College algebra by carl stitz and jeff zeager

Section 6.4: Logarithmic Equations and Inequalities, from College Algebra: Corrected Edition by Carl Stitz, Ph.D. and Jeff Zeager, Ph.D. is available under a Creative Commons Attribution- NonCommercial-ShareAlike 3.0 license. © 2013, Carl Stitz. UMGC has modified this work and it is available under the original license.

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6.4 Logarithmic Equations and Inequalities 459

6.4 Logarithmic Equations and Inequalities

In Section 6.3 we solved equations and inequalities involving exponential functions using one of two basic strategies. We now turn our attention to equations and inequalities involving logarithmic functions, and not surprisingly, there are two basic strategies to choose from. For example, suppose we wish to solve log2(x) = log2(5). Theorem 6.4 tells us that the only solution to this equation is x = 5. Now suppose we wish to solve log2(x) = 3. If we want to use Theorem 6.4, we need to rewrite 3 as a logarithm base 2. We can use Theorem 6.3 to do just that: 3 = log2

( 23 )

= log2(8). Our equation then becomes log2(x) = log2(8) so that x = 8. However, we could have arrived at the same answer, in fewer steps, by using Theorem 6.3 to rewrite the equation log2(x) = 3 as 2

3 = x, or x = 8. We summarize the two common ways to solve log equations below.

Steps for Solving an Equation involving Logarithmic Functions

1. Isolate the logarithmic function.

2. (a) If convenient, express both sides as logs with the same base and equate the arguments of the log functions.

(b) Otherwise, rewrite the log equation as an exponential equation.

Example 6.4.1. Solve the following equations. Check your solutions graphically using a calculator.

1. log117(1− 3x) = log117 ( x2 − 3

) 2. 2− ln(x− 3) = 1

3. log6(x+ 4) + log6(3− x) = 1 4. log7(1− 2x) = 1− log7(3− x)

5. log2(x+ 3) = log2(6− x) + 3 6. 1 + 2 log4(x+ 1) = 2 log2(x)

Solution.

1. Since we have the same base on both sides of the equation log117(1 − 3x) = log117 ( x2 − 3

) ,

we equate what’s inside the logs to get 1 − 3x = x2 − 3. Solving x2 + 3x − 4 = 0 gives x = −4 and x = 1. To check these answers using the calculator, we make use of the change of base formula and graph f(x) = ln(1−3x)ln(117) and g(x) =

ln(x2−3) ln(117) and we see they intersect only

at x = −4. To see what happened to the solution x = 1, we substitute it into our original equation to obtain log117(−2) = log117(−2). While these expressions look identical, neither is a real number,1 which means x = 1 is not in the domain of the original equation, and is not a solution.

2. Our first objective in solving 2−ln(x−3) = 1 is to isolate the logarithm. We get ln(x−3) = 1, which, as an exponential equation, is e1 = x − 3. We get our solution x = e + 3. On the calculator, we see the graph of f(x) = 2 − ln(x − 3) intersects the graph of g(x) = 1 at x = e+ 3 ≈ 5.718.

1They do, however, represent the same family of complex numbers. We stop ourselves at this point and refer the reader to a good course in Complex Variables.

460 Exponential and Logarithmic Functions

y = f(x) = log117(1− 3x) and y = f(x) = 2− ln(x− 3) and y = g(x) = log117

( x2 − 3

) y = g(x) = 1

3. We can start solving log6(x+4)+log6(3−x) = 1 by using the Product Rule for logarithms to rewrite the equation as log6 [(x+ 4)(3− x)] = 1. Rewriting this as an exponential equation, we get 61 = (x + 4)(3 − x). This reduces to x2 + x − 6 = 0, which gives x = −3 and x = 2. Graphing y = f(x) = ln(x+4)ln(6) +

ln(3−x) ln(6) and y = g(x) = 1, we see they intersect twice, at

x = −3 and x = 2.

y = f(x) = log6(x+ 4) + log6(3− x) and y = g(x) = 1

4. Taking a cue from the previous problem, we begin solving log7(1− 2x) = 1− log7(3− x) by first collecting the logarithms on the same side, log7(1−2x)+ log7(3−x) = 1, and then using the Product Rule to get log7[(1− 2x)(3− x)] = 1. Rewriting this as an exponential equation gives 71 = (1−2x)(3−x) which gives the quadratic equation 2x2−7x−4 = 0. Solving, we find x = −12 and x = 4. Graphing, we find y = f(x) =

ln(1−2x) ln(7) and y = g(x) = 1−

ln(3−x) ln(7) intersect

only at x = −12 . Checking x = 4 in the original equation produces log7(−7) = 1− log7(−1), which is a clear domain violation.

