Question

If $g\left(x\right)=\log \left(\frac{2+x}{2-x}\right)$ for $0<x<2$, find the value of $\frac{1}{2}g\left[\frac{8x}{4+x^2}\right]$.

Answer 1

$g\left(x\right)=\log \left(\frac{2+x}{2-x}\right)g\left(\frac{8x}{4+x^2}\right)=\log \left(\frac{2+\left(\frac{8x}{4+x^2}\right)}{2-\left(\frac{8x}{4+x^2}\right)}\right)g\left(\frac{8x}{4+x^2}\right)=\log \left(\frac{\frac{8+2x^2+8x}{4+x^2}}{\frac{8+2x^2-8x}{4+x^2}}\right)g\left(\frac{8x}{4+x^2}\right)=\log \left(\frac{2{(x+2)}^2}{2{(x-2)}^2}\right)\Rightarrow g\left(\frac{8x}{4+x^2}\right)=\log {\left(\frac{x+2}{x-2}\right)}^2=2\log \left(\frac{x+2}{x-2}\right)\Rightarrow \frac{1}{2}g\left(\frac{8x}{4+x^2}\right)=\log \left(\frac{x+2}{x-2}\right)$