Algebra can produce candidate values that do not belong to the domain of the original logarithms. In AMC 12 preparation, record the restrictions before transforming the equation, then apply them again at the end.
This is an original teaching example, not an official AMC 12 problem.
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Write the domain before the equation changes
Solve log2(x) + log2(x - 2) = 3 for real x. Each logarithm needs a positive argument: x > 0 and x - 2 > 0. Together these require x > 2.
The base is 2, which is positive and not equal to 1. These conditions make the real logarithms well defined.
Combine the logarithms on that domain
For positive arguments, the sum becomes log2[x(x - 2)] = 3. Therefore x(x - 2) = 23 = 8.
Rearrange to x2 - 2x - 8 = 0, or (x - 4)(x + 2) = 0. The algebraic candidates are 4 and -2.
Return to the original restrictions
Only 4 satisfies x > 2. Substitution confirms log2(4) + log2(2) = 2 + 1 = 3. The real solution is x = 4.
At -2, both original arguments are negative. Their product is positive, so the combined expression alone can hide the original restriction. That is why a final product condition cannot replace the two starting argument conditions.
Try a related problem and check the same issue
Solve log2(x) + log2(x - 6) = 4. The domain is x > 6. Combining gives x(x - 6) = 16, whose candidates are 8 and -2. Only 8 is valid; its logs are 3 and 1.
Keep three lines in your notes: original domain, transformed equation and final filtered solution. Before using a logarithm rule, state why its arguments meet the rule's conditions. This small habit also helps with rational and radical equations, where algebraic transformations need their own restrictions.
Continue reading: AMC 12 Trigonometry: Solve Equations Without Losing Solutions.
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