Quickie #17

**Soroban** · January 28th 2007, 01:43 PM

If $f(x)\:=\:\ln\left(\frac{1+x}{1-x}\right)$ .for $-1 < x < 1$ ,

. . express $f\left(\frac{3x+x^3}{1 + 3x^2}\right)$ in terms of $f(x)$ .

**topsquark** · January 29th 2007, 09:07 AM

Originally Posted by Soroban

If $f(x)\:=\:\ln\left(\frac{1+x}{1-x}\right)$ .for $-1 < x < 1$ ,

. . express $f\left(\frac{3x+x^3}{1 + 3x^2}\right)$ in terms of $f(x)$ .

I don't know if there is a shortcut, but if you simply plug the argument into the f(x) formula you get:
$ln \left ( \frac{ 1 + \frac{3x + x^3}{1 + 3x^2} }{ 1 - \frac{3x + x^3}{1 + 3x^2} } \right )$

With some simplification:
$= ln \left ( \left [ \frac{1 + x}{1 - x} \right ]^3 \right )$

$= 3 ln \left ( \frac{1 + x}{1 - x} \right ) = 3f(x)$

-Dan

**Soroban** · January 29th 2007, 01:19 PM

Lovely work, Dan!

That is the recommended solution.

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