Logarithmic Equations and Inequalities

A logarithmic equation has the unknown inside a logarithm, such as log₂(x + 3) = 4, and it is solved by rewriting it in exponential form or by combining logarithms, followed by a check of the domain. In brief: log₂(x + 3) = 4 means x + 3 = 2⁴ = 16, so x = 13. For logarithmic inequalities, keep the direction of the sign when the base is greater than 1 and reverse it when the base is between 0 and 1, always within the domain.

The main risks are extraneous solutions (arguments that become zero or negative) and the sign change for small bases. A reliable routine is to find the domain first, transform the equation, solve and check. This guide covers equations with one logarithm, with several logarithms, with unknown bases and the substitution method, and then inequalities, with a table of strategies and worked examples.

How do you solve a logarithmic equation step by step?

Find the domain: every argument must be positive (and every variable base positive and not 1).
Use the rules of logarithms to combine the logarithms into one on each side.
Either equate arguments (if log_b A = log_b B, then A = B), or rewrite in exponential form (log_b A = c means A = b^c).
Solve the resulting algebraic equation.
Check each solution against the domain and in the original equation.

Worked examples

Example 1: one logarithm

log₃(2x − 1) = 2. Domain: 2x − 1 > 0, so x > 1/2. Exponential form: 2x − 1 = 9, so x = 5. Check: log₃ 9 = 2 ✓.

Example 2: equal logarithms

ln(x + 4) = ln(3x − 2). Domain: x > −4 and x > 2/3, so x > 2/3. Equate arguments: x + 4 = 3x − 2, so x = 3. Check: ln 7 = ln 7 ✓.

Example 3: sum of logarithms

log₂ x + log₂(x − 6) = 4. Domain: x > 6. Combine: log₂(x(x − 6)) = 4, so x² − 6x = 16, which gives x² − 6x − 16 = 0, so (x − 8)(x + 2) = 0. The solutions are x = 8 and x = −2. Only x = 8 is in the domain. Check: log₂ 8 + log₂ 2 = 3 + 1 = 4 ✓.

Example 4: difference of logarithms

log(x + 9) − log x = 1. Domain: x > 0. Combine: log((x + 9)/x) = 1, so (x + 9)/x = 10, giving x + 9 = 10x and x = 1. Check: log 10 − log 1 = 1 − 0 = 1 ✓.

Example 5: a root outside the domain

log₂(x − 4) + log₂(x − 6) = 3. Domain: x > 6. Combine: (x − 4)(x − 6) = 8, so x² − 10x + 16 = 0, which factors as (x − 2)(x − 8) = 0. The root x = 2 makes both arguments negative, so only x = 8 is a solution. Check: log₂ 4 + log₂ 2 = 2 + 1 = 3 ✓.

Example 6: an unknown base

log_x 64 = 3 means x³ = 64, so x = 4. The base is positive and not 1.

Example 7: different bases

log₂ x = log₄(x + 2). Write log₄(x + 2) = (1/2) log₂(x + 2). Then 2 log₂ x = log₂(x + 2), so x² = x + 2, x² − x − 2 = 0, giving x = 2 or x = −1. Domain x > 0, so x = 2. Check: log₂ 2 = 1 and log₄ 4 = 1 ✓.

How do you use a substitution?

When the equation is quadratic in a logarithm, let u = log_b x.

Example 8

(log₂ x)² − 3 log₂ x + 2 = 0. Let u = log₂ x: u² − 3u + 2 = 0, so u = 1 or u = 2. Then x = 2 or x = 4.

Example 9: logarithm on both sides of an exponential equation

x^{log x} = 100x (base 10). Take log: (log x)² = 2 + log x. Let u = log x: u² − u − 2 = 0, so u = 2 or u = −1. Then x = 100 or x = 0.1. Check x = 0.1: 0.1⁻¹ = 10 and 100 · 0.1 = 10 ✓.

How do you solve logarithmic inequalities?

Find the domain.
Write both sides as logarithms of the same base.
For b > 1: log_b A > log_b B means A > B. For 0 < b < 1: it means A < B.
Solve the resulting inequality and intersect with the domain.

Example 10: log₂(x − 1) < 3

Domain: x > 1. Since 3 = log₂ 8: x − 1 < 8, so x < 9. Solution: 1 < x < 9.

Example 11: log_1/2(x) ≥ 2

Domain: x > 0. The base is less than 1, so the inequality reverses: x ≤ (1/2)² = 1/4. Solution: 0 < x ≤ 1/4. Check x = 1/8: log_1/2(1/8) = 3 ≥ 2 ✓.

Example 12: ln(x + 2) > ln(5 − x)

Domain: x > −2 and x < 5. Base e > 1: x + 2 > 5 − x, so x > 1.5. Solution: 1.5 < x < 5.

Example 13: a compound interest question

When will 1,000 dollars at 4% yearly exceed 3,000? 1000 · 1.04^t > 3000 gives t log 1.04 > log 3, so t > 0.4771/0.01703 ≈ 28.0 years. The base 1.04 is greater than 1, so the sign is kept.

Strategy table

Strategies for logarithmic equations and inequalities
Form	Strategy	Example	Solution
log_b A = c	exponential form A = b^c	log₃(2x − 1) = 2	x = 5
log_b A = log_b B	equate A = B	ln(x + 4) = ln(3x − 2)	x = 3
sum or difference of logs	combine, then exponential form	log₂ x + log₂(x − 6) = 4	x = 8
quadratic in a log	substitute u = log x	(log₂ x)² − 3 log₂ x + 2 = 0	x = 2, 4
log_b A < c, b > 1	A < b^c	log₂(x − 1) < 3	1 < x < 9
log_b A ≥ c, 0 < b < 1	A ≤ b^c	log_1/2 x ≥ 2	0 < x ≤ 1/4

Where do these methods come from?

Logarithmic equations arose as soon as logarithms were used for calculation, in the seventeenth century, for example to solve for time in compound interest problems. The practice of checking solutions against the domain became important with the function concept in the eighteenth century. The conditions for logarithmic inequalities rest on the monotonicity of the logarithm.

Where are they used today?

The time at which an investment reaches a target, the number of half-lives until a medicine drops below a threshold and the distance at which a signal falls below a level all lead to logarithmic equations or inequalities. In acoustics, the distance at which a sound drops to 60 dB solves an inequality in the decibel formula. Computer science uses them to find the smallest number of steps n with 2ⁿ > N, so n > log₂ N.

Which mistakes are common?

Skipping the domain check

In Example 3, x = −2 satisfies the quadratic equation but makes the arguments negative.

Splitting a logarithm of a sum

log(x + 3) is not log x + log 3. Only products can be split.

Not reversing the inequality for a base below 1

For log_1/2 x ≥ 2, the result is x ≤ 1/4, not x ≥ 1/4.

Forgetting to intersect with the domain

The solution of an inequality must satisfy the positivity conditions of all arguments.