Lesson 3: Solving Exponential and Logarithmic Equations

Introduction

Now that we understand the properties of exponential and logarithmic functions, we're ready to solve equations involving these functions. Solving these equations requires careful strategy selection and attention to mathematical properties.

In this lesson, you'll learn:

How to solve exponential equations using the same base method
How to solve exponential equations by taking logarithms
How to solve logarithmic equations using exponential form
How to use the one-to-one property of logarithms
How to identify and exclude extraneous solutions
How to apply these techniques to real-world problems

Strategy Overview:

For exponential equations: Try to express both sides with the same base, or take the logarithm of both sides
For logarithmic equations: Convert to exponential form, or use properties to consolidate logarithms
Always check solutions: Logarithms have domain restrictions (arguments must be positive)

Section 1: Solving Exponential Equations (Same Base Method)

When both sides of an exponential equation can be expressed with the same base, we can use the property that if b^x = b^y, then x = y (assuming b > 0 and b ≠ 1).

Same Base Property:
If b^x = b^y, then x = y

Example 1: Solve 2^x = 16

Solution:

Step 1: Express 16 as a power of 2

16 = 2⁴

Step 2: Rewrite the equation

2^x = 2⁴

Step 3: Since the bases are equal, the exponents must be equal

x = 4

Step 4: Check

2⁴ = 16

Answer: x = 4

Example 2: Solve 3^x+1 = 27

Solution:

Step 1: Express 27 as a power of 3

27 = 3³

Step 2: Rewrite the equation

3^x+1 = 3³

Step 3: Equate the exponents

x + 1 = 3

Step 4: Solve for x

x = 2

Step 5: Check

3²⁺¹ = 3³ = 27

Answer: x = 2

Example 3: Solve 4^x = 8

Solution:

Step 1: Express both 4 and 8 as powers of 2

4 = 2², so 4^x = (2²)^x = 2^2x

8 = 2³

Step 2: Rewrite the equation

2^2x = 2³

Step 3: Equate the exponents

2x = 3

Step 4: Solve for x

x = 3/2

Step 5: Check

4^3/2 = (4^1/2)³ = 2³ = 8

Answer: x = 3/2 or 1.5

Example 4: Solve (1/2)^x = 32

Solution:

Step 1: Express 1/2 and 32 as powers of 2

1/2 = 2^-1, so (1/2)^x = (2^-1)^x = 2^-x

32 = 2⁵

Step 2: Rewrite the equation

2^-x = 2⁵

Step 3: Equate the exponents

-x = 5

Step 4: Solve for x

x = -5

Step 5: Check

(1/2)^-5 = 2⁵ = 32

Answer: x = -5

Example 5: Solve 9^x = 3^x+4

Solution:

Step 1: Express 9 as a power of 3

9 = 3², so 9^x = (3²)^x = 3^2x

Step 2: Rewrite the equation

3^2x = 3^x+4

Step 3: Equate the exponents

2x = x + 4

Step 4: Solve for x

2x - x = 4

x = 4

Step 5: Check

9⁴ = 6561 and 3⁴⁺⁴ = 3⁸ = 6561

Answer: x = 4

Example 6: Solve 25^x-1 = 125

Solution:

Step 1: Express both sides as powers of 5

25 = 5², so 25^x-1 = (5²)^x-1 = 5^2(x-1) = 5^2x-2

125 = 5³

Step 2: Rewrite the equation

5^2x-2 = 5³

Step 3: Equate the exponents

2x - 2 = 3

Step 4: Solve for x

2x = 5

x = 5/2

Answer: x = 5/2 or 2.5

Section 2: Solving Exponential Equations (Different Bases)

When the bases cannot easily be expressed as powers of the same number, we take the logarithm of both sides. You can use either common logarithms (base 10) or natural logarithms (base e).