5. Starting with log2(x + 3) = log2(6 − x) + 3, we gather the logarithms to one side and get log2(x+ 3)− log2(6− x) = 3. We then use the Quotient Rule and convert to an exponential equation

log2

( x+ 3

6− x

) = 3 ⇐⇒ 23 = x+ 3

6− x

This reduces to the linear equation 8(6− x) = x+ 3, which gives us x = 5. When we graph f(x) = ln(x+3)ln(2) and g(x) =

ln(6−x) ln(2) + 3, we find they intersect at x = 5.

6.4 Logarithmic Equations and Inequalities 461

y = f(x) = log7(1− 2x) and y = f(x) = log2(x+ 3) and y = g(x) = 1− log7(3− x) y = g(x) = log2(6− x) + 3

6. Starting with 1 + 2 log4(x+ 1) = 2 log2(x), we gather the logs to one side to get the equation 1 = 2 log2(x) − 2 log4(x + 1). Before we can combine the logarithms, however, we need a common base. Since 4 is a power of 2, we use change of base to convert

log4(x+ 1) = log2(x+ 1)

log2(4) =

1

2 log2(x+ 1)

Hence, our original equation becomes

1 = 2 log2(x)− 2 (

1 2 log2(x+ 1)

) 1 = 2 log2(x)− log2(x+ 1) 1 = log2

( x2 ) − log2(x+ 1) Power Rule

1 = log2

( x2

x+ 1

) Quotient Rule

Rewriting this in exponential form, we get x 2

x+1 = 2 or x 2 − 2x− 2 = 0. Using the quadratic

formula, we get x = 1 ± √

3. Graphing f(x) = 1 + 2 ln(x+1)ln(4) and g(x) = 2 ln(x) ln(2) , we see the

graphs intersect only at x = 1 + √

3 ≈ 2.732. The solution x = 1 − √

3 < 0, which means if substituted into the original equation, the term 2 log2

( 1− √

3 )

is undefined.

y = f(x) = 1 + 2 log4(x+ 1) and y = g(x) = 2 log2(x)

462 Exponential and Logarithmic Functions

If nothing else, Example 6.4.1 demonstrates the importance of checking for extraneous solutions2

when solving equations involving logarithms. Even though we checked our answers graphically, extraneous solutions are easy to spot - any supposed solution which causes a negative number inside a logarithm needs to be discarded. As with the equations in Example 6.3.1, much can be learned from checking all of the answers in Example 6.4.1 analytically. We leave this to the reader and turn our attention to inequalities involving logarithmic functions. Since logarithmic functions are continuous on their domains, we can use sign diagrams.

Example 6.4.2. Solve the following inequalities. Check your answer graphically using a calculator.

1. 1

ln(x) + 1 ≤ 1 2. (log2(x))

2 < 2 log2(x) + 3 3. x log(x+ 1) ≥ x

Solution.

1. We start solving 1ln(x)+1 ≤ 1 by getting 0 on one side of the inequality: 1

ln(x)+1 − 1 ≤ 0. Getting a common denominator yields 1ln(x)+1 −

ln(x)+1 ln(x)+1 ≤ 0 which reduces to

− ln(x) ln(x)+1 ≤ 0,

or ln(x)ln(x)+1 ≥ 0. We define r(x) = ln(x)

ln(x)+1 and set about finding the domain and the zeros

of r. Due to the appearance of the term ln(x), we require x > 0. In order to keep the denominator away from zero, we solve ln(x) + 1 = 0 so ln(x) = −1, so x = e−1 = 1e . Hence, the domain of r is

( 0, 1e ) ∪ (

1 e ,∞

) . To find the zeros of r, we set r(x) = ln(x)ln(x)+1 = 0 so that

ln(x) = 0, and we find x = e0 = 1. In order to determine test values for r without resorting to the calculator, we need to find numbers between 0, 1e , and 1 which have a base of e. Since e ≈ 2.718 > 1, 0 < 1

e2 < 1e <

1√ e < 1 < e. To determine the sign of r

( 1 e2

) , we use the fact that

ln (

1 e2

) = ln

( e−2 )

= −2, and find r (

1 e2

) = −2−2+1 = 2, which is (+). The rest of the test values

are determined similarly. From our sign diagram, we find the solution to be ( 0, 1e ) ∪ [1,∞).

Graphing f(x) = 1ln(x)+1 and g(x) = 1, we see the graph of f is below the graph of g on the solution intervals, and that the graphs intersect at x = 1.

0

(+)

1 e

” (−)

1

0 (+)

y = f(x) = 1ln(x)+1 and y = g(x) = 1

2Recall that an extraneous solution is an answer obtained analytically which does not satisfy the original equation.

