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Section 6.3: Exponential 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.

448 Exponential and Logarithmic Functions

6.3 Exponential Equations and Inequalities

In this section we will develop techniques for solving equations involving exponential functions. Suppose, for instance, we wanted to solve the equation 2x = 128. After a moment’s calculation, we find 128 = 27, so we have 2x = 27. The one-to-one property of exponential functions, detailed in Theorem 6.4, tells us that 2x = 27 if and only if x = 7. This means that not only is x = 7 a solution to 2x = 27, it is the only solution. Now suppose we change the problem ever so slightly to 2x = 129. We could use one of the inverse properties of exponentials and logarithms listed in Theorem 6.3 to write 129 = 2log2(129). We’d then have 2x = 2log2(129), which means our solution is x = log2(129). This makes sense because, after all, the definition of log2(129) is ‘the exponent we put on 2 to get 129.’ Indeed we could have obtained this solution directly by rewriting the equation 2x = 129 in its logarithmic form log2(129) = x. Either way, in order to get a reasonable decimal approximation to this number, we’d use the change of base formula, Theorem 6.7, to give us something more calculator friendly,1 say log2(129) =

ln(129) ln(2)

. Another way to arrive at this answer is as follows

2x = 129 ln (2x) = ln(129) Take the natural log of both sides. x ln(2) = ln(129) Power Rule

x = ln(129)

ln(2)

‘Taking the natural log’ of both sides is akin to squaring both sides: since f(x) = ln(x) is a function, as long as two quantities are equal, their natural logs are equal.2 Also note that we treat ln(2) as any other non-zero real number and divide it through3 to isolate the variable x. We summarize below the two common ways to solve exponential equations, motivated by our examples.

Steps for Solving an Equation involving Exponential Functions

1. Isolate the exponential function.

2. (a) If convenient, express both sides with a common base and equate the exponents.

(b) Otherwise, take the natural log of both sides of the equation and use the Power Rule.

Example 6.3.1. Solve the following equations. Check your answer graphically using a calculator.

1. 23x = 161−x 2. 2000 = 1000 · 3−0.1t 3. 9 · 3x = 72x

4. 75 = 100 1+3e−2t

5. 25x = 5x + 6 6. e x−e−x

2 = 5

Solution.

1You can use natural logs or common logs. We choose natural logs. (In Calculus, you’ll learn these are the most ‘mathy’ of the logarithms.)

2This is also the ‘if’ part of the statement logb(u) = logb(w) if and only if u = w in Theorem 6.4. 3Please resist the temptation to divide both sides by ‘ln’ instead of ln(2). Just like it wouldn’t make sense to

divide both sides by the square root symbol ‘ √

’ when solving x √

2 = 5, it makes no sense to divide by ‘ln’.

6.3 Exponential Equations and Inequalities 449

1. Since 16 is a power of 2, we can rewrite 23x = 161−x as 23x = ( 24 )1−x

. Using properties of

exponents, we get 23x = 24(1−x). Using the one-to-one property of exponential functions, we get 3x = 4(1−x) which gives x = 4

7 . To check graphically, we set f(x) = 23x and g(x) = 161−x

and see that they intersect at x = 4 7 ≈ 0.5714.

2. We begin solving 2000 = 1000·3−0.1t by dividing both sides by 1000 to isolate the exponential which yields 3−0.1t = 2. Since it is inconvenient to write 2 as a power of 3, we use the natural log to get ln

( 3−0.1t

) = ln(2). Using the Power Rule, we get −0.1t ln(3) = ln(2), so we

divide both sides by −0.1 ln(3) to get t = − ln(2) 0.1 ln(3)

= −10 ln(2) ln(3)

. On the calculator, we graph

f(x) = 2000 and g(x) = 1000 ·3−0.1x and find that they intersect at x = −10 ln(2) ln(3)

≈−6.3093.

y = f(x) = 23x and y = f(x) = 2000 and y = g(x) = 161−x y = g(x) = 1000 · 3−0.1x

3. We first note that we can rewrite the equation 9·3x = 72x as 32·3x = 72x to obtain 3x+2 = 72x. Since it is not convenient to express both sides as a power of 3 (or 7 for that matter) we use the natural log: ln

( 3x+2

) = ln

( 72x ) . The power rule gives (x + 2) ln(3) = 2x ln(7). Even

though this equation appears very complicated, keep in mind that ln(3) and ln(7) are just constants. The equation (x + 2) ln(3) = 2x ln(7) is actually a linear equation and as such we gather all of the terms with x on one side, and the constants on the other. We then divide both sides by the coefficient of x, which we obtain by factoring.