Logarithm Method:
To solve b^x = a:
1. Take log of both sides: log(b^x) = log(a)
2. Use power rule: x·log(b) = log(a)
3. Solve for x: x = log(a) / log(b)

Example 7: Solve 2^x = 10

Solution:

Step 1: Take the logarithm of both sides (we'll use common log)

log(2^x) = log(10)

Step 2: Apply the power rule

x·log(2) = log(10)

Step 3: Note that log(10) = 1

x·log(2) = 1

Step 4: Solve for x

x = 1 / log(2)

x ≈ 1 / 0.3010

x ≈ 3.322

Answer: x = 1/log(2) ≈ 3.322

Example 8: Solve 5^x = 30

Solution:

Step 1: Take the natural logarithm of both sides

ln(5^x) = ln(30)

Step 2: Apply the power rule

x·ln(5) = ln(30)

Step 3: Solve for x

x = ln(30) / ln(5)

x ≈ 3.4012 / 1.6094

x ≈ 2.113

Answer: x = ln(30)/ln(5) ≈ 2.113

Example 9: Solve 3·2^x = 24

Solution:

Step 1: Isolate the exponential expression first

2^x = 24/3

2^x = 8

Step 2: Recognize that 8 = 2³

2^x = 2³

Step 3: Equate exponents

x = 3

Alternative method using logarithms:

Take log of both sides: log(2^x) = log(8)

x·log(2) = log(8)

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

Answer: x = 3

Example 10: Solve 2^x-1 = 7

Solution:

Step 1: Take the logarithm of both sides

log(2^x-1) = log(7)

Step 2: Apply the power rule

(x - 1)·log(2) = log(7)

Step 3: Solve for x

x - 1 = log(7) / log(2)

x = log(7) / log(2) + 1

x ≈ 2.807 + 1

x ≈ 3.807

Answer: x = log(7)/log(2) + 1 ≈ 3.807

Example 11: Solve 7^2x+1 = 50

Solution:

Step 1: Take the natural logarithm of both sides

ln(7^2x+1) = ln(50)

Step 2: Apply the power rule

(2x + 1)·ln(7) = ln(50)

Step 3: Distribute

2x·ln(7) + ln(7) = ln(50)

Step 4: Isolate the term with x

2x·ln(7) = ln(50) - ln(7)

Step 5: Solve for x

x = [ln(50) - ln(7)] / [2·ln(7)]

x ≈ [3.912 - 1.946] / [2·1.946]

x ≈ 1.966 / 3.892

x ≈ 0.505

Answer: x = [ln(50) - ln(7)] / [2·ln(7)] ≈ 0.505

Example 12: Solve 100·(1.05)^x = 200

Solution:

Step 1: Isolate the exponential expression

(1.05)^x = 200/100

(1.05)^x = 2

Step 2: Take the logarithm of both sides

log(1.05^x) = log(2)

Step 3: Apply the power rule

x·log(1.05) = log(2)

Step 4: Solve for x

x = log(2) / log(1.05)

x ≈ 0.3010 / 0.0212

x ≈ 14.21

This type of equation appears in compound interest problems.

Answer: x = log(2)/log(1.05) ≈ 14.21

Section 3: Solving Exponential Equations with e

When the base is e (Euler's number), we use the natural logarithm as the inverse operation. Remember that ln(e) = 1 and ln(e^x) = x.

Natural Logarithm Properties:
ln(e) = 1
ln(e^x) = x
e^ln(x) = x

Example 13: Solve e^x = 15

Solution:

Step 1: Take the natural logarithm of both sides

ln(e^x) = ln(15)

Step 2: Simplify using ln(e^x) = x

x = ln(15)

x ≈ 2.708

Step 3: Check

e^2.708 ≈ 15

Answer: x = ln(15) ≈ 2.708

Example 14: Solve 50e^0.08t = 200

Solution:

Step 1: Isolate the exponential expression

e^0.08t = 200/50

e^0.08t = 4

Step 2: Take the natural logarithm of both sides

ln(e^0.08t) = ln(4)

Step 3: Simplify the left side

0.08t = ln(4)

Step 4: Solve for t

t = ln(4) / 0.08

t ≈ 1.3863 / 0.08

t ≈ 17.33

This type of equation models continuous exponential growth.

Answer: t = ln(4)/0.08 ≈ 17.33

Example 15: Solve 5e^2x-1 = 30

Solution:

Step 1: Isolate the exponential expression

e^2x-1 = 30/5

e^2x-1 = 6

Step 2: Take the natural logarithm of both sides

ln(e^2x-1) = ln(6)

Step 3: Simplify

2x - 1 = ln(6)

Step 4: Solve for x

2x = ln(6) + 1

x = [ln(6) + 1] / 2

x ≈ [1.792 + 1] / 2

x ≈ 1.396

Answer: x = [ln(6) + 1]/2 ≈ 1.396

Example 16: Solve A = Pe^rt for t

Solution:

This is the continuous compound interest formula. We need to isolate t.