6.4 Logarithmic Equations and Inequalities 463

2. Moving all of the nonzero terms of (log2(x)) 2 < 2 log2(x) + 3 to one side of the inequality,

we have (log2(x)) 2 − 2 log2(x) − 3 < 0. Defining r(x) = (log2(x))

2 − 2 log2(x) − 3, we get the domain of r is (0,∞), due to the presence of the logarithm. To find the zeros of r, we set r(x) = (log2(x))

2 − 2 log2(x) − 3 = 0 which results in a ‘quadratic in disguise.’ We set u = log2(x) so our equation becomes u

2− 2u− 3 = 0 which gives us u = −1 and u = 3. Since u = log2(x), we get log2(x) = −1, which gives us x = 2−1 = 12 , and log2(x) = 3, which yields x = 23 = 8. We use test values which are powers of 2: 0 < 14 <

1 2 < 1 < 8 < 16, and from our

sign diagram, we see r(x) < 0 on (

1 2 , 8 ) . Geometrically, we see the graph of f(x) =

( ln(x) ln(2)

)2 is below the graph of y = g(x) = 2 ln(x)ln(2) + 3 on the solution interval.

0

(+)

1 2

0 (−)

8

0 (+)

y = f(x) = (log2(x)) 2 and y = g(x) = 2 log2(x) + 3

3. We begin to solve x log(x+1) ≥ x by subtracting x from both sides to get x log(x+1)−x ≥ 0. We define r(x) = x log(x+1)−x and due to the presence of the logarithm, we require x+1 > 0, or x > −1. To find the zeros of r, we set r(x) = x log(x + 1) − x = 0. Factoring, we get x (log(x+ 1)− 1) = 0, which gives x = 0 or log(x+1)−1 = 0. The latter gives log(x+1) = 1, or x + 1 = 101, which admits x = 9. We select test values x so that x + 1 is a power of 10, and we obtain −1 < −0.9 < 0 <

√ 10 − 1 < 9 < 99. Our sign diagram gives the solution to

be (−1, 0] ∪ [9,∞). The calculator indicates the graph of y = f(x) = x log(x + 1) is above y = g(x) = x on the solution intervals, and the graphs intersect at x = 0 and x = 9.

−1

(+)

0

0 (−)

9

0 (+)

y = f(x) = x log(x+ 1) and y = g(x) = x

464 Exponential and Logarithmic Functions

Our next example revisits the concept of pH first seen in Exercise 77 in Section 6.1.

Example 6.4.3. In order to successfully breed Ippizuti fish the pH of a freshwater tank must be at least 7.8 but can be no more than 8.5. Determine the corresponding range of hydrogen ion concentration, and check your answer using a calculator.

Solution. Recall from Exercise 77 in Section 6.1 that pH = − log[H+] where [H+] is the hydrogen ion concentration in moles per liter. We require 7.8 ≤ − log[H+] ≤ 8.5 or −7.8 ≥ log[H+] ≥ −8.5. To solve this compound inequality we solve −7.8 ≥ log[H+] and log[H+] ≥ −8.5 and take the intersection of the solution sets.3 The former inequality yields 0 < [H+] ≤ 10−7.8 and the latter yields [H+] ≥ 10−8.5. Taking the intersection gives us our final answer 10−8.5 ≤ [H+] ≤ 10−7.8. (Your Chemistry professor may want the answer written as 3.16 × 10−9 ≤ [H+] ≤ 1.58 × 10−8.) After carefully adjusting the viewing window on the graphing calculator we see that the graph of f(x) = − log(x) lies between the lines y = 7.8 and y = 8.5 on the interval [3.16×10−9, 1.58×10−8].

The graphs of y = f(x) = − log(x), y = 7.8 and y = 8.5

We close this section by finding an inverse of a one-to-one function which involves logarithms.

Example 6.4.4. The function f(x) = log(x)

1− log(x) is one-to-one. Find a formula for f−1(x) and

check your answer graphically using your calculator. Solution. We first write y = f(x) then interchange the x and y and solve for y.

y = f(x)

y = log(x)

1− log(x)

x = log(y)

1− log(y) Interchange x and y.

x (1− log(y)) = log(y) x− x log(y) = log(y)

x = x log(y) + log(y) x = (x+ 1) log(y)

x

x+ 1 = log(y)

y = 10 x x+1 Rewrite as an exponential equation.

3Refer to page 4 for a discussion of what this means.

6.4 Logarithmic Equations and Inequalities 465

We have f−1(x) = 10 x x+1 . Graphing f and f−1 on the same viewing window yields

y = f(x) = log(x)

1− log(x) and y = g(x) = 10

x x+1

466 Exponential and Logarithmic Functions

6.4.1 Exercises

In Exercises 1 - 24, solve the equation analytically.

