To solve a logarithmic equation, get the logarithm alone on one side, convert it to exponential form, solve the resulting equation, and then check the domain — any candidate that makes a log’s argument zero or negative must be rejected. That last step is where most students lose marks: the algebra can hand you two answers while the original equation only allows one. This guide walks through the full 4-step method with worked examples, including the classic domain-trap question examiners love.
What Counts as a Logarithmic Equation?
A logarithmic equation is any equation where the unknown appears inside a logarithm — for example log₂(x) = 5, log(x − 1) = 2, or log₂(x) + log₂(x − 2) = 3. These look intimidating, but every one of them is solved by the same idea: a logarithm is just an exponent in disguise, since log_b(x) = y means exactly b^y = x. Converting from logarithmic form to exponential form turns the unfamiliar log problem into an ordinary equation you already know how to solve. If you want to double-check your answers, Calcrio’s free logarithm calculator solves log equations step by step, with automatic domain checks.
How Do You Solve a Logarithmic Equation? (The 4-Step Method)
Follow these four steps in order, every time:
- Isolate or combine the logarithm. Get a single logarithm equal to a number. If the equation has two logs, combine them with the log rules: a sum becomes one log of a product (log_b(x) + log_b(y) = log_b(xy)), a difference becomes one log of a quotient (log_b(x) − log_b(y) = log_b(x/y)), and a coefficient becomes an exponent (n · log_b(x) = log_b(x^n)). Keep going until you have the form log_b(something) = c.
- Exponentiate — convert to exponential form. log_b(something) = c becomes something = b^c. This is the step that actually removes the logarithm.
- Solve the resulting equation. This will usually be linear or quadratic. For a quadratic like x² − 2x − 8 = 0, factor or use the formula (Calcrio’s quadratic equation solver shows every step if you need it).
- Check the domain and reject extraneous roots. Plug every candidate back into the original equation. A logarithm’s argument must be positive — if any candidate makes any argument zero or negative, it is not a solution. This is not optional; skipping it is the most common way to get the question wrong.
Worked Example 1: The Domain Trap (Exam Classic)
Solve: log₂(x) + log₂(x − 2) = 3
This is the question that appears on algebra exams precisely because it traps students who skip step 4.
Step 1 — Combine. The left side is a sum of two logs with the same base, so use the product rule: log₂(x) + log₂(x − 2) = log₂(x(x − 2)). The equation is now log₂(x(x − 2)) = 3.
Step 2 — Exponentiate. Convert to exponential form: x(x − 2) = 2³ = 8.
Step 3 — Solve. Expand: x² − 2x − 8 = 0. Factor: (x − 4)(x + 2) = 0, so x = 4 or x = −2. Both check out in this reduced equation: 4 × 2 = 8 ✓ and (−2) × (−4) = 8 ✓.
Step 4 — Check the domain. Here is the trap. The original equation needs log₂(x) and log₂(x − 2) to exist, which requires x > 0 and x − 2 > 0, i.e. x > 2. x = 4 passes: log₂(4) + log₂(2) = 2 + 1 = 3 ✓. x = −2 fails: log₂(−2) is undefined, since no exponent on 2 produces a negative number. The answer is x = 4 only.
Notice what happened: the algebra in step 3 was perfectly correct — −2 really does satisfy x(x − 2) = 8. But squaring-like operations (combining logs, clearing denominators) can create extraneous solutions, so the domain check is what separates the true answer from the impostor.
Worked Example 2: Two Logs, One Rejected Answer
Solve: ln(x) + ln(x − 3) = ln(28)
Step 1 — Combine: ln(x(x − 3)) = ln(28).
Step 2 — Exponentiate (shortcut): when both sides are logs with the same base, you can drop the logs and equate the arguments directly: x(x − 3) = 28. (This is the same as exponentiating both sides with base e.)
Step 3 — Solve: x² − 3x − 28 = 0 → (x − 7)(x + 4) = 0 → x = 7 or x = −4.
Step 4 — Check the domain. The arguments are x and x − 3, so we need x > 0 and x > 3 — i.e. x > 3. x = 7 works: ln(7) + ln(4) = ln(28) ✓. x = −4 gives ln(−4), which is undefined. Answer: x = 7 only.
Worked Example 3: A Single Log (The Easy Case)
Solve: log(x) = 3
When the log is already isolated, steps 1 and 3 are trivial. A bare “log” means base 10, so convert to exponential form: x = 10³ = 1000. Domain check: 1000 > 0 ✓. Answer: x = 1000.
Similarly, log₂(x) = 5 gives x = 2⁵ = 32 (verified: log₂(32) = 5 because 2⁵ = 32). The pattern never changes: exponentiate, solve, verify the domain.
What If the Bases Don’t Match?
You can only combine logarithms that share the same base. If you meet something like log₂(x) + log₁₀(x) = 5, rewrite one of the logs with the change of base formula — log_b(x) = ln(x)/ln(b) — so both share a base, then proceed with the 4-step method. Calcrio’s logarithm calculator handles any base automatically, and the exponential equation solver covers the mirror-image case where the unknown sits in the exponent.
FAQ
What is the first step in solving a logarithmic equation?
Get the logarithm alone (or combined into a single logarithm) on one side of the equation. Use the product, quotient, and power rules to combine multiple logs into one before converting to exponential form.
Why do you have to check the domain after solving?
Because combining logarithms and exponentiating can create extraneous solutions — values that satisfy the reduced equation but make a logarithm’s argument zero or negative in the original equation. Those values must be rejected, as the x = −2 case above shows.
What does “extraneous solution” mean?
A number that emerges from correct algebra but does not satisfy the original equation. In log equations it happens whenever a candidate makes an argument non-positive; log₂(−2) is undefined, so any such candidate is discarded.
Can a logarithmic equation have no solution?
Yes. If every candidate fails the domain check — for example, solving leads only to values that make an argument zero or negative — the equation has no (real) solution.
What is log₂(x) + log₂(x − 2) = 3 solved?
x = 4 only. Combining gives log₂(x(x − 2)) = 3, so x(x − 2) = 8, giving x = 4 or x = −2 — but x = −2 makes both log arguments negative and must be rejected.
How is solving log_b(x) = c different from solving b^x = c?
They are mirror images: log_b(x) = c converts to x = b^c (a power), while b^x = c converts to x = log_b(c) (a logarithm). Both use the same log_b(x) = y ⟺ b^y = x relationship, just applied in opposite directions.