Step 1: Divide both sides by P

A/P = e^rt

Step 2: Take the natural logarithm of both sides

ln(A/P) = ln(e^rt)

Step 3: Simplify the right side

ln(A/P) = rt

Step 4: Solve for t

t = ln(A/P) / r

This formula tells us how long it takes for an investment to grow from P to A at rate r.

Answer: t = ln(A/P) / r

Section 4: Solving Logarithmic Equations

To solve logarithmic equations, we can either convert to exponential form or use properties of logarithms to consolidate into a single logarithm. Always check for extraneous solutions since logarithms are only defined for positive arguments.

Strategies for Logarithmic Equations:
1. Convert to exponential form: If log_b(x) = y, then b^y = x
2. One-to-one property: If log_b(M) = log_b(N), then M = N
3. Always check that solutions make arguments positive

Example 17: Solve log₂(x) = 5

Solution:

Step 1: Convert to exponential form

If log₂(x) = 5, then 2⁵ = x

Step 2: Evaluate

x = 32

Step 3: Check (is x > 0?)

Yes, 32 > 0

log₂(32) = log₂(2⁵) = 5

Answer: x = 32

Example 18: Solve log(x) = 3

Solution:

Note: log(x) means log₁₀(x)

Step 1: Convert to exponential form

If log₁₀(x) = 3, then 10³ = x

Step 2: Evaluate

x = 1000

Step 3: Check

log(1000) = log(10³) = 3

Answer: x = 1000

Example 19: Solve ln(x) = 2

Solution:

Step 1: Convert to exponential form

If ln(x) = 2, then e² = x

Step 2: Evaluate

x = e² ≈ 7.389

Step 3: Check

ln(e²) = 2

Answer: x = e² ≈ 7.389

Example 20: Solve log₃(2x - 1) = 4

Solution:

Step 1: Convert to exponential form

If log₃(2x - 1) = 4, then 3⁴ = 2x - 1

Step 2: Evaluate 3⁴

81 = 2x - 1

Step 3: Solve for x

82 = 2x

x = 41

Step 4: Check (is 2x - 1 > 0?)

2(41) - 1 = 81 > 0

log₃(81) = log₃(3⁴) = 4

Answer: x = 41

Example 21: Solve log(x) + log(x - 3) = 1

Solution:

Step 1: Use the product rule to combine logarithms

log[x(x - 3)] = 1

log(x² - 3x) = 1

Step 2: Convert to exponential form

10¹ = x² - 3x

10 = x² - 3x

Step 3: Rearrange to standard form

x² - 3x - 10 = 0

Step 4: Factor

(x - 5)(x + 2) = 0

Step 5: Solve

x = 5 or x = -2

Step 6: Check domain restrictions

For x = 5: log(5) and log(2) are both defined

For x = -2: log(-2) is undefined

x = -2 is an extraneous solution.

Answer: x = 5

Example 22: Solve log₂(x + 2) - log₂(x) = 3

Solution:

Step 1: Use the quotient rule to combine logarithms

log₂[(x + 2)/x] = 3

Step 2: Convert to exponential form

2³ = (x + 2)/x

8 = (x + 2)/x

Step 3: Multiply both sides by x

8x = x + 2

Step 4: Solve for x

7x = 2

x = 2/7

Step 5: Check domain (x > 0 and x + 2 > 0)

2/7 > 0 and 2/7 + 2 > 0

Answer: x = 2/7

Example 23: Solve log₅(x² - 4) = 2

Solution:

Step 1: Convert to exponential form

5² = x² - 4

25 = x² - 4

Step 2: Solve for x²

x² = 29

Step 3: Take square root of both sides

x = ±√29

Step 4: Check domain (x² - 4 > 0)

For x = √29: (√29)² - 4 = 29 - 4 = 25 > 0

For x = -√29: (-√29)² - 4 = 29 - 4 = 25 > 0

Both solutions are valid since x² - 4 > 0 for both.

