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Unit 4 · Chapter 4.6

4.6Exponential and Logarithmic Equations

Solve exponential equations using the one-to-one property or by taking logarithms of both sides. Solve logarithmic equations by condensing and converting to exponential form — always checking for extraneous solutions.

Solving exponential and logarithmic equations is the core skill behind compound interest calculations, radioactive decay problems, pH chemistry, and any real-world model where a quantity grows or decays exponentially.

Essential Question

Why do we take the logarithm of both sides to solve an exponential equation — and why must we always check solutions to logarithmic equations for extraneous roots?

Lesson Overview

Exponential and logarithmic equations are inverses of each other — solving one type often requires converting to the other. There are two main strategies: the one-to-one property (rewrite both sides with the same base, then set exponents equal) and the logarithm method (take log of both sides of an exponential equation, or convert a condensed log equation to exponential form). Logarithmic equations require an extra step: always check solutions, because the domain restriction of logs can produce extraneous roots.

Exponential Equations — Two Methods

  • Same base: bˣ = bʸ → x = y (one-to-one property)
  • Different base: take ln or log of both sides, apply power rule, solve
  • Prefer ln when base is e; prefer log₁₀ otherwise
  • No extraneous solutions for exponential equations

Logarithmic Equations — Two Methods

  • Same base: log_b(x) = log_b(y) → x = y
  • Condense → convert: get one log, convert to exponential form, solve
  • Always check: all arguments must be positive in original equation
  • Extraneous solutions are common — never skip the check
Which Strategy?Same basepossible?YESOne-to-Onebˣ = bʸ → x=yNOExponentialequation?YESTake logboth sidesNOLog equationCondense → convert to exp⚠ Always check for extraneous solutions!One-to-One PropertiesExponentialbˣ = bʸ⟺x = yLogarithmiclog_b(x) = log_b(y)⟺x = ySolving Exponential Equations — Step Pattern1Isolate the exponential expression3·2ˣ = 24 → 2ˣ = 82Take log of both sides (use ln or log)ln(2ˣ) = ln(8)3Apply power rule: bring exponent downx·ln(2) = ln(8)4Divide to isolate xx = ln(8)/ln(2) = 3Solving Logarithmic Equations — Step Pattern1Condense all logs to one sidelog₂(x) + log₂(x−2) = 32Apply product/quotient ruleslog₂(x(x−2)) = 33Convert to exponential formx(x−2) = 2³ = 84Solve and check for extraneousx²−2x−8=0 → x=4 ✓ x=−2 ✗

Worked Examples

Example 1

Solve: 4^(x+1) = 64

Rewrite 64 as a power of 4: 64 = 4³

Equation becomes: 4^(x+1) = 4³

One-to-one property (same base): x + 1 = 3

Solve: x = 2

Check: 4^(2+1) = 4³ = 64 ✓

Answer:x = 2
Example 2

Solve: 5^x = 18 (round to four decimal places)

Bases cannot be made equal, so take ln of both sides: ln(5^x) = ln(18)

Power rule: x·ln(5) = ln(18)

Divide: x = ln(18)/ln(5)

Calculate: x = 2.8904/1.6094 ≈ 1.7959

Check: 5^1.7959 ≈ 18 ✓

Answer:x ≈ 1.7959
Example 3

Solve: 3·e^(2x) − 5 = 22

Isolate the exponential: 3·e^(2x) = 27 → e^(2x) = 9

Take ln of both sides: ln(e^(2x)) = ln(9)

Simplify left side (ln and e cancel): 2x = ln(9)

Divide: x = ln(9)/2

Calculate: x = 2.1972/2 ≈ 1.0986

Answer:x = ln(9)/2 ≈ 1.0986
Example 4

Solve: log₃(x + 4) = 2

The log is already isolated. Convert to exponential form: 3² = x + 4

9 = x + 4

x = 5

Check: log₃(5 + 4) = log₃(9) = 2 ✓ (argument 9 > 0 ✓)

Answer:x = 5
Example 5

Solve: log₂(x) + log₂(x − 2) = 3

Condense left side using product rule: log₂(x(x−2)) = 3

Convert to exponential form: x(x−2) = 2³ = 8

Expand: x² − 2x = 8 → x² − 2x − 8 = 0

Factor: (x − 4)(x + 2) = 0 → x = 4 or x = −2

Check x = 4: log₂(4) + log₂(2) = 2 + 1 = 3 ✓

Check x = −2: log₂(−2) is undefined (argument negative) ✗ — extraneous!