(x + 2) ln(3) = 2x ln(7) x ln(3) + 2 ln(3) = 2x ln(7)

2 ln(3) = 2x ln(7) −x ln(3) 2 ln(3) = x(2 ln(7) − ln(3)) Factor.

x = 2 ln(3)

2 ln(7)−ln(3)

Graphing f(x) = 9·3x and g(x) = 72x on the calculator, we see that these two graphs intersect at x =

2 ln(3) 2 ln(7)−ln(3) ≈ 0.7866.

4. Our objective in solving 75 = 100 1+3e−2t

is to first isolate the exponential. To that end, we

clear denominators and get 75 ( 1 + 3e−2t

) = 100. From this we get 75 + 225e−2t = 100,

which leads to 225e−2t = 25, and finally, e−2t = 1 9 . Taking the natural log of both sides

450 Exponential and Logarithmic Functions

gives ln ( e−2t

) = ln

( 1 9

) . Since natural log is log base e, ln

( e−2t

) = −2t. We can also use

the Power Rule to write ln (

1 9

) = − ln(9). Putting these two steps together, we simplify

ln ( e−2t

) = ln

( 1 9

) to −2t = − ln(9). We arrive at our solution, t = ln(9)

2 which simplifies to

t = ln(3). (Can you explain why?) The calculator confirms the graphs of f(x) = 75 and g(x) = 100

1+3e−2x intersect at x = ln(3) ≈ 1.099.

y = f(x) = 9 · 3x and y = f(x) = 75 and y = g(x) = 72x y = g(x) = 100

1+3e−2x

5. We start solving 25x = 5x + 6 by rewriting 25 = 52 so that we have ( 52 )x

= 5x + 6, or 52x = 5x + 6. Even though we have a common base, having two terms on the right hand side of the equation foils our plan of equating exponents or taking logs. If we stare at this long enough, we notice that we have three terms with the exponent on one term exactly twice that of another. To our surprise and delight, we have a ‘quadratic in disguise’. Letting u = 5x, we have u2 = (5x)

2 = 52x so the equation 52x = 5x + 6 becomes u2 = u + 6. Solving this as

u2 − u − 6 = 0 gives u = −2 or u = 3. Since u = 5x, we have 5x = −2 or 5x = 3. Since 5x = −2 has no real solution, (Why not?) we focus on 5x = 3. Since it isn’t convenient to express 3 as a power of 5, we take natural logs and get ln (5x) = ln(3) so that x ln(5) = ln(3)

or x = ln(3) ln(5)

. On the calculator, we see the graphs of f(x) = 25x and g(x) = 5x + 6 intersect

at x = ln(3) ln(5)

≈ 0.6826.

6. At first, it’s unclear how to proceed with e x−e−x

2 = 5, besides clearing the denominator to

obtain ex−e−x = 10. Of course, if we rewrite e−x = 1 ex

, we see we have another denominator lurking in the problem: ex − 1

ex = 10. Clearing this denominator gives us e2x − 1 = 10ex,

and once again, we have an equation with three terms where the exponent on one term is exactly twice that of another - a ‘quadratic in disguise.’ If we let u = ex, then u2 = e2x so the equation e2x − 1 = 10ex can be viewed as u2 − 1 = 10u. Solving u2 − 10u− 1 = 0, we obtain by the quadratic formula u = 5±

√ 26. From this, we have ex = 5±

√ 26. Since 5−

√ 26 < 0,

we get no real solution to ex = 5− √

26, but for ex = 5 + √

26, we take natural logs to obtain x = ln

( 5 + √

26 ) . If we graph f(x) = e

x−e−x 2

and g(x) = 5, we see that the graphs intersect

at x = ln ( 5 + √

26 ) ≈ 2.312

6.3 Exponential Equations and Inequalities 451

y = f(x) = 25x and y = f(x) = e x−e−x

2 and

y = g(x) = 5x + 6 y = g(x) = 5

The authors would be remiss not to mention that Example 6.3.1 still holds great educational value. Much can be learned about logarithms and exponentials by verifying the solutions obtained in Example 6.3.1 analytically. For example, to verify our solution to 2000 = 1000 · 3−0.1t, we substitute t = −10 ln(2)

ln(3) and obtain

2000 ? = 1000 · 3−0.1

( −10 ln(2)

ln(3)

) 2000

? = 1000 · 3

ln(2) ln(3)

2000 ? = 1000 · 3log3(2) Change of Base

2000 ? = 1000 · 2 Inverse Property

2000 X = 2000

The other solutions can be verified by using a combination of log and inverse properties. Some fall out quite quickly, while others are more involved. We leave them to the reader.

