You will get two roots. Only one is valid.

Solve for x:
log₂(x) + log₂(x − 2) = 3

A) x = 4 only
B) x = −2 only
C) x = 4 and x = −2
D) No solution

The algebra is not the difficult part. The trap is stopping as soon as the quadratic gives you two roots.

Step 1 — Write the domain first.

A logarithm is defined only when its argument is positive.

The original equation contains both log₂(x) and log₂(x − 2), so:

• x > 0
• x − 2 > 0, which means x > 2

Both conditions must hold. Therefore:

x > 2

Keep that condition beside your working.

Step 2 — Combine the logarithms.

The logarithms have the same base, so:

log₂(x) + log₂(x − 2) = log₂(x(x − 2))

Therefore:

log₂(x(x − 2)) = 3

Step 3 — Convert to exponential form.

log₂(something) = 3 means that the quantity inside the logarithm equals 2³:

x(x − 2) = 8

Step 4 — Solve the quadratic.

x² − 2x − 8 = 0
(x − 4)(x + 2) = 0

So:

x = 4 or x = −2

This is where option C becomes tempting. But the question is not finished.

Step 5 — Check the roots against the original domain.

• x = 4 satisfies x > 2, so it is valid.
• x = −2 does not satisfy x > 2. It would also make log₂(x) undefined.

Answer: A — x = 4 only.

Why did x = −2 appear?

The logarithm rule is correct, but it is valid here only while the original domain condition is kept. If you solve the combined equation over a wider set of values and forget that x must be greater than 2, the quadratic can produce an extra root.

That extra value is called an extraneous root: it solves one of the later equations in your working, but it does not solve the original equation.

The habit that saves the mark

For every logarithmic equation, write the domain before doing any algebra. Here it takes only a few seconds:

x > 2

Then, when the quadratic gives you two roots, you already know which one must be rejected.