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Combine logs: \( \log_2(x(x - 2)) = 3 \) → \( x(x - 2) = 2^3 = 8 \).
So, \( x^2 - 2x - 8 = 0 \) → factor: \( (x - 4)(x + 2) = 0 \).
So \( x = 4 \) or \( x = -2 \). But \( x > 2 \) for log to be defined, so \( x = 4 \).

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