Question

If $f\left(x\right)={\log }_e\left(\frac{1+x}{1-x}\right)$, then $f\left(\frac{2x}{1+x^2}\right)$ is equal to

(a) [f(x)]²
(b) [f(x)]³
(c) 2f(x)
(d) 3f(x)

Answer 1

(c) 2f(x)

$f\left(x\right)=\log \left(\frac{1+x}{1-x}\right)$
$$\begin{gathered} \mathrm{Then},f\left(\frac{2x}{1+x^2}\right) \\ =\log \left(\frac{1+{\displaystyle \frac{2x}{1+x^2}}}{1-\frac{2x}{1+x^2}}\right) \\ =\log \left(\frac{{\displaystyle \frac{1+x^2+2x}{1+x^2}}}{{\displaystyle \frac{1+x^2-2x}{1+x^2}}}\right) \\ =\log \left(\frac{(1+x{)}^2}{(1-x{)}^2}\right) \\ =2\log \left(\frac{1+x}{1-x}\right) \\ =2\left(f\left(x\right)\right) \end{gathered}$$