Since exponential functions are continuous on their domains, the Intermediate Value Theorem 3.1 applies. As with the algebraic functions in Section 5.3, this allows us to solve inequalities using sign diagrams as demonstrated below.

Example 6.3.2. Solve the following inequalities. Check your answer graphically using a calculator.

1. 2x 2−3x − 16 ≥ 0 2.

ex

ex − 4 ≤ 3 3. xe

2x < 4x

Solution.

1. Since we already have 0 on one side of the inequality, we set r(x) = 2x 2−3x−16. The domain

of r is all real numbers, so in order to construct our sign diagram, we need to find the zeros of r. Setting r(x) = 0 gives 2x

2−3x −16 = 0 or 2x 2−3x = 16. Since 16 = 24 we have 2x

2−3x = 24, so by the one-to-one property of exponential functions, x2 − 3x = 4. Solving x2 − 3x− 4 = 0 gives x = 4 and x = −1. From the sign diagram, we see r(x) ≥ 0 on (−∞,−1]∪[4,∞), which corresponds to where the graph of y = r(x) = 2x

2−3x − 16, is on or above the x-axis.

452 Exponential and Logarithmic Functions

(+)

−1

0 (−)

4

0 (+)

y = r(x) = 2x 2−3x − 16

2. The first step we need to take to solve e x

ex−4 ≤ 3 is to get 0 on one side of the inequality. To that end, we subtract 3 from both sides and get a common denominator

ex

ex − 4 ≤ 3

ex

ex − 4 − 3 ≤ 0

ex

ex − 4 −

3 (ex − 4) ex − 4

≤ 0 Common denomintors.

12 − 2ex

ex − 4 ≤ 0

We set r(x) = 12−2e x

ex−4 and we note that r is undefined when its denominator e x − 4 = 0, or

when ex = 4. Solving this gives x = ln(4), so the domain of r is (−∞, ln(4)) ∪ (ln(4),∞). To find the zeros of r, we solve r(x) = 0 and obtain 12−2ex = 0. Solving for ex, we find ex = 6, or x = ln(6). When we build our sign diagram, finding test values may be a little tricky since we need to check values around ln(4) and ln(6). Recall that the function ln(x) is increasing4

which means ln(3) < ln(4) < ln(5) < ln(6) < ln(7). While the prospect of determining the sign of r (ln(3)) may be very unsettling, remember that eln(3) = 3, so

r (ln(3)) = 12 − 2eln(3)

eln(3) − 4 =

12 − 2(3) 3 − 4

= −6

We determine the signs of r (ln(5)) and r (ln(7)) similarly.5 From the sign diagram, we find our answer to be (−∞, ln(4)) ∪ [ln(6),∞). Using the calculator, we see the graph of f(x) = e

x

ex−4 is below the graph of g(x) = 3 on (−∞, ln(4)) ∪ (ln(6),∞), and they intersect at x = ln(6) ≈ 1.792.

4This is because the base of ln(x) is e > 1. If the base b were in the interval 0 < b < 1, then logb(x) would decreasing.

5We could, of course, use the calculator, but what fun would that be?

6.3 Exponential Equations and Inequalities 453

(−)

ln(4)

� (+)

ln(6)

0 (−)

y = f(x) = e x

ex−4 y = g(x) = 3

3. As before, we start solving xe2x < 4x by getting 0 on one side of the inequality, xe2x−4x < 0. We set r(x) = xe2x − 4x and since there are no denominators, even-indexed radicals, or logs, the domain of r is all real numbers. Setting r(x) = 0 produces xe2x − 4x = 0. We factor to get x

( e2x − 4

) = 0 which gives x = 0 or e2x − 4 = 0. To solve the latter, we isolate the

exponential and take logs to get 2x = ln(4), or x = ln(4)

2 = ln(2). (Can you explain the last

equality using properties of logs?) As in the previous example, we need to be careful about choosing test values. Since ln(1) = 0, we choose ln

( 1 2

) , ln

( 3 2

) and ln(3). Evaluating,6 we get

r ( ln (

1 2

)) = ln

( 1 2

) e2 ln(

1 2 ) − 4 ln

( 1 2

) = ln

( 1 2

) eln(

1 2 )

2

− 4 ln (

1 2

) Power Rule

= ln (

1 2

) eln(

1 4 ) − 4 ln

( 1 2

) = 1

4 ln (

1 2

) − 4 ln

( 1 2

) = −15

4 ln (

1 2

) Since 1

2 < 1, ln

( 1 2

) < 0 and we get r(ln

( 1 2

) ) is (+), so r(x) < 0 on (0, ln(2)). The calculator

confirms that the graph of f(x) = xe2x is below the graph of g(x) = 4x on these intervals.7

(+)

0

0 (−)

ln(2)

0 (+)

y = f(x) = xe2x and y = g(x) = 4x

6A calculator can be used at this point. As usual, we proceed without apologies, with the analytical method. 7Note: ln(2) ≈ 0.693.

