Question

If $e^{f\left(x\right)}=\frac{10+x}{10-x},x\in \left(-10,10\right)$ and $f\left(x\right)=kf\left(\frac{200x}{100+x^2}\right)$ then $k=$

Answer 1

Answer: 0.5

$e^{f\left(x\right)}=\frac{10+x}{10-x},x\in \left(-10,10\right)$
$\Rightarrow f\left(x\right)=log\left(\frac{10+x}{10-x}\right)$
$\Rightarrow f\left(\frac{200x}{100+x^2}\right)=log\left[\frac{10+\frac{200x}{100+x^2}}{10-\frac{200x}{100+x^2}}\right]$
$=log\left[\frac{10{\left(10+x\right)}^2}{10{\left(10-x\right)}^2}\right]=2$ log$\left(\frac{10+x}{10-x}\right)=2f\left(x\right)$
$\therefore f\left(x\right)=\frac{1}{2}f\left(\frac{200x}{100+x^2}\right)$
$\Rightarrow k=\frac{1}{2}=0.5$