Answer: x = ±√29 (approximately ±5.385)

Example 24: Solve ln(x) + ln(x - 1) = ln(12)

Solution:

Step 1: Use the product rule on the left side

ln[x(x - 1)] = ln(12)

ln(x² - x) = ln(12)

Step 2: Use the one-to-one property

If ln(A) = ln(B), then A = B

x² - x = 12

Step 3: Rearrange to standard form

x² - x - 12 = 0

Step 4: Factor

(x - 4)(x + 3) = 0

Step 5: Solve

x = 4 or x = -3

Step 6: Check domain (x > 0 and x - 1 > 0, so x > 1)

For x = 4: 4 > 1

For x = -3: -3 is not > 1

Answer: x = 4

Section 5: Equations with Both Exponentials and Logarithms

Some equations involve both exponential and logarithmic expressions. Use the inverse relationship between these functions to simplify.

Example 25: Solve ln(e^x) = 4

Solution:

Step 1: Use the property ln(e^x) = x

x = 4

Step 2: Check

ln(e⁴) = 4

Answer: x = 4

Example 26: Solve log(10^2x) = 6

Solution:

Step 1: Use the power rule

2x·log(10) = 6

Step 2: Since log(10) = 1

2x·(1) = 6

2x = 6

Step 3: Solve for x

x = 3

Answer: x = 3

Example 27: Solve e^ln(x) = 20

Solution:

Step 1: Use the property e^ln(x) = x

x = 20

Step 2: Check domain (x must be positive for ln(x) to be defined)

20 > 0

Step 3: Verify

e^ln(20) = 20

Answer: x = 20

Section 6: Applications of Exponential and Logarithmic Equations

Exponential and logarithmic equations arise naturally in many real-world contexts, including finance, population growth, radioactive decay, and chemistry.

Example 28: How long will it take for an investment to double if it earns 6% annual interest compounded continuously?

Solution:

Use the continuous compound interest formula: A = Pe^rt

Given information:

A = 2P (double the principal)

r = 0.06

Find: t

Step 1: Substitute into the formula

2P = Pe^0.06t

Step 2: Divide both sides by P

2 = e^0.06t

Step 3: Take the natural logarithm of both sides

ln(2) = ln(e^0.06t)

ln(2) = 0.06t

Step 4: Solve for t

t = ln(2) / 0.06

t ≈ 0.693 / 0.06

t ≈ 11.55 years

Answer: Approximately 11.55 years

Example 29: A bacteria population grows according to P(t) = 500·2^t/3, where t is in hours. When will the population reach 5000?

Solution:

Given: P(t) = 500·2^t/3

Find t when P(t) = 5000

Step 1: Set up the equation

5000 = 500·2^t/3

Step 2: Divide both sides by 500

10 = 2^t/3

Step 3: Take log of both sides

log(10) = log(2^t/3)

1 = (t/3)·log(2)

Step 4: Solve for t

t/3 = 1/log(2)

t = 3/log(2)

t ≈ 3/0.301

t ≈ 9.97 hours

Answer: Approximately 10 hours

Example 30: The half-life of a radioactive substance is 50 years. If we start with 100 grams, the amount remaining after t years is A(t) = 100·(1/2)^t/50. How long until only 20 grams remain?

Solution:

Given: A(t) = 100·(1/2)^t/50

Find t when A(t) = 20

Step 1: Set up the equation

20 = 100·(1/2)^t/50

Step 2: Divide both sides by 100

0.2 = (1/2)^t/50

Step 3: Take log of both sides

log(0.2) = log[(1/2)^t/50]

log(0.2) = (t/50)·log(1/2)

Step 4: Solve for t

t/50 = log(0.2) / log(0.5)

t = 50·[log(0.2) / log(0.5)]

t ≈ 50·[-0.699 / -0.301]

t ≈ 50·(2.322)

t ≈ 116.1 years

Answer: Approximately 116 years

Example 31: The pH of a solution is given by pH = -log[H⁺], where [H⁺] is the hydrogen ion concentration. If a solution has pH = 4.5, what is the hydrogen ion concentration?

Solution:

Given: pH = -log[H⁺] and pH = 4.5

Find: [H⁺]

Step 1: Substitute pH = 4.5

4.5 = -log[H⁺]

Step 2: Multiply both sides by -1

-4.5 = log[H⁺]

Step 3: Convert to exponential form

[H⁺] = 10^-4.5

[H⁺] ≈ 3.16 × 10^-5

This is approximately 0.0000316 moles per liter.

Answer: [H⁺] = 10^-4.5 ≈ 3.16 × 10^-5 mol/L

Example 32: Newton's Law of Cooling states that T(t) = T_s + (T₀ - T_s)e^-kt, where T(t) is temperature at time t, T_s is surrounding temperature, and T₀ is initial temperature. A cup of coffee at 95°C is placed in a room at 20°C. After 10 minutes, the temperature is 70°C. How long until the coffee reaches 50°C?