454 Exponential and Logarithmic Functions

Example 6.3.3. Recall from Example 6.1.2 that the temperature of coffee T (in degrees Fahren- heit) t minutes after it is served can be modeled by T(t) = 70 + 90e−0.1t. When will the coffee be warmer than 100◦F?

Solution. We need to find when T(t) > 100, or in other words, we need to solve the inequality 70 + 90e−0.1t > 100. Getting 0 on one side of the inequality, we have 90e−0.1t − 30 > 0, and we set r(t) = 90e−0.1t − 30. The domain of r is artificially restricted due to the context of the problem to [0,∞), so we proceed to find the zeros of r. Solving 90e−0.1t − 30 = 0 results in e−0.1t = 1

3 so that t = −10 ln

( 1 3

) which, after a quick application of the Power Rule leaves us with

t = 10 ln(3). If we wish to avoid using the calculator to choose test values, we note that since 1 < 3, 0 = ln(1) < ln(3) so that 10 ln(3) > 0. So we choose t = 0 as a test value in [0, 10 ln(3)). Since 3 < 4, 10 ln(3) < 10 ln(4), so the latter is our choice of a test value for the interval (10 ln(3),∞). Our sign diagram is below, and next to it is our graph of y = T(t) from Example 6.1.2 with the horizontal line y = 100.

0

(+)

10 ln(3)

0 (−)

H.A. y = 70

y = 100

t

y

2 4 6 8 10 12 14 16 18 20

20

40

60

80

120

140

160

180

y = T(t)

In order to interpret what this means in the context of the real world, we need a reasonable approximation of the number 10 ln(3) ≈ 10.986. This means it takes approximately 11 minutes for the coffee to cool to 100◦F. Until then, the coffee is warmer than that.8

We close this section by finding the inverse of a function which is a composition of a rational function with an exponential function.

Example 6.3.4. The function f(x) = 5ex

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

your answer graphically using your calculator.

Solution. We start by writing y = f(x), and interchange the roles of x and y. To solve for y, we first clear denominators and then isolate the exponential function.

8Critics may point out that since we needed to use the calculator to interpret our answer anyway, why not use it earlier to simplify the computations? It is a fair question which we answer unfairly: it’s our book.

6.3 Exponential Equations and Inequalities 455

y = 5ex

ex + 1

x = 5ey

ey + 1 Switch x and y

x (ey + 1) = 5ey

xey + x = 5ey

x = 5ey −xey

x = ey(5 −x)

ey = x

5 −x

ln (ey) = ln

( x

5 −x

) y = ln

( x

5 −x

) We claim f−1(x) = ln

( x

5−x

) . To verify this analytically, we would need to verify the compositions(

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. We leave this to the reader. To verify our solution graphically, we graph y = f(x) = 5e x

ex+1

and y = g(x) = ln (

x 5−x

) on the same set of axes and observe the symmetry about the line y = x.

Note the domain of f is the range of g and vice-versa.

y = f(x) = 5e x

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

( x

5−x

)

456 Exponential and Logarithmic Functions

6.3.1 Exercises

In Exercises 1 - 33, solve the equation analytically.

1. 24x = 8 2. 3(x−1) = 27 3. 52x−1 = 125

4. 42x = 1 2

5. 8x = 1 128

6. 2(x 3−x) = 1

7. 37x = 814−2x 8. 9 · 37x = (

1 9

)2x 9. 32x = 5

10. 5−x = 2 11. 5x = −2 12. 3(x−1) = 29

13. (1.005)12x = 3 14. e−5730k = 1 2

15. 2000e0.1t = 4000

16. 500 ( 1 −e2x

) = 250 17. 70 + 90e−0.1t = 75 18. 30 − 6e−0.1x = 20

19. 100ex

ex + 2 = 50 20.

5000

1 + 2e−3t = 2500 21.

