What is an exponential equation?
In an exponential equation the unknown x sits in the exponent (the power): 5^x = 125, 3^(x+1) = 27, 2^x = 7.
Key fact: for a base b > 0 and b ≠ 1, the function y = b^x is one-to-one. Each output comes from only one input. So if b^p = b^q, then p = q. This is why we can compare powers.
Also b^x is always positive. So 2^x = −4 or 2^x = 0 have no solution.
Method 1: make the bases the same
Write both sides as powers of one base, then set the exponents equal.
- 9^x = 27 → 3^(2x) = 3³ → 2x = 3 → x = 1.5
- (1/2)^x = 16 → 2^(−x) = 2⁴ → x = −4
- 5^(x−1) = 1 → 5^(x−1) = 5⁰ → x = 1
Useful powers to know: 2, 4, 8, 16, 32, 64; 3, 9, 27, 81; 5, 25, 125.
Common factor
2^(x+2) + 2^x = 40 → 2^x(4 + 1) = 40 → 2^x = 8 → x = 3.
Method 2: substitution (hidden quadratic)
If you see b^(2x) and b^x together, let t = b^x. Then b^(2x) = t².
Example: 4^x − 6·2^x + 8 = 0. Since 4^x = (2^x)², let t = 2^x: t² − 6t + 8 = 0 → (t − 2)(t − 4) = 0 → t = 2 or 4 → x = 1 or x = 2.
Always check t > 0. If you get t = −3, reject it, because 2^x can never be negative.
Method 3: take logarithms
When the numbers are not powers of one base, take the log of both sides and use log(b^x) = x·log b.
3^x = 20 → x log 3 = log 20 → x = log 20 / log 3 ≈ 1.301 / 0.477 ≈ 2.73.
Two different bases: 2^x = 5^(x−1) → x log 2 = (x − 1) log 5 → x(log 5 − log 2) = log 5 → x = log 5 / log 2.5 ≈ 1.76.
Natural logs (ln) work just as well. In growth problems with e, like e^(0.05t) = 2, ln gives t = ln 2 / 0.05 ≈ 13.9.
Exponential inequalities and systems
Make the same base, then compare exponents:
- If b > 1 the curve rises, so the sign stays: 2^x > 8 → x > 3.
- If 0 < b < 1 the curve falls, so the sign flips: (1/3)^x > 9 → (1/3)^x > (1/3)^(−2) → x < −2.
Systems
Turn each equation into a simple one. 2^x · 2^y = 32 and 3^(x−y) = 3 give x + y = 5 and x − y = 1, so x = 3, y = 2.
Try it: predict, then check
Fold a sheet of paper in half again and again and count the layers: 2, 4, 8… After how many folds would you get 64 layers? Write 2^n = 64 and solve. Then, in the 3D free play, set base 2 and value 16: predict x first, then check.
Key formulas and definitions
- b^p = b^q ⇔ p = q (b > 0, b ≠ 1)
- b^x = c ⇒ x = log c / log b (c > 0)
- b^x > b^k: x > k if b > 1, x < k if 0 < b < 1
- t = b^x ⇒ b^(2x) = t², t > 0
- log(b^x) = x·log b
Worked examples
1. Solve 3^(x+1) = 81.
81 = 3⁴, so x + 1 = 4, x = 3.
2. Solve 8^x = 32.
2^(3x) = 2⁵ → 3x = 5 → x = 5/3.
3. Solve (1/5)^x = 125.
5^(−x) = 5³ → x = −3.
4. Solve 3^(x+1) + 3^x = 36.
3^x(3 + 1) = 36 → 3^x = 9 → x = 2.
5. Solve 9^x − 4·3^x + 3 = 0.
t = 3^x: t² − 4t + 3 = 0 → t = 1 or 3 → x = 0 or 1.
6. Solve 5^x = 12 (to 2 d.p.).
x = log 12 / log 5 = 1.0792 / 0.6990 ≈ 1.54.
7. Solve 2^(x−3) < 1/4.
2^(x−3) < 2^(−2); base 2 > 1 so x − 3 < −2 → x < 1.
8. Solve (0.5)^(2x) ≥ 0.125.
0.125 = 0.5³; base 0.5 < 1 so flip: 2x ≤ 3 → x ≤ 1.5.
Common mistakes
- Taking 4^x = 2^(x+2) and setting x = x + 2 without first making the bases the same.
- Keeping a negative value of t = b^x in substitution problems. b^x is always positive.
- Forgetting to flip the inequality sign when the base is between 0 and 1.
- Writing log 20 / log 3 as log(20/3). The quotient of logs is not the log of a quotient.