To solve log₂(x) + log₂(x - 2) = 3, we use the product rule of logarithms: logₐ(m) + logₐ(n) = logₐ(m*n). This gives log₂(x(x - 2)) = 3. Converting this to exponential form, we get x(x - 2) = 2³, which simplifies to x² - 2x = 8, or x² - 2x - 8 = 0. Factoring the quadratic equation gives (x - 4)(x + 2) = 0, resulting in x = 4 or x = -2. However, the definition of a logarithm logₐ(y) requires y > 0. For log₂(x) to be defined, x > 0, and for log₂(x - 2) to be defined, x > 2. Therefore, x = -2 is an extraneous solution and must be rejected. The only valid solution is x = 4.