150

1 + 29e−0.8t = 75

22. 25 (

4 5

)x = 10 23. e2x = 2ex 24. 7e2x = 28e−6x

25. 3(x−1) = 2x 26. 3(x−1) = (

1 2

)(x+5) 27. 73+7x = 34−2x

28. e2x − 3ex − 10 = 0 29. e2x = ex + 6 30. 4x + 2x = 12

31. ex − 3e−x = 2 32. ex + 15e−x = 8 33. 3x + 25 · 3−x = 10

In Exercises 34 - 39, solve the inequality analytically.

34. ex > 53 35. 1000 (1.005) 12t ≥ 3000

36. 2(x 3−x) < 1 37. 25

( 4 5

)x ≥ 10 38.

150

1 + 29e−0.8t ≤ 130 39. 70 + 90e−0.1t ≤ 75

In Exercises 40 - 45, use your calculator to help you solve the equation or inequality.

40. 2x = x2 41. ex = ln(x) + 5 42. e √ x = x + 1

43. e−x −xe−x ≥ 0 44. 3(x−1) < 2x 45. ex < x3 −x

46. Since f(x) = ln(x) is a strictly increasing function, if 0 < a < b then ln(a) < ln(b). Use this fact to solve the inequality e(3x−1) > 6 without a sign diagram. Use this technique to solve the inequalities in Exercises 34 - 39. (NOTE: Isolate the exponential function first!)

47. Compute the inverse of f(x) = ex −e−x

2 . State the domain and range of both f and f−1.

6.3 Exponential Equations and Inequalities 457

48. In Example 6.3.4, we found that the inverse of f(x) = 5ex

ex + 1 was f−1(x) = ln

( x

5 −x

) but

we left a few loose ends for you to tie up.

(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) = 5x

x + 1 and h(x) = ex. Show that f = g ◦h and that (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.)

49. With the help of your classmates, solve the inequality ex > xn for a variety of natural numbers n. What might you conjecture about the “speed” at which f(x) = ex grows versus any polynomial?

458 Exponential and Logarithmic Functions

6.3.2 Answers

1. x = 3 4

2. x = 4 3. x = 2

4. x = −1 4

5. x = −7 3

6. x = −1, 0, 1

7. x = 16 15

8. x = − 2 11 9. x =

ln(5) 2 ln(3)

10. x = − ln(2) ln(5)

11. No solution. 12. x = ln(29)+ln(3)

ln(3)

13. x = ln(3)

12 ln(1.005) 14. k = ln( 12 ) −5730 =

ln(2) 5730

15. t = ln(2) 0.1

= 10 ln(2)

16. x = 1 2

ln (

1 2

) = −1

2 ln(2) 17. t =

ln( 118 ) −0.1 = 10 ln(18)

18. x = −10 ln (

5 3

) = 10 ln

( 3 5

) 19. x = ln(2)

20. t = 1 3

ln(2) 21. t = ln( 129 ) −0.8 =

5 4

ln(29)

22. x = ln( 25 ) ln( 45 )

= ln(2)−ln(5) ln(4)−ln(5)

23. x = ln(2)

24. x = −1 8

ln (

1 4

) = 1

4 ln(2) 25. x =

ln(3) ln(3)−ln(2)

26. x = ln(3)+5 ln( 12 ) ln(3)−ln( 12 )

= ln(3)−5 ln(2) ln(3)+ln(2)

27. x = 4 ln(3)−3 ln(7) 7 ln(7)+2 ln(3)

28. x = ln(5) 29. x = ln(3) 30. x = ln(3) ln(2)

31. x = ln(3) 32. x = ln(3), ln(5) 33. x = ln(5) ln(3)

34. (ln(53),∞) 35. [

ln(3) 12 ln(1.005)

,∞ )

36. (−∞,−1) ∪ (0, 1) 37. ( −∞,

ln( 25 ) ln( 45 )

] = ( −∞, ln(2)−ln(5)

ln(4)−ln(5)

] 38.

( −∞,

ln( 2377 ) −0.8

] = ( −∞, 5

4 ln (

377 2

)] 39.

[ ln( 118 ) −0.1 ,∞

) = [10 ln(18),∞)

40. x ≈−0.76666, x = 2, x = 4 41. x ≈ 0.01866, x ≈ 1.7115

42. x = 0 43. (−∞, 1]

44. ≈ (−∞, 2.7095) 45. ≈ (2.3217, 4.3717)

46. x > 1 3 (ln(6) + 1)

47. f−1 = ln ( x + √ x2 + 1

) . Both f and f−1 have domain (−∞,∞) and range (−∞,∞).