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Mathematics LibreTexts

9.4: Solve Exponential and Logarithmic Equations

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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

Before you get started, take this readiness quiz.

  1. Solve x2=16.
  2. Solve x25x+6=0.
  3. Solve x(x+6)=2x+5.

Solve Logarithmic Equations Using the Properties of Logarithms

In the section on logarithms, 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=logaN then it is true that 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, and a1 is any real number,

if logaM=logaN, then M=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 9.4.1

Solve 2log5x=log581.

Solution
  2log5x=log581
Use the Power Property. log5x2=log581
Use the One-to-One Property, if logaM=logaN, then M=N. x2=81
Solve using the Square Root Property. x=±9
We eliminate x=9 as we cannot take the logarithm of a negative number. x=9 or x=9
Check. x=9 2log5x=log5812log59?=log581log592?=log581log581=log581
Try It 9.4.2

Solve 2log3x=log336.

Answer

x=6

Try It 9.4.3

Solve 3logx=log64.

Answer

x=4

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

Example 9.4.4

Solve log3x+log3(x8)=2.

Solution
 
 
  log3x+log3(x8)=2

Use the Product Property, logaM+logaN=logaMN.

log3x(x8)=2
Rewrite in exponential form. 32=x(x8)
Simplify.  
Subtract 9 from each side.  
Factor. 0=(x9)(x+1)
Use the Zero-Product-Property x9=0 or x+1=0
Solve each equation. x=9,x=1
Check.

x=1

log3x+log3(x8)=2log3(1)+log3(18)?=2

We cannot take the log of a negative number.

x=9

log3x+log3(x8)=2log39+log3(98)?=22+0?=22=2

   

 

Try It 9.4.5

Solve log2x+log2(x2)=3.

Answer

x=4

Try It 9.4.6

Solve log2x+log2(x6)=4,

Answer

x=8

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

Example 9.4.7

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

Solution
  log4(x+6)log4(2x+5)=log4x
Use the Quotient Property on the left side and the PowerProperty on the right.

log4(x+62x+5)=log4x1

Rewrite x1=1x. log4(x+62x+5)=log41x

Use the One-to-One Property, if logaM=logaN, then M=N.

x+62x+5=1x
Solve the rational equation. x(x+6)=2x+5
Distribute. x2+6x=2x+5
Write in standard form. x2+4x5=0
Factor. (x+5)(x1)=0
Use the Zero-Product-Property. x+5=0 or x1=0
Solve each equation. x=5 or x=1
Check. We leave the check for you.
Try It 9.4.8

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

Answer

x=3

Try It 9.4.9

Solve log(x2)log(4x+16)=log(1x).

Answer

x=8

Example 9.4.10

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

Solution
  5x=11
Since the exponential is isolated, take the logarithm of both sides. log5x=log11

Use the Power Property to get the x as a factor, not an exponent.

xlog5=log11
Solve for x. Find the exact answer. x=log11log5
Approximate the answer.

x1.490

Since 51=5 and 52=25, does it makes sense that 51.49011?

Try It 9.4.11

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

Answer

x=log43log71.933

Try It 9.4.12

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

Answer

x=log98log82.205

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 9.4.13

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

Solution
  3ex+2=24
Isolate the exponential by dividing both sides by 3. ex+2=8
Take the natural logarithm of both sides. lnex+2=ln8

Use the Power Property to get the x as a factor, not an exponent.

(x+2)lne=ln8
Use the property lne=1 to simplify. x+2=ln8

Solve the equation. Find the exact answer.

x=ln82
Approximate the answer. x0.079
Try It 9.4.14

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

Answer

x=ln9+24.197

Try It 9.4.15

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

Answer

x=ln520.805

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)nt when compounded n times a year. A=Pert when compounded continuously. 

Example 9.4.16

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?

Solution
Identify the variables in the formula. A=$50,000P=$10,000r=?t=17 years A=Pert
Substitute the values into the formula. 50,000=10,000er17
Solve for r. Divide each side by 10,000. 5=e17r
Take the natural log of each side. ln5=lne17r
Use the Power Property. ln5=17rlne
Simplify. ln5=17r
Divide each side by 17. ln517=r
Approximate the answer. r0.095
Convert to a percentage. r9.5%
  They need the rate of growth to be approximately 9.5%.
Try It 9.4.17

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

Answer

r9.3%

Try It 9.4.18

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

Answer

r11.9%

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

Exponential Growth and Decay

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

A=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 9.4.19

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?

Solution

This problem requires two main steps. First we must find the unknown rate, k. Then we use that value of k to help us find the unknown number of bacteria.

Identify the variables in the formula.

A=300A0=100k=?t=3 hours A=A0ekt

Substitute the values in the formula. 300=100ek3
Solve for k. Divide each side by 100. 3=e3k
Take the natural log of each side. ln3=lne3k
Use the Power Property. ln3=3klne
Simplify. ln3=3k
Divide each side by 3. ln33=k
Approximate the answer. k0.366
We use this rate of growth to predict the number of bacteria there will be in 24 hours. A=?A0=100k=ln33t=24 hours A=A0ekt
Substitute in the values.

