 4 ) log ( 5 x + 12 ) = log x

301.

log ( x 1 ) log ( x + 3 ) = log 1 x log ( x 1 ) log ( x + 3 ) = log 1 x

302.

log 5 ( x + 3 ) + log 5 ( x 6 ) = log 5 10 log 5 ( x + 3 ) + log 5 ( x 6 ) = log 5 10

303.

log 5 ( x + 1 ) + log 5 ( x 5 ) = log 5 7 log 5 ( x + 1 ) + log 5 ( x 5 ) = log 5 7

304.

log 3 ( 2 x 1 ) = log 3 ( x + 3 ) + log 3 3 log 3 ( 2 x 1 ) = log 3 ( x + 3 ) + log 3 3

305.

log ( 5 x + 1 ) = log ( x + 3 ) + log 2 log ( 5 x + 1 ) = log ( x + 3 ) + log 2

Solve Exponential Equations Using Logarithms

In the following exercises, solve each exponential equation. Find the exact answer and then approximate it to three decimal places.

306.

3 x = 89 3 x = 89

307.

2 x = 74 2 x = 74

308.

5 x = 110 5 x = 110

309.

4 x = 112 4 x = 112

310.

e x = 16 e x = 16

311.

e x = 8 e x = 8

312.

( 1 2 ) x = 6 ( 1 2 ) x = 6

313.

( 1 3 ) x = 8 ( 1 3 ) x = 8

314.

4 e x + 1 = 16 4 e x + 1 = 16

315.

3 e x + 2 = 9 3 e x + 2 = 9

316.

6 e 2 x = 24 6 e 2 x = 24

317.

2 e 3 x = 32 2 e 3 x = 32

318.

1 4 e x = 3 1 4 e x = 3

319.

1 3 e x = 2 1 3 e x = 2

320.

e x + 1 + 2 = 16 e x + 1 + 2 = 16

321.

e x 1 + 4 = 12 e x 1 + 4 = 12

In the following exercises, solve each equation.

322.

3 3 x + 1 = 81 3 3 x + 1 = 81

323.

6 4 x 17 = 216 6 4 x 17 = 216

324.

e x 2 e 14 = e 5 x e x 2 e 14 = e 5 x

325.

e x 2 e x = e 20 e x 2 e x = e 20

326.

log a 64 = 2 log a 64 = 2

327.

log a 81 = 4 log a 81 = 4

328.

ln x = −8 ln x = −8

329.

ln x = 9 ln x = 9

330.

log 5 ( 3 x 8 ) = 2 log 5 ( 3 x 8 ) = 2

331.

log 4 ( 7 x + 15 ) = 3 log 4 ( 7 x + 15 ) = 3

332.

ln e 5 x = 30 ln e 5 x = 30

333.

ln e 6 x = 18 ln e 6 x = 18

334.

3 log x = log 125 3 log x = log 125

335.

7 log 3 x = log 3 128 7 log 3 x = log 3 128

336.

log 6 x + log 6 ( x 5 ) = log 6 24 log 6 x + log 6 ( x 5 ) = log 6 24

337.

log 9 x + log 9 ( x 4 ) = log 9 12 log 9 x + log 9 ( x 4 ) = log 9 12

338.

log 2 ( x + 2 ) log 2 ( 2 x + 9 ) = log 2 x log 2 ( x + 2 ) log 2 ( 2 x + 9 ) = log 2 x

339.

log 6 ( x + 1 ) log 6 ( 4 x + 10 ) = log 6 1 x log 6 ( x + 1 ) log 6 ( 4 x + 10 ) = log 6 1 x

In the following exercises, solve for x, giving an exact answer as well as an approximation to three decimal places.

340.

6 x = 91 6 x = 91

341.

( 1 2 ) x = 10 ( 1 2 ) x = 10

342.

7 e x 3 = 35 7 e x 3 = 35

343.

8 e x + 5 = 56 8 e x + 5 = 56

Use Exponential Models in Applications

In the following exercises, solve.

344.

Sung Lee invests $5,000$5,000 at age 18. He hopes the investments will be worth $10,000$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?

345.

Alice invests $15,000$15,000 at age 30 from the signing bonus of her new job. She hopes the investments will be worth $30,000$30,000 when she turns 40. If the interest compounds continuously, approximately what rate of growth will she need to achieve her goal?

346.

Coralee invests $5,000$5,000 in an account that compounds interest monthly and earns 7%.7%. How long will it take for her money to double?

347.

Simone invests $8,000$8,000 in an account that compounds interest quarterly and earns 5%.5%. How long will it take for his money to double?

348.

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?

349.

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?

350.

A virus takes 6 days to double its original population (A=2A0).(A=2A0). How long will it take to triple its population?

351.

A bacteria doubles its original population in 24 hours (A=2A0).(A=2A0). How big will its population be in 72 hours?

352.

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?

353.

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?

Writing Exercises

354.

Explain the method you would use to solve these equations: 3x+1=81,3x+1=81, 3x+1=75.3x+1=75. Does your method require logarithms for both equations? Why or why not?

355.

What is the difference between the equation for exponential growth versus the equation for exponential decay?

Self Check

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has four rows and four columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads solve logarithmic equations using the properties of logarithms, solve exponential equations using logarithms, and use exponential models in applications. The rest of the cells are blank.

After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

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