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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 $x$.2012-05-30

4 Answers 4

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

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

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

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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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    @AD. That's why the put it between " "2012-05-31