Q.Solve for $x$: $\log x + \log(x-3) = 1$ (common logarithm, base 10).

Combine the two log terms into one using the product law, convert to exponential form, then check both roots of the resulting quadratic against the original equation's domain.

Step 1 — combine the two logarithms (product law):

Step 2 — convert to exponential form:

Step 3 — solve the resulting quadratic:

Step 4 — check BOTH roots against the ORIGINAL equation's domain (every log argument must be positive):

• $x=5$: $\log 5$ (argument $5>0$, valid) and $\log(5-3)=\log2$ (argument $2>0$, valid). Both defined — $x=5$ is a genuine solution. …

Solving the quadratic correctly but reporting BOTH roots ($5$ and $-2$) as valid solutions, without checking either one against the domain of the original logarithmic equation — a step that is …