Logarithm Equation Calculator

Raise the base to both sides: x = b^c. Example: log₂(x) = 5 becomes x = 2^5 = 32. This calculator also verifies the answer by evaluating log_b(x) back to c.

A number that solves the combined polynomial but makes a log argument zero or negative, where logarithms are undefined. Example: log(x) + log(x−3) = 1 combines to x(x−3) = 10, giving candidates x = 5 and x = −2. But log(−2) and log(−5) do not exist, so x = −2 is extraneous and must be rejected; only x = 5 is valid. This calculator rejects extraneous roots automatically and tells you exactly why.

Use the product rule: log_b(M) + log_b(N) = log_b(M·N). So log(x) + log(x−3) = log(x(x−3)), and the equation log(x) + log(x−3) = 1 becomes log(x(x−3)) = 1, then x(x−3) = 10^1 = 10, i.e. x² − 3x − 10 = 0.

Use the quotient rule: log_b(M) − log_b(N) = log_b(M/N). So log₃((x+1)/(x−1)) = 1, giving (x+1)/(x−1) = 3^1 = 3, hence x + 1 = 3x − 3 and x = 2. Domain check: x + 1 > 0 and x − 1 > 0 both hold at x = 2, so it is valid.

log_b(x) = ln(x) / ln(b) = log₁₀(x) / log₁₀(b). Example: log₂(32) = ln(32) / ln(2) = 3.4657359028 / 0.69314718056 = 5. The change-of-base tab on this page shows every step of the natural-log version.

Logarithms are only defined for positive arguments. When you combine several logs with the product or quotient rule, the resulting polynomial can have roots that make an original argument zero or negative — those are extraneous and must be thrown out. Every candidate this calculator finds is checked against each original argument before being reported as a solution.

That tool converts between the forms b^y = x and log_b(x) = y, evaluates logarithms, and solves exponential equations like b^x = c. This page does the complementary job: it solves logarithmic equations — including multi-term equations such as log(x) + log(x−3) = 1 — with automatic domain validation that rejects extraneous roots. Neither function exists on the other tool.

Yes — use base e (≈ 2.7182818285). Example: ln(x) + ln(x−2) = ln(8) combines to x(x−2) = 8, giving candidates x = 4 and x = −2; the domain check keeps x = 4 and rejects x = −2 because ln(−2) is undefined.

It means every candidate root failed the domain check, and the tool says so honestly instead of showing a fake answer. Example: log(x−5) − log(x−1) = 1 has only one candidate, x ≈ 0.555555555556, but both arguments are negative there — so the equation has no (real) solution.

Any positive base except 1: integers, fractions, and e all work. For example, log base 1/2 of x = 2 gives x = (1/2)^2 = 0.25. Base 1 and non-positive bases are rejected with a clear error, because the logarithm is not defined for them.