Solution:

Given: T_s = 20, T₀ = 95

T(10) = 70

Step 1: Find k using T(10) = 70

70 = 20 + (95 - 20)e^-10k

50 = 75e^-10k

2/3 = e^-10k

ln(2/3) = -10k

k = -ln(2/3)/10 ≈ 0.0405

Step 2: Find t when T(t) = 50

50 = 20 + 75e^-0.0405t

30 = 75e^-0.0405t

0.4 = e^-0.0405t

ln(0.4) = -0.0405t

t = ln(0.4) / -0.0405

t ≈ 22.6 minutes

Answer: Approximately 22.6 minutes

Check Your Understanding

Try these problems to test your knowledge. Click to reveal the solution after you've attempted each problem.

Problem 1: Solve 5^x = 125

Show Solution

Express 125 as a power of 5: 125 = 5³

5^x = 5³

x = 3

Answer: x = 3

Problem 2: Solve 3^x = 20

Show Solution

Take log of both sides: log(3^x) = log(20)

x·log(3) = log(20)

x = log(20)/log(3) ≈ 2.727

Answer: x = log(20)/log(3) ≈ 2.727

Problem 3: Solve e^2x = 50

Show Solution

Take natural log: ln(e^2x) = ln(50)

2x = ln(50)

x = ln(50)/2 ≈ 1.956

Answer: x = ln(50)/2 ≈ 1.956

Problem 4: Solve log₄(x) = 3

Show Solution

Convert to exponential form: 4³ = x

x = 64

Check: log₄(64) = log₄(4³) = 3

Answer: x = 64

Problem 5: Solve log(x) + log(x + 21) = 2

Show Solution

Combine logs: log[x(x + 21)] = 2

Convert to exponential: x(x + 21) = 10² = 100

x² + 21x - 100 = 0

Factor: (x + 25)(x - 4) = 0

x = -25 or x = 4

Check domain: -25 makes log(x) undefined

Answer: x = 4

Problem 6: Solve ln(x + 1) - ln(x) = 2

Show Solution

Use quotient rule: ln[(x+1)/x] = 2

Convert to exponential: (x+1)/x = e²

x + 1 = xe²

1 = x(e² - 1)

x = 1/(e² - 1) ≈ 0.157

Answer: x = 1/(e² - 1) ≈ 0.157

Problem 7: Solve 2·5^x+1 = 100

Show Solution

Divide by 2: 5^x+1 = 50

Take log: log(5^x+1) = log(50)

(x+1)·log(5) = log(50)

x + 1 = log(50)/log(5)

x = log(50)/log(5) - 1 ≈ 1.431 - 1 = 0.431

Answer: x ≈ 0.431

Problem 8: Solve log₂(x² - 5) = 3

Show Solution

Convert to exponential: x² - 5 = 2³ = 8

x² = 13

x = ±√13

Check domain: Both make x² - 5 = 13 - 5 = 8 > 0

Answer: x = ±√13 ≈ ±3.606

Problem 9: An investment grows according to A = 1000(1.08)^t. How long until it reaches $2000?

Show Solution

Set up equation: 2000 = 1000(1.08)^t

Divide by 1000: 2 = 1.08^t

Take log: log(2) = t·log(1.08)

t = log(2)/log(1.08) ≈ 9.01 years

Answer: Approximately 9 years

Problem 10: Solve 4^x = 2^x+3

Show Solution

Express 4 as 2²: (2²)^x = 2^x+3

Simplify: 2^2x = 2^x+3

Equate exponents: 2x = x + 3

Solve: x = 3

Answer: x = 3

Key Takeaways

For exponential equations with the same base: Express both sides with the same base and equate exponents
For exponential equations with different bases: Take the logarithm of both sides
When the base is e, use natural logarithms (ln) for efficiency
For logarithmic equations: Convert to exponential form or use the one-to-one property
Always check solutions to ensure logarithm arguments are positive
Extraneous solutions can arise when solving logarithmic equations - always verify!
These techniques are essential for solving real-world problems in finance, science, and engineering
Remember: ln(e^x) = x and e^ln(x) = x
The change of base formula can help: log_b(a) = log(a)/log(b) = ln(a)/ln(b)