Question

If $x$ satisfies the equation ${\log }^2(1+\frac{4}{x})+{\log }^2(1-\frac{4}{x+4})=2{\log }^2(\frac{2}{x-1}-1)$, then find the sum of squares of values of $x$.

Answer 1

${\log }^2\left(\frac{x+4}{x}\right)+{\log }^2\left(\frac{x}{x+4}\right)=2{\log }^2\left(\frac{3-x}{x-1}\right)$
$\Rightarrow 2{\log }^2\left(\frac{x+4}{x}\right)=2{\log }^2\left(\frac{3-x}{x-1}\right)$
$\Rightarrow \log \left(\frac{x+4}{x}\right)=\pm \log \left(\frac{3-x}{x-1}\right)$
Case 1:
$\Rightarrow \frac{x+4}{x}=\frac{3-x}{x-1}$
$\Rightarrow x=\pm \sqrt{2}$
$\Rightarrow x=\sqrt{2}$ is the only solution since it lies the domain.
Case 2:
$\frac{x+4}{x}=\frac{x-1}{3-x}$
$\Rightarrow x=\pm \sqrt{6}$
$\Rightarrow x=\sqrt{6}$, lies in the domain.

Required Answer $={\sqrt{2}}^2+{\sqrt{6}}^2=8$