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
