Question

Solve for $:{\log }_2(x-1)+{\log }_2(x+1)={\log }_{2}1$

Answer 1

$\log (x-1)+\log (x+1)={\log }_{2}1$

$\Rightarrow \log (x^2-1)={\log }_{2}1$

$\Rightarrow (x^2-1)=1$

$\Rightarrow x=\pm \sqrt{2}$

As $x\ne -\sqrt{2}$, since value inside log become negative

Hence, $x=\sqrt{2}$