Log Equations: The Answer You Must Reject

The question

Solve $\log_2 x + \log_2(x - 2) = 3$ .

The common wrong answer

Most students handle the algebra well:

\log_2[x(x - 2)] = 3 \quad\Rightarrow\quad x(x - 2) = 8 \quad\Rightarrow\quad x^2 - 2x - 8 = 0

(x - 4)(x + 2) = 0 \quad\Rightarrow\quad x = 4 \quad\text{or}\quad x = -2

and then write both values as the final answer. The working is fine; the mistake is the last line.

Substituting $x = -2$ into the original equation gives $\log_2(-2) + \log_2(-4)$ , and neither of these exists: you can only take the log of a positive number. Writing $x = -2$ as an answer usually loses the final accuracy mark.

The correct working

Use the law $\log a + \log b = \log ab$ :

\log_2[x(x - 2)] = 3 \quad\Rightarrow\quad x(x - 2) = 2^3 = 8

x^2 - 2x - 8 = 0 \quad\Rightarrow\quad (x - 4)(x + 2) = 0 \quad\Rightarrow\quad x = 4 \text{ or } x = -2

Now check each value in the original equation:

$x = 4$ : $\log_2 4 + \log_2 2 = 2 + 1 = 3$ ✓
$x = -2$ : $\log_2(-2)$ is undefined ✗

So the only answer is $x = 4$ .

Why it works

$\log_2 x$ asks “2 to what power gives $x$ ?” Any power of 2 is positive ( $2^3 = 8$ , $2^{-1} = 0.5$ ), so there is no power that gives a negative number or zero. That’s why every log in the equation needs a positive argument.

Combining the logs into $\log_2[x(x - 2)]$ hides this. The product $x(x - 2)$ is positive for $x = -2$ , because two negatives multiply to a positive, so the combined equation accepts $-2$ even though the original doesn’t. The new equation has an extra solution the original never had, and only checking in the original equation removes it.

How to spot it next time

After solving any log equation, substitute each answer back into the original equation and check every log has a positive argument.
A related trap: $\log(a + b)$ is not $\log a + \log b$ . The law only works for a product inside one log.