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Exponential and Logarithmic Functions
Copyright © Cengage Learning. All rights reserved.

2. 4.5 Exponential and Logarithmic Equations
Copyright © Cengage Learning. All rights reserved.

3. Objectives Exponential Equations Logarithmic Equations
Compound Interest

4. Exponential Equations

5. Exponential Equations
An exponential equation is one in which the variable occurs in the exponent. For example,
2x = 7
The variable x presents a difficulty because it is in the exponent.
To deal with this difficulty, we take the logarithm of each side and then use the Laws of Logarithms to “bring down x” from the exponent.

6. Exponential Equations
ln 2x = ln 7
x ln 2 = ln 7
 2.807
Given Equation
Take In of each side
Law 3 (Bring down exponent)
Solve for x
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7. Exponential Equations
Recall that Law 3 of the Laws of Logarithms says that
loga AC = C loga A.
The method that we used to solve 2x = 7 is typical of how we solve exponential equations in general.

8. Example 1 – Solving an Exponential Equation
Find the solution of the equation 3x + 2 = 7, rounded to six decimal places.
Solution:
We take the common logarithm of each side and use Law 3.
3x + 2 = 7
log(3x + 2) = log 7
Given Equation
Take log of each side

9. Example 1 – Solution (x + 2)log 3 = log 7 x + 2 =  –0.228756 cont’d
Law 3 (bring down exponent)
Divide by log 3
Subtract 2
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10. Example 1 – Solution Check Your Answer
cont’d
Check Your Answer
Substituting x = – into the original equation and using a calculator, we get
3(– ) + 2  7

11. Example 4 – An Exponential Equation of Quadratic Type
Solve the equation e2x – ex – 6 = 0.
Solution:
To isolate the exponential term, we factor.
e2x – ex – 6 = 0
(ex)2 – ex – 6 = 0
Given Equation
Law of Exponents

12. Example 4 – Solution (ex – 3)(ex + 2) = 0 ex – 3 = 0 or ex + 2 = 0
cont’d
(ex – 3)(ex + 2) = 0
ex – 3 = or ex + 2 = 0
ex = ex = –2
The equation ex = 3 leads to x = ln 3.
But the equation ex = –2 has no solution because ex > 0 for all x.
Thus, x = ln 3  is the only solution.
Factor (a quadratic in ex)
Zero-Product Property

13. Logarithmic Equations

14. Logarithmic Equations
A logarithmic equation is one in which a logarithm of the variable occurs.
For example,
log2(x + 2) = 5
To solve for x, we write the equation in exponential form.
x + 2 = 25
x = 32 – 2 = 30
Exponential form
Solve for x

15. Logarithmic Equations
Another way of looking at the first step is to raise the base, 2, to each side of the equation.
2log2(x + 2) = 25
x + 2 = 25
x = 32 – 2 = 30
Raise 2 to each side
Property of logarithms
Solve for x

16. Logarithmic Equations
The method used to solve this simple problem is typical. We summarize the steps as follows.

17. Example 6 – Solving Logarithmic Equations
Solve each equation for x.
(a) ln x = 8
(b) log2(25 – x) = 3
Solution:
(a) ln x = 8
x = e8
Therefore, x = e8  2981.
Given equation
Exponential form

18. Example 6 – Solution We can also solve this problem another way:
cont’d
We can also solve this problem another way:
ln x = 8
eln x = e8
x = e8
Given equation
Raise e to each side
Property of ln

19. Example 6 – Solution
cont’d
(b) The first step is to rewrite the equation in exponential form.
log2(25 – x) = 3
25 – x = 23
25 – x = 8
x = 25 – 8 = 17
Given equation
Exponential form (or raise 2 to each side)

20. Example 6 – Solution Check Your Answer If x = 17, we get
cont’d
Check Your Answer
If x = 17, we get
log2(25 – 17) = log2 8 = 3

21. Logarithmic Equations
Logarithmic equations are used in determining the amount of light that reaches various depths in a lake. (This information helps biologists to determine the types of life a lake can support.)
As light passes through water (or other transparent materials such as glass or plastic), some of the light is absorbed.
It’s easy to see that the murkier the water, the more light is absorbed. The exact relationship between light absorption and the distance light travels in a material is described in the next example.

22. Example 10 – Transparency of a Lake
If I0 and I denote the intensity of light before and after going through a material and x is the distance (in feet) the light travels in the material, then according to the Beer Lambert Law,
where k is a constant depending on the type of material.

23. Example 10 – Transparency of a Lake
cont’d
(a) Solve the equation for I.
(b) For a certain lake k = 0.025, and the light intensity is I0 = 14 lumens (lm).
Find the light intensity at a depth of 20 ft.

24. Example 10 – Solution (a) We first isolate the logarithmic term.
I = I0e–kx
Given equation
Multiply by –k
Exponential form
Multiply by I0

25. Example 10 – Solution (b) We find I using the formula from part (a).
cont’d
(b) We find I using the formula from part (a).
I = I0e–kx
= 14e(–0.025)(20)
 8.49
The light intensity at a depth of 20 ft is about 8.5 lm.
From part (a)
I0 = 14, k = 0.025, x = 20
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26. Compound Interest

27. Compound Interest
If a principal P is invested at an interest rate r for a period of t years, then the amount A of the investment is given by
A = P(1 + r)
A(t) = Pert
We can use logarithms to determine the time it takes for the principal to increase to a given amount.
Simple interest (for one year)
Interest compounded n times per year
Interest compounded continuously

28. Example 11 – Finding the Term for an Investment to Double
A sum of $5000 is invested at an interest rate of 5% per year.
Find the time required for the money to double if the interest is compounded according to the following method.
(a) Semiannually (b) Continuously

29. Example 11 – Solution
(a) We use the formula for compound interest with P = $5000, A(t) = $10,000, r = 0.05, and n = 2 and solve the resulting exponential equation for t.
(1.025)2t = 2
log t = log 2
2t log = log 2
Divide by 5000
Take log of each side
Law 3 (bring down the exponent)

30. Example 11 – Solution t  14.04 The money will double in 14.04 years.
cont’d
t  14.04
The money will double in years.
Divide by 2 log 1.025
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31. Example 11 – Solution
cont’d
(b) We use the formula for continuously compounded interest with P = $5000, A(t) = $10,000, and r = 0.05 and solve the resulting exponential equation for t.
5000e0.05t = 10,000
e0.05t = 2
ln e0.05t = ln 2
0.05t = ln 2
Pert = A
Divide by 5000
Take ln of each side
Property of ln

32. Example 11 – Solution t  13.86 The money will double in 13.86 years.
cont’d
t  13.86
The money will double in years.
Divide by 0.05
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