# How to solve $y = \ln\frac{1+x}{1-x}$?

Can the following equation be solved for x?

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

*This is not actually homework - it comes from part of the rate law for a particular chemical reaction I am studying - but like a homework question, I'd be happy with pointers / strategies for the solution. I ought to be able to do this on my own, but today I'm making no progress!

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Take exponentials of both sides. Then multiple across by $1-x$ and solve for $x$. – copper.hat May 30 '12 at 19:27

$$e^y=\frac{1+x}{1-x}$$

$$e^y-1=(e^y+1)x$$

$$x=\frac{e^{y}-1}{e^{y}+1}$$

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Taking the exponential, we have \begin{align*} y &= \ln\left(\frac{1+x}{1-x}\right)\\ e^y &= \frac{1+x}{1-x}\\ e^y(1-x) &= 1+x\\ e^y -1 &= x + e^yx\\ e^y - 1 &= x(1+e^y)\\ \frac{e^y - 1}{e^y+1} &= x. \end{align*}

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There is the "special" function

$$\tanh^{-1} x = \frac 1 2 \log \frac{1+x}{1-x}$$

which is the inverse of

$$\tanh x = \frac{e^{x}-e^{-x}}{e^x+e^{-x}}$$

So what you want is

$$2\tanh^{-1} x =y$$

or

$$x=\tanh \frac{y}{2}$$

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As I see it $\operatorname{tanh}^{-1}=\operatorname{arctanh}$ is en elementary function. But sure every function is "Special" in a weak sense. :) – AD. May 31 '12 at 6:41
@AD. That's why the put it between " " – Pedro Tamaroff May 31 '12 at 9:16

Just consider the inverse function of $\frac{1+x}{1-x}$ that is $\frac{x-1}{x+1}$. After taking exponential of both sides, place each of the sides in the inverse function and get the final result. $$\frac{e^y - 1}{e^y+1} = x$$ The proof is complete.

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