Question

Solve: ${\left(log(1+\frac{4}{x})\right)}^2+{\left(log(1-\frac{4}{x+4})\right)}^2$

Answer 1

${\left(log(1+\frac{4}{x})\right)}^2+{\left(log(1-\frac{4}{x+4})\right)}^2$

$={\left(log\left(\frac{x+4}{x}\right)\right)}^2+log{\left(\frac{x+4-4}{x+4}\right)}^2$

$={\left(log\left(\frac{x+4}{x}\right)\right)}^2+log{\left(\frac{x}{x+4}\right)}^2$

$={\{log\left(\frac{x+4}{x}\right)+log\left(\frac{x}{x+4}\right)\}}^2-2log\left(\frac{x+4}{x}\right)log\left(\frac{x}{x+4}\right)$

$={\left(log(\frac{x+4}{x}\cdot \frac{x}{x+4})\right)}^2+2log\left(\frac{x+4}{x}\right)(-log\left(\frac{x}{x+4}\right))$

$=(log1{)}^2+2log\left(\frac{x+4}{x}\right)log\left(\frac{x+4}{x}\right)$ $[\because logx=log\frac{1}{x}]$

$=0+2{\left(log\left(\frac{x+4}{x}\right)\right)}^2$ $[\because log1=0]$

$=2{\left(log\left(\frac{x+4}{x}\right)\right)}^2=2{\left(log(1+\frac{4}{x})\right)}^2$.