Logarithmic Equations and Inequalities

Domain

The expression under the logarithm must be positive, and the base must be positive and not equal to 1. First write down the domain, then check the roots against this condition.

• For logₐ f(x), we need f(x) > 0.
• If x is in the base: x > 0, x ≠ 1.
• When the base < 1, the inequality sign flips, but the domain is still required.

Basic Equations

• logₐ f = c ⇒ f = aᶜ
• logₐ f = logₐ g ⇒ f = g, f > 0, g > 0
• logₐ f + logₐ g = logₐ(fg), but only if both f and g are positive.

Inequalities

Base	Solution of logₐ f > logₐ g
a > 1	f > g > 0 (the sign is kept)
0 < a < 1	0 < f < g (the sign is reversed)

In both cases, the condition that the expressions are positive is added.

1. Write down the domain
2. Reduce the logarithms to one base, on one side
3. Solve in the form f = g (or the inequality)
4. Intersect the answer with the domain

Substitution Method

If lg x or logₐx appears in a quadratic form, make the substitution t = logₐx. Do not forget to find x = aᵗ from t.

Examples

• If log₂(x + 3) = 3, then x + 3 = 8, x = 5.
• log₃(x − 1) + log₃(x + 1) = 1: log₃(x² − 1) = 1, x² = 4. The domain is x > 1, so x = 2 (−2 is not in the domain).
• log₂(2x − 1) < 3: 0 < 2x − 1 < 8, that is, 0,5 < x < 4,5.
• The logarithm of x to the base 1/2 is greater than 3: the base is less than 1, so the sign flips, 0 < x < (1/2)³, that is, 0 < x < 1/8.
• (lg x)² − 3lg x + 2 = 0: t = lg x, t = 1 or t = 2, so x = 10 or x = 100.