# Quickie #17

• Jan 28th 2007, 01:43 PM
Soroban
Quickie #17

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

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

• Jan 29th 2007, 09:07 AM
topsquark
Quote:

Originally Posted by Soroban

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

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

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

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

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

-Dan
• Jan 29th 2007, 01:19 PM
Soroban
Lovely work, Dan!

That is the recommended solution.