Solve the logarithmic equation: \(log_2 (6 - x) = 3 - log_2 x\)

A.

\(x\) = 4 or 2

B.

\(x\) = -4 or -2

C.

\(x\) = -4 or 2

D.

\(x\) = 4 or -2

Correct answer is A

\(log_2 (6 - x) = 3 - log_2 x\)

⇒ \(log_2 (6 - x) = 3 log_2 2 - log_2 x\) (since \(log_2\) 2 = 1)

⇒ \(log_2 (6 - x) = log_2 2^3 - log_2 x\) \((a log\) c = \(log\) c\(^a)\)

⇒ \(log_2 (6 - x) = log_2 8 - log_2 x\)

⇒\(log_2 (6 - x) = log_2 \frac{8}{x}\) (\(log\) a - \(log\) b = \(log \frac{a}{b})\)

⇒ \(6 - x = \frac{8}{x}\)

⇒ \(x (6 - x) = 8\)

⇒ \(6x - x^2 = 8\)

⇒ \(x^2 - 6x + 8 = 0\)

⇒ \(x^2 - 4x - 2x + 8 = 0\)

⇒ \(x (x - 4) - 2(x - 4) = 0\)

⇒ \((x - 4)(x - 2) = 0\)

⇒ \(x - 4 = 0 or x - 2 = 0\)

∴ x = 4 or 2