Question:
If $f(x)=\log \left(\frac{1+x}{1-x}\right)$, then $f\left(\frac{2 x}{1+x^{2}}\right)$ is equal to
(a) $\{f(x)\}^{2}$
(b) $\{f(x)\}^{3}$
(c) $2 f(x)$
(d) $3 f(x)$
Solution:
(c) $2 f(x)$
$f(x)=\log \left(\frac{1+x}{1-x}\right)$
Then, $f\left(\frac{2 x}{1+x^{2}}\right)=\log \left(\frac{1+\frac{2 x}{1+x^{2}}}{1-\frac{2 x}{1+x^{2}}}\right)$
$=\log \left(\frac{\frac{1+x^{2}+2 x}{1+x^{2}}}{\frac{1+x^{2}-2 x}{1+x^{2}}}\right)$
$=\log \left(\frac{(1+x)^{2}}{(1-x)^{2}}\right)$
$=2 \log \left(\frac{1+x}{1-x}\right)$
$=2(f(x))$
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