If $f\left(a\right)=\log \frac{2+a}{2-a}$ for $0<a<2$, then $\frac{1}{2}f\left(\frac{{\displaystyle 8a}}{{\displaystyle 4+a^2}}\right)=$

$f\left(a\right)=\log \frac{2+a}{2-a}0<a<2$

$\frac{1}{2}f\left(\frac{8a}{4+a^2}\right)=\frac{1}{2}\left[\log \frac{{\displaystyle 2+\frac{{\displaystyle 8a}}{{\displaystyle 4+a^2}}}}{{\displaystyle 2-\frac{{\displaystyle 8a}}{{\displaystyle 4+a^2}}}}\right]$

$=\frac{1}{2}\log \left[\frac{{\displaystyle 8+2a^2+8a}}{{\displaystyle 8+2a^2-8a}}\right]$$=\frac{1}{2}\log \left(\frac{{\displaystyle 4+a^2+4a}}{{\displaystyle 4+a^2-4a}}\right)$

$=\frac{1}{2}\log {\left(\frac{{\displaystyle 2+a}}{{\displaystyle 2-a}}\right)}^2$$=\log \left(\frac{{\displaystyle 2+a}}{{\displaystyle 2-a}}\right)$

$=f\left(a\right)$