Skip to content

Solving Exponential Equations

Using properties and logarithms to solve

Written and reviewed by the Study Mondo Education TeamLast updated
🎯⭐ INTERACTIVE LESSON

Try the Interactive Version!

Learn step-by-step with practice exercises built right in.

Start Interactive Lesson →

Solving Exponential Equations

Strategy 1: Same Base

If you can write both sides with the same base, set exponents equal.

Example: 2x=82^x = 8 2x=232^x = 2^3 x=3x = 3

Strategy 2: Take Logarithms

When bases can't match, use logarithms:

Example: 3x=73^x = 7 ln⁡(3x)=ln⁡(7)\ln(3^x) = \ln(7) xln⁡(3)=ln⁡(7)x \ln(3) = \ln(7) x=ln⁡(7)ln⁡(3)x = \frac{\ln(7)}{\ln(3)}

Properties Used

Power Property: log⁡(ab)=blog⁡(a)\log(a^b) = b \log(a)

One-to-One Property: If bx=byb^x = b^y, then x=yx = y

Common Equations

Form: a⋅bcx=da \cdot b^{cx} = d

Steps:

  1. Isolate the exponential term
  2. Take log of both sides
  3. Use power property
  4. Solve for xx

Example: 5⋅23x=405 \cdot 2^{3x} = 40 23x=82^{3x} = 8 23x=232^{3x} = 2^3 3x=33x = 3 x=1x = 1

📚 Practice Problems

1Problem 1easy

❓ Question:

Solve: 2ˣ = 32

💡 Show Solution

Step 1: Express both sides with the same base: 32 = 2⁵

Step 2: Rewrite the equation: 2ˣ = 2⁵

Step 3: Since bases are equal, exponents must be equal: x = 5

Step 4: Check: 2⁵ = 32 ✓

Answer: x = 5

2Problem 2easy

❓ Question:

Solve: 5x=1255^x = 125

💡 Show Solution

Write 125 as a power of 5: 5x=535^x = 5^3

Since the bases are equal: x=3x = 3

Answer: x=3x = 3

3Problem 3easy

❓ Question:

Solve: 3ˣ⁺¹ = 81

💡 Show Solution

Step 1: Express 81 as a power of 3: 81 = 3⁴

Step 2: Rewrite the equation: 3ˣ⁺¹ = 3⁴

Step 3: Set exponents equal: x + 1 = 4

Step 4: Solve for x: x = 3

Step 5: Check: 3³⁺¹ = 3⁴ = 81 ✓

Answer: x = 3

4Problem 4medium

❓ Question:

Solve: 4x=204^x = 20

💡 Show Solution

The bases don't match easily, so use logarithms:

ln⁡(4x)=ln⁡(20)\ln(4^x) = \ln(20)

Use power property: xln⁡(4)=ln⁡(20)x \ln(4) = \ln(20)

Solve for xx: x=ln⁡(20)ln⁡(4)≈2.161x = \frac{\ln(20)}{\ln(4)} \approx 2.161

Answer: x=ln⁡(20)ln⁡(4)x = \frac{\ln(20)}{\ln(4)} or approximately 2.1612.161

5Problem 5medium

❓ Question:

Solve: 5²ˣ = 125ˣ⁻¹

💡 Show Solution

Step 1: Express 125 as a power of 5: 125 = 5³

Step 2: Rewrite the equation: 5²ˣ = (5³)ˣ⁻¹

Step 3: Apply power rule (bᵐ)ⁿ = bᵐⁿ: 5²ˣ = 5³⁽ˣ⁻¹⁾ 5²ˣ = 5³ˣ⁻³

Step 4: Set exponents equal: 2x = 3x - 3

Step 5: Solve for x: 2x - 3x = -3 -x = -3 x = 3

Step 6: Check: Left: 5²⁽³⁾ = 5⁶ Right: 125³⁻¹ = 125² = (5³)² = 5⁶ ✓

Answer: x = 3

6Problem 6medium

❓ Question:

