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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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I formatted your question. I assume you meant $\ln(\frac{1+x}{1-x})$ and not $\ln(1+\frac x1-x)$ or $\ln(1+\frac x{1-x})$. –  Hagen von Eitzen Oct 4 '12 at 11:22
    
@HagenvonEitzen correct, thanks. –  Chris Oct 4 '12 at 11:22
    
Hint: $\frac{1+x}{1-x}=\frac2{1-x}-1$. –  Hagen von Eitzen Oct 4 '12 at 11:23
    
What is $f$ here? –  lhf Oct 4 '12 at 11:55
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2 Answers

up vote 1 down vote accepted

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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Thanks for your answer, but I do not get it.Could you explain it a little bit more? thanks –  Chris Oct 4 '12 at 11:29
    
@Chris Are you familiar with exponentiation? $y=\ln x\iff e^y=x$ –  user39572 Oct 4 '12 at 11:36
    
i figured it out a few minutes ago . Thank you ! –  Chris Oct 4 '12 at 11:40
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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 $x=\frac{e^{y}-1}{e^{y}+1}$.

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