Question

If $e^{f\left(x\right)}=\frac{10+x}{10-x},x\in \left(-10,0\right)$ and $f\left(x\right)=k.f\left(\frac{200x}{100+x^2}\right)$ then $k$ =

Answer 1

$e^{f\left(x\right)}=\frac{10+x}{10-x},xϵ(-10,0)$

$\Rightarrow f\left(x\right)=log\left(\frac{10+x}{10-x}\right)...\left(1\right)$

and $f\left(x\right)=k.f\left(\frac{200x}{100+x^2}\right)$

so $f\left(\frac{200x}{100+x^2}\right)=log\left(\frac{10+\frac{200x}{100+x^2}}{10-\frac{200x}{100+x^2}}\right)$

$=log\left(\frac{1000+10x^2+200x}{1000+10x^2-200x}\right)$

$=log\left(\frac{100+20x+x^2}{100-20x+x^2}\right)$

$=log\left\{{\left(\frac{10+x}{10-x}\right)}^2\right\}$

$\therefore f\left(x\right)=k.log\left\{{\left(\frac{10+x}{10-x}\right)}^2\right\}=log{\left\{{\left(\frac{10+x}{10-x}\right)}^2\right\}}^k$

$\Rightarrow log\left(\frac{10+x}{10-x}\right)=log{\left(\frac{10+x}{10-x}\right)}^{2k}$

$\Rightarrow 2k=1$

$\Rightarrow k=\frac{1}{2}$