Question

Solve the following equation:
$\log \sqrt{1+x}+3\log \sqrt{1-x}=\log \sqrt{1-x^2}$

Answer 1

$\log \sqrt{1+x}+3\log \sqrt{1-x}=\log \sqrt{1-x^2}\Rightarrow \log \left(\sqrt{1+x}\right){(\sqrt{1-x)}}^3=\log \sqrt{1-x^2}\Rightarrow (1-x)\sqrt{1-x^2}=\sqrt{1-x^2}\Rightarrow x\sqrt{1-x^2}=0$

As $\log \left(\sqrt{1-x^2}\right)$ is not valid at $\sqrt{1-x^2}=0.$

Therefore, $x=0.$