Solve log2(x)+log2(x−2)=3 (Extraneous Root)

This equation is famous among teachers for one reason: it punishes students who solve correctly but forget to check. The algebra flows beautifully, the quadratic gives two clean answers, and one of them is a lie. If you take only one lesson from this page, make it this: in logarithmic equations, the check step is not optional decoration; it is part of the solution.

We will solve log₂(x) + log₂(x−2) = 3, find the two algebraic candidates x = 4 and x = −2, reject −2 as extraneous, and confirm that the only true answer is x = 4.

Worked solution

Step 1: Note the domain before touching anything.

A logarithm only accepts positive arguments. So we need:

x > 0 (from log₂(x))
x − 2 > 0, i.e. x > 2 (from log₂(x−2))

Combined: any valid solution must satisfy x > 2. Write this down now; it is the net that will catch the extraneous root later.

Step 2: Combine the logarithms.

The sum of two logs with the same base is the log of the product:

log₂(x) + log₂(x−2) = log₂(x·(x−2))

So the equation becomes:

log₂(x(x−2)) = 3

Step 3: Convert to exponential form.

log₂(something) = 3 means something = 2³ = 8:

x(x − 2) = 8

Step 4: Expand and rearrange into a quadratic.

x² − 2x = 8
x² − 2x − 8 = 0

Step 5: Factor the quadratic.

We need two numbers that multiply to −8 and add to −2: −4 and +2.

(x − 4)(x + 2) = 0

So x = 4 or x = −2. These are candidates, not answers yet.

Step 6: Test each candidate against the domain (the crucial step).

x = 4: is 4 > 2? Yes ✓. Both logarithms are defined: log₂(4) and log₂(2). Candidate survives.
x = −2: is −2 > 2? No ✗. Worse, log₂(−2) and log₂(−4) do not exist at all. This candidate is extraneous: it satisfies the quadratic we derived, but it never satisfied the original equation.

Reject x = −2. The only solution is x = 4.

Step 7: Check (never skip): substitute x = 4 into the original equation.

log₂(4) + log₂(4−2) = log₂(4) + log₂(2) = 2 + 1 = 3 ✓

The left side equals the right side. The answer is confirmed: x = 4, and only x = 4.

Why this method works

Every step is reversible except one: going from "log₂(x) + log₂(x−2) = 3" to "x(x−2) = 8" silently assumes the logarithms exist. The quadratic x² − 2x − 8 = 0 does not know about that assumption, so it happily produces x = −2, a number that was never allowed in the door. Extraneous roots are not arithmetic mistakes; they are stowaways smuggled in by a non-reversible step. The domain restriction x > 2, noted in Step 1, is the filter that catches them. This is why logarithmic (and square-root) equations demand a check: the method itself can manufacture fake answers.

Common mistakes

Accepting both roots without checking. Writing "x = 4 or x = −2" as the final answer earns zero marks from any careful teacher, because x = −2 makes the original equation meaningless. The check is where the real mathematics happens in this problem.
Checking in the quadratic instead of the original. Plugging x = −2 into x² − 2x − 8 = 0 gives 0 = 0, which "works" and fools you. Always check in the original logarithmic equation, where the domain restriction lives.
Combining logs with different bases. The product rule log(x) + log(y) = log(xy) only works when the bases match. Here both are base 2, so we are safe. If one were log₂ and the other log₁₀, you would need a change of base first.

FAQ

Q: What exactly is an extraneous root?
A: A value that emerges from your algebra but does not satisfy the original equation. Here x = −2 solves the derived quadratic x² − 2x − 8 = 0, but log₂(−2) is undefined, so −2 was never a solution of log₂(x) + log₂(x−2) = 3. It is a byproduct of the method, not a mistake in arithmetic.

Q: Can I just check the domain at the start and skip checking at the end?
A: The domain note (x > 2) already rejects −2, so strictly you could stop there. But substituting x = 4 into the original equation also guards against plain arithmetic slips in Steps 2 to 5. Do both: domain first, substitution last. It takes ten seconds.

Q: Does every log equation have an extraneous root?
A: No. Many have none, and some have two valid solutions. You cannot know until you check. The habit to build is: every logarithmic equation ends with every candidate tested in the original equation. No exceptions.

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