\(log_3 x + log_3 (x - 8) = 2\) \(log_3 x(x - 8) = log_39\) since 2 = \(log_39\) \(log_3\) cancels out ⇒ x(x - 8) = 9 ⇒ \(x^2 - 8x = 9\) ⇒ \(x^2 - 8x - 9 = 0\) ⇒ \(x^2 - 9x + x - 9 = 0\) ⇒ x(x - 9) + 1(x - 9) = 0 ⇒ (x - 9)(x + 1) = 0 ⇒ x = 9 or x = -1 Since we can't have a log of negative numbers, ∴ x = 9.