Solve: \(log_3 x + log_3 (x - 8) = 2\)
A.
8
B.
6
C.
9
D.
7
Correct answer is C
\(log_3 x + log_3 (x - 8) = 2\)
\(log_3 x(x - 8) = log_39\) 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.
