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Math Help - Quickie #17

  1. #1
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    Quickie #17


    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).

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    Quote Originally Posted by Soroban View Post

    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
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    Lovely work, Dan!

    That is the recommended solution.

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