How does one prove $f(g(x))=3f(x)$ in this case?

Question

Prove $f(g(x))=3f(x)$ for:

$f(x)=\log_e \frac{1+x}{1-x}$ and $g(x)=\frac{3x+x^3}{1+3(x^2)}$

I have tried to use formula of $3\log_e x=\log_e x^3$

Answer 1

It is simple.
\begin{align*} f(g(x))&=log_e\left[\frac{1+\frac{3x+x^3}{1+3x^2}}{1-\frac{3x+x^3}{1+3x^2}}\right] \\ &=log_e\left[\frac{x^3+3x^2+3x+1}{-x^3+3x^2-3x+1}\right]\\ &=log_e\left[\frac{(x+1)^3}{(1-x)^3}\right]\\ &= 3\,log_e\left[\frac{1+x}{1-x}\right]=3\,f(x) \end{align*}