1. log(3x− 1) = log(4− x) 2. log2 ( x3 )

= log2(x)

3. ln ( 8− x2

) = ln(2− x) 4. log5

( 18− x2

) = log5(6− x)

5. log3(7− 2x) = 2 6. log 1 2 (2x− 1) = −3

7. ln ( x2 − 99

) = 0 8. log(x2 − 3x) = 1

9. log125

( 3x− 2 2x+ 3

) =

1

3 10. log

( x 10−3

) = 4.7

11. − log(x) = 5.4 12. 10 log ( x

10−12

) = 150

13. 6− 3 log5(2x) = 0 14. 3 ln(x)− 2 = 1− ln(x)

15. log3(x− 4) + log3(x+ 4) = 2 16. log5(2x+ 1) + log5(x+ 2) = 1

17. log169(3x+ 7)− log169(5x− 9) = 1

2 18. ln(x+ 1)− ln(x) = 3

19. 2 log7(x) = log7(2) + log7(x+ 12) 20. log(x)− log(2) = log(x+ 8)− log(x+ 2)

21. log3(x) = log 1 3 (x) + 8 22. ln(ln(x)) = 3

23. (log(x))2 = 2 log(x) + 15 24. ln(x2) = (ln(x))2

In Exercises 25 - 30, solve the inequality analytically.

25. 1− ln(x)

x2 < 0 26. x ln(x)− x > 0

27. 10 log ( x

10−12

) ≥ 90 28. 5.6 ≤ log

( x 10−3

) ≤ 7.1

29. 2.3 < − log(x) < 5.4 30. ln(x2) ≤ (ln(x))2

In Exercises 31 - 34, use your calculator to help you solve the equation or inequality.

31. ln(x) = e−x 32. ln(x) = 4 √ x

33. ln(x2 + 1) ≥ 5 34. ln(−2x3 − x2 + 13x− 6) < 0

6.4 Logarithmic Equations and Inequalities 467

35. Since f(x) = ex is a strictly increasing function, if a < b then ea < eb. Use this fact to solve the inequality ln(2x + 1) < 3 without a sign diagram. Use this technique to solve the inequalities in Exercises 27 - 29. (Compare this to Exercise 46 in Section 6.3.)

36. Solve ln(3− y)− ln(y) = 2x+ ln(5) for y.

37. In Example 6.4.4 we found the inverse of f(x) = log(x)

1− log(x) to be f−1(x) = 10

x x+1 .

(a) Show that ( f−1 ◦ f

) (x) = x for all x in the domain of f and that

( f ◦ f−1

) (x) = x for

all x in the domain of f−1.

(b) Find the range of f by finding the domain of f−1.

(c) Let g(x) = x

1− x and h(x) = log(x). Show that f = g ◦ h and (g ◦ h)−1 = h−1 ◦ g−1.

(We know this is true in general by Exercise 31 in Section 5.2, but it’s nice to see a specific example of the property.)

38. Let f(x) = 1

2 ln

( 1 + x

1− x

) . Compute f−1(x) and find its domain and range.

39. Explain the equation in Exercise 10 and the inequality in Exercise 28 above in terms of the Richter scale for earthquake magnitude. (See Exercise 75 in Section 6.1.)

40. Explain the equation in Exercise 12 and the inequality in Exercise 27 above in terms of sound intensity level as measured in decibels. (See Exercise 76 in Section 6.1.)

41. Explain the equation in Exercise 11 and the inequality in Exercise 29 above in terms of the pH of a solution. (See Exercise 77 in Section 6.1.)

42. With the help of your classmates, solve the inequality n √ x > ln(x) for a variety of natural

numbers n. What might you conjecture about the “speed” at which f(x) = ln(x) grows versus any principal nth root function?

468 Exponential and Logarithmic Functions

6.4.2 Answers

1. x = 54 2. x = 1 3. x = −2

4. x = −3, 4 5. x = −1 6. x = 92

7. x = ±10 8. x = −2, 5 9. x = −177

10. x = 101.7 11. x = 10−5.4 12. x = 103

13. x = 252 14. x = e 3/4 15. x = 5

16. x = 12 17. x = 2 18. x = 1

e3−1

19. x = 6 20. x = 4 21. x = 81

22. x = ee 3

23. x = 10−3, 105 24. x = 1, x = e2

25. (e,∞) 26. (e,∞) 27. [ 10−3,∞

) 28.

[ 102.6, 104.1

] 29.

( 10−5.4, 10−2.3

) 30. (0, 1] ∪ [e2,∞)

31. x ≈ 1.3098 32. x ≈ 4.177, x ≈ 5503.665

33. ≈ (−∞,−12.1414) ∪ (12.1414,∞) 34. ≈ (−3.0281,−3)∪(0.5, 0.5991)∪(1.9299, 2)

35. −1 2 < x <

e3 − 1 2

36. y = 3

5e2x + 1

38. f−1(x) = e2x − 1 e2x + 1

= ex − e−x

ex + e−x . (To see why we rewrite this in this form, see Exercise 51 in

Section 11.10.) The domain of f−1 is (−∞,∞) and its range is the same as the domain of f , namely (−1, 1).