Why is x = -2 not a solution if it solves the quadratic?

The quadratic came from combining the logs, which is only valid when x > 0 and x - 2 > 0. At x = -2 the original terms log2(-2) and log2(-4) do not exist, so -2 is an extraneous root.

What is the fastest way to check on the ACT?

Plug each answer choice into the original equation. log2(4) = 2 and log2(2) = 1 add to 3 in a few seconds, and any choice that makes an argument zero or negative is out immediately.

ACT Math: Solve log₂(x) + log₂(x − 2) = 3 (Why x = −2 Is Rejected)

The Problem

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

Log equations like this show up in the harder half of the ACT Math section and in every Algebra 2 / Precalculus unit on logarithms. The math is short; the point of the question is whether you remember to check the domain.

Step 1: Write Down the Domain First

Both arguments must be positive:

$x > 0$
$x - 2 > 0 \implies x > 2$

So any valid answer must satisfy $x > 2$.

Step 2: Combine the Logs (Product Rule)

$$ \log_b M + \log_b N = \log_b(MN) $$

$$ \log_2\big(x(x - 2)\big) = 3 $$

Step 3: Rewrite in Exponential Form

$\log_2(\text{something}) = 3$ means $\text{something} = 2^3$:

$$ x(x - 2) = 8 $$

Step 4: Solve the Quadratic

$$ x^2 - 2x - 8 = 0 $$

$$ (x - 4)(x + 2) = 0 \implies x = 4 \ \text{or}\ x = -2 $$

Step 5: Reject the Extraneous Root

$x = -2$ fails the domain $x > 2$ (it would require $\log_2(-2)$, which is undefined).

Answer: $\mathbf{x = 4}$

Step 6: Verify

$$ \log_2(4) + \log_2(4 - 2) = \log_2 4 + \log_2 2 = 2 + 1 = 3 \ \checkmark $$

Traps on the Answer Choices

"−2 and 4": the classic trap for students who skip the domain check.
"x = 5": from the wrong move $\log_2 x + \log_2(x-2) = \log_2(2x - 2)$, then $2x - 2 = 8$. Adding logs multiplies the arguments; it never adds them.
"x = 3": from forgetting to exponentiate and writing $x(x - 2) = 3$ instead of $2^3$; that gives $x^2 - 2x - 3 = 0$, so $x = 3$ or $x = -1$. Check: $\log_2 3 + \log_2 1 = \log_2 3 \approx 1.58$, not 3.
"x = 2": makes $\log_2(0)$, which is undefined.

Two More to Practice (Same Method)

$\log_2(x + 4) + \log_2(x - 3) = 3$. Combine: $(x + 4)(x - 3) = 8$, so $x^2 + x - 20 = 0$, $(x + 5)(x - 4) = 0$. Domain needs $x > 3$, so $x = 4$.
$\log(x) + \log(x - 3) = 1$ (base 10). Combine: $x(x - 3) = 10$, so $x^2 - 3x - 10 = 0$, $(x - 5)(x + 2) = 0$. Domain needs $x > 3$, so $x = 5$.

15-Second Calculator Check

Most graphing calculators (and Desmos) accept $\log_2$. Graph $y = \log_2(x) + \log_2(x - 2)$ and $y = 3$. They cross only once, at $x = 4$. The graph starts at $x = 2$, which shows you visually why $-2$ cannot be an answer.

For another ACT Math question type that hides a cycle trick, see powers of i on the ACT.