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If $\mathrm{f}(\mathrm{x})=\log \left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)$, then what is $\mathrm{f}\left(\frac{2 \mathrm{x}}{1-\mathrm{x}^{2}}\right)$ equal to ?
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The correct answer is:
$2 \mathrm{f}(\mathrm{x})$
Given that $\mathrm{f}(\mathrm{x})=\log \left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)$
So, $f\left(\frac{2 x}{1+x^{2}}\right)=\log \left(\frac{1+\frac{2 x}{1+x^{2}}}{1-\frac{2 x}{1+x^{2}}}\right)$
$=\log \left(\frac{1+x^{2}+2 x}{1+x^{2}-2 x}\right)=\log \left(\frac{(1+x)^{2}}{(1-x)^{2}}\right)$
$=\log \left(\frac{1+x}{1-x}\right)^{2}=2 \log \left(\frac{1+x}{1-x}\right)$
$=2 \mathrm{f}(\mathrm{x})$
$\left[\right.$ since $\left.f(x)=\log \left(\frac{1+x}{1-x}\right)\right]$
So, $f\left(\frac{2 x}{1+x^{2}}\right)=\log \left(\frac{1+\frac{2 x}{1+x^{2}}}{1-\frac{2 x}{1+x^{2}}}\right)$
$=\log \left(\frac{1+x^{2}+2 x}{1+x^{2}-2 x}\right)=\log \left(\frac{(1+x)^{2}}{(1-x)^{2}}\right)$
$=\log \left(\frac{1+x}{1-x}\right)^{2}=2 \log \left(\frac{1+x}{1-x}\right)$
$=2 \mathrm{f}(\mathrm{x})$
$\left[\right.$ since $\left.f(x)=\log \left(\frac{1+x}{1-x}\right)\right]$
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