\(log_3 x + log_3 (x - 8) = 2\)

\(log_3 x(x - 8) = log_39\) 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.