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
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
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}. $
AS $y=\ln\frac{1+x}{1-x}$, by exponential map, we can get that $e^{y}=\frac{1+x}{1-x}$, then $x=\frac{e^{y}-1}{e^{y}+1}$.