Exponential equations and inequalities

In an exponential equation the unknown is in the exponent: 2ˣ = 8, 4ˣ − 3·2ˣ + 2 = 0. Such equations describe compound interest, the growth of bacteria and radioactive decay, and every year the exam contains several questions on this topic. The main idea is to bring both sides to the same base or to introduce a new variable. All of this rests on the laws of indices and on the monotonicity of the exponential function.

§1. Laws of indices (revision)

To solve exponential equations you need to use the laws of indices fluently. For a > 0, b > 0:

These laws let us write every power in an equation with one base: 8 = 2³, 1/4 = 2⁻², √2 = 2^(1/2), 0.125 = 2⁻³, 1/27 = 3⁻³.
Note that 2ˣ + 2ˣ = 2 · 2ˣ = 2ˣ⁺¹, not 4ˣ. There is no special rule for a sum of powers — only for products and quotients.

§2. The exponential function

• The domain is all real numbers; the range is (0; +∞): aˣ is always positive.
• For a > 1 it is increasing, for 0 < a < 1 decreasing.
• The graph passes through (0; 1); the x-axis is a horizontal asymptote.

The base must satisfy a > 0, because with a negative base, for example, (−4)^(1/2) is not a real number. And a = 1 is excluded, since 1ˣ = 1 is constant.
The function is strictly monotonic, so it takes each value only once. This gives the main rule:

Also, the equation aˣ = b has exactly one solution when b > 0 (x = logₐ b) and none when b ≤ 0. For example, 3ˣ = −9 has no roots.

§3. Bringing to the same base

Write both sides as powers with the same base, then equate the exponents.

• 9ˣ = 27 → 3²ˣ = 3³ → 2x = 3 → x = 3/2.
• (1/4)ˣ = 32 → 2⁻²ˣ = 2⁵ → x = −5/2.
• 2^(x² − x) = 4 → x² − x = 2 → x² − x − 2 = 0 → x = 2 or x = −1.

Different bases, same exponent: aˣ = bˣ → (a/b)ˣ = 1 → x = 0. Combining a product also helps: 3ˣ·2ˣ = 6ˣ, so 3ˣ·2ˣ = 36 → 6ˣ = 6² → x = 2.
If the sides cannot be written with the same base, use a logarithm: 2ˣ = 5 → x = log₂ 5.

§4. Taking out a common factor and substitution

Common factor. If the exponents differ by a number (x + 1, x, x − 2), take the smallest power outside the bracket.

Worked example

Solve 5ˣ⁺¹ − 5ˣ = 100.

1. 5ˣ⁺¹ = 5 · 5ˣ, so take out 5ˣ: 5ˣ(5 − 1) = 100.
2. 4 · 5ˣ = 100 → 5ˣ = 25.
3. 5ˣ = 5² → x = 2.

Substitution. If the equation contains a²ˣ and aˣ, put t = aˣ, since a²ˣ = (aˣ)² = t². This gives a quadratic equation.

For example, 4ˣ − 3·2ˣ + 2 = 0 → t² − 3t + 2 = 0 → t = 1; 2 → x = 0; 1.

Worked example (official sample)

The sum of the roots of 3³ˣ − 2·3²ˣ + 9·3ˣ⁻² = 0?

1. 9·3ˣ⁻² = 3²·3ˣ⁻² = 3ˣ.
2. t = 3ˣ > 0: t³ − 2t² + t = 0 → t(t − 1)² = 0.
3. t = 0 is rejected, t = 1 → 3ˣ = 1 → x = 0.
4. Answer: 0.

§5. Homogeneous exponential equations

An equation of the form A·a²ˣ + B·(ab)ˣ + C·b²ˣ = 0 is called homogeneous. For example, 4ˣ, 6ˣ and 9ˣ appear together: 4ˣ = (2ˣ)², 9ˣ = (3ˣ)², 6ˣ = 2ˣ·3ˣ. Divide both sides by b²ˣ (for example by 9ˣ) — it is never zero, so the division is safe — and substitute t = (a/b)ˣ.

Worked example

Solve 4ˣ + 6ˣ = 2·9ˣ.

1. Divide both sides by 9ˣ > 0: (4/9)ˣ + (6/9)ˣ = 2.
2. t = (2/3)ˣ > 0: (4/9)ˣ = t², (6/9)ˣ = t, so t² + t − 2 = 0.
3. t = 1 or t = −2; t = −2 is rejected.
4. (2/3)ˣ = 1 → x = 0. Answer: x = 0.

§6. Exponential inequalities

In an inequality we also bring both sides to the same base, but now the size of the base is decisive:

The reason is monotonicity: for an increasing function a larger argument gives a larger value; for a decreasing one it is the other way round.

• (1/2)ˣ > 8 → (1/2)ˣ > (1/2)⁻³ → x < −3.
• (1/3)ˣ ≥ 9 → (1/3)ˣ ≥ (1/3)⁻² → x ≤ −2.
• 2ˣ > −5 holds for every x, because 2ˣ > 0.

Worked example

Solve 4ˣ − 5·2ˣ + 4 < 0.

1. t = 2ˣ > 0: t² − 5t + 4 < 0.
2. (t − 1)(t − 4) < 0 → 1 < t < 4.
3. 1 < 2ˣ < 4 → 2⁰ < 2ˣ < 2²; the base 2 > 1, so the sign is kept.
4. Answer: 0 < x < 2.

Check yourself

1. Solve 2ˣ⁺¹ = 16.

2. Solve 9ˣ = 27.

3. Solve 4ˣ − 3·2ˣ + 2 = 0.

4. Solve (1/2)ˣ > 8.

5. Solve 3ˣ·2ˣ = 36.

6. Solve 2ˣ⁺³ + 2ˣ = 72.

7. The sum of the roots of 3²ˣ⁺¹ − 10·3ˣ + 3 = 0.

8. How many roots does 3ˣ = −9 have?

Common mistakes

• Not reversing the inequality sign when the base is 0 < a < 1.
• Keeping a negative or zero t after the substitution t = aˣ.
• Writing 2ˣ + 2ˣ = 4ˣ (correct: 2·2ˣ = 2ˣ⁺¹).
• Looking for solutions of aˣ = 0 or aˣ = −b (the exponential function takes only positive values).
• Finding t and forgetting to go back to x.