A=100eln3324

Evaluate. A656,100
  At this rate of growth, they can expect 656,100 bacteria.
Try It 9.4.20

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?

Answer

There will be 62,500 bacteria.

Try It 9.4.21

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?

Answer

There will be 5,870,061 bacteria.

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 9.4.22

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

Solution

This problem requires two main steps. First we must find the decay constant k. If we start with 100-mg, at the half-life there will be 50-mg remaining. We will use this information to find k. Then we use that value of k to help us find the amount of sample that will be left in 500 years.

Identify the variables in the formula. A=50A0=100k=?t=1590 years A=A0ekt
Substitute the values in the formula. 50=100ek1590
Solve for k. Divide each side by 100. 0.5=e1590k

Take the natural log of each side.

ln0.5=lne1590k

Use the Power Property. ln0.5=1590klne
Simplify. ln0.5=1590k
Divide each side by 1590. ln0.51590=k exact answer
We use this rate of growth to predict the amount that will be left in 500 years. A=?A0=100k=ln0.51590t=500yearsA=A0ekt
Substitute in the values. A=100e1n0.51500500
Evaluate. A80.4mg
  In 500 years there would be approximately 80.4 mg remaining.
Try It 9.4.23

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

Answer

There will be 6.43 mg left.

Try It 9.4.24

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

Answer

There will be 31.5 mg left.

Key Concepts

  • One-to-One Property of Logarithmic Equations: For M>0,N>0,a>0, and a1 is any real number:

    if logaM=logaN, then M=N

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

    A=P(1+rn)nt when compounded n times a year. A=Pert when compounded continuously. 

  • Exponential Growth and Decay: For an original amount, A0 that grows or decays at a rate, r, for a certain time t, the final amount, A, is A=A0ert.

Practice Makes Perfect

Note that even-numbered answers are given in this section.

(Optional)  Solve Logarithmic Equations Using the Properties of Logarithms

In the following exercises, solve for x.

  1. log464=2log4x
  2. log49=2logx
  3. 3log3x=log327
  4. 3log6x=log664
  5. log5(4x2)=log510
  6. log3(x2+3)=log34x
Answer

2. x=7

4. x=4

6. x=1,x=3

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.

  1. 3x=89
  2. 2x=74
  3. 5x=110
  4. 4x=112
  5. ex=16
  6. ex=8
  7. (12)x=6
  8. (13)x=8
  9. 4ex+1=16
  10. 3ex+2=9
  11. 6e2x=24
  12. 2e3x=32
  13. 14ex=3
  14. 13ex=2
  15. ex+1+2=16
  16. ex1+4=12
Answer

8. x=log74log26.209

10. x=log112log43.404

12. x=ln82.079

14. x=log8log131.893

16. x=ln320.901

18. x=ln1630.924

20. x=ln61.792

22. x=ln8+13.079

Solve Exponential Equations Using Logarithms

In the following exercises, solve each equation.

  1. 33x+1=81
  2. 64x17=216
  3. ex2e14=e5x
  4. ex2ex=e20
  5. loga64=2
  6. loga81=4
  7. lnx=8
  8. lnx=9
  9. log5(3x8)=2
  10. log4(7x+15)=3
  11. lne5x=30
  12. lne6x=18
  13. 3logx=log125
  14. 7log3x=log3128
Answer

24. x=5

26. x=4,x=5

28. a=3

30. x=e9

32. x=7

34. x=3

36. x=2

Solve Exponential Equations Using Logarithms

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

  1. 6x=91
  2. (12)x=10
  3. 7ex3=35
  4. 8ex+5=56
Answer

38. x=ln10ln23.322

40. x=ln753.054

Use Exponential Models in Applications

In the following exercises, solve.

  1. 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?
  2. 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?
  3. Coralee invests $5,000 in an account that compounds interest monthly and earns 7%. How long will it take for her money to double?
  4. Simone invests $8,000 in an account that compounds interest quarterly and earns 5%. How long will it take for his money to double?
  5. 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?
  6. 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?
  7. A virus takes 6 days to double its original population (A=2A0). How long will it take to triple its population?
  8. A bacteria doubles its original population in 24 hours (A=2A0). How big will its population be in 72 hours?
  9. 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?
  10. 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

42. 6.9%

44. 13.9 years

46. 122,070 bacteria

48. 8 times as large as the original population

50. 0.03 mL

Writing Exercises
  1. Explain the method you would use to solve these equations: 3x+1=81, 3x+1=75. Does your method require logarithms for both equations? Why or why not?
  2. What is the difference between the equation for exponential growth versus the equation for exponential decay?
Answer

52. Answers will vary.

 

Additional Exercises

53.  Solve for x:

a) log4x+log4x=3

b) log3x+log3(x+6)=3

c) logx+log(x15)=2

Answer

a) 8, b) 3, c) 20

 

Self Check

a. 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.
Figure 10.5.1

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

 


This page titled 9.4: Solve Exponential and Logarithmic Equations is shared under a CC BY license and was authored, remixed, and/or curated by OpenStax.

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