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I have to solve the following

$y = f(x)$

$y = \ln(\frac{1+x}{1-x}) $

I need to find $x$.

It is ruining my life for the past 30 minutes ... and i am sure it is really easy

Thank you in advance

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    What is $f$ here?2012-10-04

2 Answers 2

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Well, by definition of logarithm, $ \frac{1+x}{1-x} = e^y, $ or $ 1+x=(1-x)e^y. $ Hence $ (1+e^y)x=e^y-1, $ that is $ x = \frac{e^y-1}{e^y+1}. $

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    i figured it out a few minutes ago . Thank you !2012-10-04
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AS $y=\ln\frac{1+x}{1-x}$, by exponential map, we can get that $e^{y}=\frac{1+x}{1-x}$, then $x=\frac{e^{y}-1}{e^{y}+1}$.