Question

Rezolvati in multimea numerelor reale ecuatia $log_{2} \frac{x-1}{x+1} +log_{2} (x^2-1)=4$

Answer 1

$log_{2}(\frac{x-1}{x+1}) + log_{2}({x}^{2}-1) = 4 \\ log_{2}(\frac{x-1}{x+1}.{x}^{2}-1) =log_{2}({2}^{4}) \\ log_{2}(\frac{(x-1)({x}^{2}-1)}{x+1}) =log_{2}(16) \\ \frac{(x-1)({x}^{2}-1)}{x+1}= 16 \\ \frac{(x-1)(x+1)(x-1)}{x+1} = 16 \\ {(x-1)}^{2} = 16 \\ x = - 3 \\x = 5$