A negative candidate is not automatically wrong
log(x2) = 2 → x2 = 102 = 100 → x = ±10
x = 10 → argument x2 = 100✓ keep
x = −10 → argument x2 = 100✓ keep
Both work: log(100) = 2 either way. The test is never “is x negative?” — it is “is every argument positive?” Rejecting negatives on sight is a habit, not a rule, and it will cost you a real solution.
The whole discipline in three lines: (1) before you solve, write down what each argument requires; (2) solve; (3) test every candidate in the original equation, not in the line you rewrote. Step 3 is the one people skip, and it is the only one that catches the impostor.