Logarithmic Equations & Extraneous Solutions

You’ll be able to

Solve logarithmic equations by condensing to a single logarithm and exponentiating
Check every candidate solution against the domain of the original equation
Explain why exponentiating can introduce solutions that do not satisfy the original equation

Condense, then exponentiate

The standard method has three steps. Condense the logarithms into one using the product, quotient and power rules. Exponentiate both sides to remove the logarithm. Check every candidate against the original equation. That third step is not a formality — it is where roughly half of these problems are actually decided.

The three log rules used to condense

log a + log b = log(ab) · log a − log b = log(a/b) · n·log a = log(aⁿ)

Every one of these requires a > 0 and b > 0. The rules are identities only on the domain where both sides are defined — which is precisely why extraneous solutions appear.

Where the false solutions come from

log x is defined only for x > 0, so the equation log(x − 3) + log(x + 1) = log 5 silently demands x > 3. Condensing to log[(x − 3)(x + 1)] = log 5 loses that information, because the product (x − 3)(x + 1) is positive when both factors are negative too. The condensed equation therefore has a larger domain than the original, and any solution living in the extra territory is extraneous — an artifact of the algebra, not a solution.

Worked example

Solve log₂(x) + log₂(x − 2) = 3.

1.Domain first: need x > 0 and x − 2 > 0, so x > 2.
2.Condense: log₂[x(x − 2)] = 3.
3.Exponentiate: x(x − 2) = 2³ = 8.
4.Expand and solve: x² − 2x − 8 = 0, so (x − 4)(x + 2) = 0, giving x = 4 or x = −2.
5.Check against x > 2: x = 4 qualifies; x = −2 does not.
6.Verify x = 4: log₂ 4 + log₂ 2 = 2 + 1 = 3. ✓

Answer: x = 4 is the only solution. x = −2 is extraneous — it satisfies the quadratic but makes both original logarithms undefined. Stating the domain before starting is what makes the rejection immediate rather than a matter of luck.

Watch out

Check candidates in the original equation, not the condensed one. The condensed form has a wider domain by construction, so it will happily accept the extraneous root and give you no warning at all.

Checkpoint

Solve log(x + 3) − log(x) = 1.

When both sides carry a logarithm

If the equation reduces to log_b(A) = log_b(B) with the same base, then A = B — because a logarithmic function is one-to-one. This is often faster than exponentiating. But the domain requirement is unchanged: both A and B must be positive at the solution, and the deduction A = B does not enforce that on its own.

Worked example

Solve ln(x + 6) = ln(x) + ln(4).

1.Domain: need x + 6 > 0 and x > 0, so x > 0 governs.
2.Condense the right side: ln(x) + ln(4) = ln(4x).
3.Now ln(x + 6) = ln(4x), and ln is one-to-one, so x + 6 = 4x.
4.Solve: 6 = 3x, so x = 2.
5.Check the domain: 2 > 0. ✓ Verify: ln 8 = ln 2 + ln 4 = ln 8. ✓

Answer: x = 2. Here no extraneous solution appeared, which is worth noticing — the domain check is not a ritual that always discards something. It is a genuine test, and sometimes everything passes.

On the exam

On a free-response question, showing the domain restriction and the rejection of an extraneous root earns credit. Writing only the surviving answer can lose a point even when that answer is right, because the reasoning that eliminated the other root is not on the page.

Checkpoint

Why must candidate solutions to a logarithmic equation be checked?

Answer the 2 checkpoints as you read.