Exponential Equations and Inequalities

An exponential equation has the unknown in the exponent, such as 2^x = 32, and it is solved by writing both sides with the same base or by taking logarithms. In brief: 3^{x + 1} = 81 becomes 3^{x + 1} = 3⁴, so x + 1 = 4 and x = 3. For 2^x = 7, take logarithms: x = log₂ 7 = ln 7/ln 2 ≈ 2.807.

The key fact is that an exponential function with base b > 0, b ≠ 1, is one-to-one: b^u = b^v implies u = v. For inequalities, the direction depends on whether the base is greater or less than 1. This guide shows three solution methods, how to solve inequalities and equations of quadratic type and gives a table of the methods.

How do you solve by writing the same base?

If both sides can be written as powers of the same base, set the exponents equal.

1. Rewrite each side as a power of a common base, such as 2, 3 or 10.
2. Use b^u = b^v ⇒ u = v.
3. Solve the resulting equation.

Example 1: 4^x = 32

4^x = (2²)^x = 2^2x and 32 = 2⁵. So 2x = 5 and x = 5/2.

Example 2: (1/3)^{x − 2} = 27

(1/3)^{x − 2} = 3^{−(x − 2)} = 3^{2 − x}. And 27 = 3³. So 2 − x = 3 and x = −1. Check: (1/3)⁻³ = 27 ✓.

Example 3: 9^x = 27^{x − 1}

3^2x = 3^{3(x − 1)}, so 2x = 3x − 3 and x = 3. Check: 9³ = 729 and 27² = 729 ✓.

How do you solve with logarithms?

When the bases cannot be matched, take a logarithm of both sides. By the power rule, log(b^x) = x · log b.

Example 4: 5^x = 40

x = log₅ 40 = ln 40/ln 5 ≈ 3.689/1.609 ≈ 2.292. Check: 5^2.292 ≈ 40 ✓.

Example 5: 3 · 2^x − 5 = 19

Isolate the power: 3 · 2^x = 24, so 2^x = 8 and x = 3.

Example 6: unknown on both sides

2^x = 3^{x − 1}. Take ln: x ln 2 = (x − 1) ln 3. So x ln 2 − x ln 3 = −ln 3, giving x = ln 3/(ln 3 − ln 2) = ln 3/ln 1.5 ≈ 1.0986/0.4055 ≈ 2.710. Check: 2^2.71 ≈ 6.54 and 3^1.71 ≈ 6.54 ✓.

How do you solve equations that need a substitution?

When the equation contains b^x and b^2x, let u = b^x > 0 and solve the quadratic equation.

Example 7: 4^x − 3 · 2^x − 4 = 0

Let u = 2^x. Then u² − 3u − 4 = 0, so (u − 4)(u + 1) = 0 and u = 4 or u = −1. Since 2^x > 0, reject u = −1. From 2^x = 4, x = 2.

How do you solve exponential inequalities?

The function b^x is increasing for b > 1, so b^u < b^v means u < v (the inequality is kept). For 0 < b < 1 it is decreasing, so b^u < b^v means u > v (the inequality is reversed).

Example 8: 2^x > 16

2^x > 2⁴, so x > 4.

Example 9: (1/2)^x ≤ 1/8

(1/2)^x ≤ (1/2)³. The base is less than 1, so the inequality reverses: x ≥ 3.

Example 10: 3^{2x − 1} < 27

3^{2x − 1} < 3³, so 2x − 1 < 3 and x < 2.

Example 11: with logarithms

1.05^t > 2 asks when an amount growing by 5% per year doubles. Since 1.05 > 1, t > ln 2/ln 1.05 ≈ 0.6931/0.04879 ≈ 14.21. So after about 14.2 years.

Method table

Where do these methods come from?

Babylonian scribes solved questions such as “how long until an investment doubles”, using tables of powers. The systematic solution by logarithms became possible after Napier published his tables in 1614, and Henry Briggs soon made base-10 tables. Before calculators, logarithm tables were the standard tool for solving such equations.

Where are they used today?

Finance uses exponential equations to find how long a loan takes to double. In medicine, the time at which a drug level falls below a threshold solves an inequality such as 0.85^t < 0.1. Population models, cooling processes and the spread of a rumor lead to equations of this type, and calculators and spreadsheets evaluate the logarithms.

Which mistakes are common?

Taking the logarithm of a sum

ln(2^x + 3) is not x ln 2 + ln 3. Isolate the power before taking logarithms.

Forgetting to reverse the inequality for bases below 1

For (1/2)^x, the function decreases and the sign reverses.

Keeping a negative solution for u

With u = b^x, the value u must be positive. Reject u ≤ 0.

Making errors with the sign of exponents

(1/3)^x = 3^−x. Rewrite carefully before equating exponents.