Solve using logarithms: 2ˣ = 15

💡 Show Solution

Step 1: Take logarithm of both sides: We can use any base, but log₁₀ or ln are common log(2ˣ) = log(15)

Step 2: Apply power rule: x log(2) = log(15)

Step 3: Solve for x: x = log(15)/log(2)

Step 4: Calculate (using calculator): log(15) ≈ 1.1761 log(2) ≈ 0.3010 x ≈ 1.1761/0.3010 x ≈ 3.907

Step 5: Check: 2³·⁹⁰⁷ ≈ 15.00 ✓

Answer: x = log(15)/log(2) ≈ 3.907

7Problem 7hard

❓ Question:

Solve: 3⋅2x+1=483 \cdot 2^{x+1} = 48

💡 Show Solution

Step 1: Isolate the exponential 2x+1=162^{x+1} = 16

Step 2: Write 16 as a power of 2 2x+1=242^{x+1} = 2^4

Step 3: Set exponents equal x+1=4x + 1 = 4

Step 4: Solve x=3x = 3

Check: 3⋅23+1=3⋅24=3⋅16=483 \cdot 2^{3+1} = 3 \cdot 2^4 = 3 \cdot 16 = 48 ✓

Answer: x=3x = 3

8Problem 8hard

❓ Question:

Solve: 4ˣ - 2ˣ⁺¹ - 8 = 0

💡 Show Solution

Step 1: Express 4ˣ in terms of 2ˣ: 4ˣ = (2²)ˣ = 2²ˣ = (2ˣ)²

Step 2: Express 2ˣ⁺¹: 2ˣ⁺¹ = 2ˣ · 2¹ = 2 · 2ˣ

Step 3: Let u = 2ˣ, then substitute: (2ˣ)² - 2 · 2ˣ - 8 = 0 u² - 2u - 8 = 0

Step 4: Factor the quadratic: (u - 4)(u + 2) = 0

Step 5: Solve for u: u = 4 or u = -2

Step 6: Substitute back 2ˣ for u: 2ˣ = 4 or 2ˣ = -2

Step 7: Solve each equation: 2ˣ = 4 → 2ˣ = 2² → x = 2 ✓ 2ˣ = -2 → No solution (2ˣ is always positive)

Step 8: Check x = 2: 4² - 2²⁺¹ - 8 = 16 - 8 - 8 = 0 ✓

Answer: x = 2

Explain using:

⚠️ Common Mistakes: Solving Exponential Equations

Avoid these 3 frequent errors

🌍 Real-World Applications: Solving Exponential Equations

See how this math is used in the real world

📝 Worked Example: Solving a Quadratic by Factoring

Problem:

Solve x2−5x+6=0x^2 - 5x + 6 = 0.

2Factor the quadratic
3Set each factor equal to zero
← Previous Topic
Logarithmic Functions
Next Topic →
Solving Logarithmic Equations
🎴

Practice with Flashcards

Rate this topic's cards with spaced repetition. Cards join your deck when you finish a topic's lesson and take its exit quiz.

📖

Browse All Topics

Explore more Algebra 2 topics

📌 Related Topics in Exponential and Logarithmic Functions

Exponential FunctionsLogarithmic FunctionsSolving Logarithmic Equations

❓ Frequently Asked Questions

What is Solving Exponential Equations?▾
Using properties and logarithms to solve
How can I study Solving Exponential Equations effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 8 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Solving Exponential Equations study guide free?▾
Yes — all study notes, flashcards, and practice problems for Solving Exponential Equations on Study Mondo are free to access. No account is needed.
What course covers Solving Exponential Equations?▾
Solving Exponential Equations is part of the Algebra 2 course on Study Mondo, specifically in the Exponential and Logarithmic Functions section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Solving Exponential Equations?▾
Yes, this page includes 8 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.