Answer:x = 4 (x = −2 is extraneous)

Guided Practice

Guided Problem 1

Solve: 2^(x−3) = 32

Hint: Rewrite 32 as a power of 2. Then use the one-to-one property to set exponents equal.

Guided Problem 2

Solve: 7^x = 50 (round to four decimal places)

Hint: Take ln of both sides. Apply the power rule to bring x down. Then divide by ln(7).

Guided Problem 3

Solve: 4·e^(x+1) = 60

Hint: Isolate e^(x+1) first by dividing by 4. Then take ln of both sides.

Guided Problem 4

Solve: log₅(2x − 1) = 2

Hint: Convert directly to exponential form: 5² = 2x − 1. Then solve for x and check.

Guided Problem 5

Solve: log(x + 3) + log(x − 2) = 1

Hint: Condense using the product rule. Convert to exponential form (base 10). Solve the quadratic and check both solutions.

Key Vocabulary

One-to-One Property (Exponential)

If bˣ = bʸ, then x = y. Works when both sides can be written as the same base. The most efficient method when applicable.

One-to-One Property (Logarithmic)

If log_b(x) = log_b(y), then x = y. Both sides must have the same base before applying.

Logarithm Method

Taking the logarithm of both sides of an exponential equation. The power rule then brings the variable exponent down as a coefficient.

Extraneous Solution

A value that satisfies the transformed equation but not the original. In log equations, any solution that makes an argument ≤ 0 is extraneous.

Isolate the Exponential

Before taking a log, rewrite the equation so the exponential expression stands alone on one side.

Convert to Exponential Form

Rewriting log_b(x) = y as b^y = x. Used to solve log equations after condensing to a single log.

Condensing

Combining multiple logarithms into one using product, quotient, and power rules. Required before converting a log equation to exponential form.

Natural Logarithm Method

Using ln (base e) to solve equations with base e. Since ln(eˣ) = x, the exponential and log cancel cleanly.

Check Your Understanding

Interactive Practice — 5 Questions

1

Solve: 3^(x+2) = 27

2

Solve: 2^x = 10 (to four decimal places)

3

Solve: log₄(x − 1) = 3

4

Solve: ln(x) + ln(x + 2) = ln(8). Which value is the solution?

5

Solve: e^(3x) = 20 (to four decimal places)

Independent Practice

Independent Practice

1

Solve: 5^(x−1) = 125

2

Solve: 6^x = 40 (round to four decimal places)

3

Solve: 2·e^(x−3) = 14 (round to four decimal places)

4

Solve: log₂(3x + 1) = 4

5

Solve: log₃(x) + log₃(x + 6) = 3. Check for extraneous solutions.

⚠ Extraneous Solutions in Log EquationsWhen you solve log(x) = something, the solution must makeALL arguments of ALL logs in the original equation positive.Example: log(x+2)+log(x−1)=1Solutions: x=3 and x=−4x=−4 is extraneous (log(−2) undefined)Always check:Substitute back into theORIGINAL equation

Extraneous solutions arise when a candidate makes a log argument non-positive

⚠️

Common Mistakes

❌

Forgetting to check for extraneous solutions in log equations — accepting x = −2 without verifying.

✅

Always substitute every solution back into the ORIGINAL equation. If any log argument is ≤ 0, that solution is extraneous and must be rejected.

❌

Taking log of both sides before isolating the exponential: log(3·2^x) = log(24) → 3x·log(2) = log(24).

✅

Isolate first: 3·2^x = 24 → 2^x = 8. Now either use one-to-one (2³ = 8, x = 3) or take log: x·log(2) = log(8).

❌

Applying the one-to-one property when bases are different: 2^x = 3^x → x = x.

✅

One-to-one only works when BOTH sides have the SAME base. For different bases, take log of both sides.

❌

Solving log₂(x) + log₂(x−2) = 3 by converting each log separately: 2^3 = x and 2^3 = x−2.

✅

Condense FIRST using the product rule: log₂(x(x−2)) = 3. Then convert the single log to exponential form.

💡

Math Tips

📌

For exponential equations: try same-base first (fastest). If impossible, take ln (for base e) or log₁₀ (for other bases).

📌

Isolate the exponential BEFORE taking a log. Extra coefficients or constants must be moved first.

📌

For log equations: condense to ONE log, then convert to exponential form. Never convert multiple logs separately.

⚠️

Always check log equation solutions. Substitute back into the ORIGINAL equation — not the simplified version.

📌

ln(eˣ) = x and e^(ln x) = x. These cancellations make solving equations with base e especially clean.