Solving for x in $\frac{1+y}{1-y}=(1+x)^{1+x}(1-x)^{1-x}$ 
How can I solve for x?
$$\frac{1+y}{1-y}=(1+x)^{1+x}(1-x)^{1-x}$$

I have tried a few methods, but to be honest, I do not think I have what it takes to find the solutions. I'm hoping someone has has experience with these sorts of equations and can help me out!
 A: Let ${y+1\over y-1}=M$
Then $M-1={2\over y-1}\implies y={2\over M-1}+1$ or $y={M+1\over M-1}={(1+x)^{1+x}(1-x)^{1-x}+1 \over (1+x)^{1+x}(1-x)^{1-x}-1}$
A: IfI properly understand, what you are asked is for find, for a given value of $y$ the corresponding $x$.
This equation is very highly transcendental and you will need some numerical method and, for sure, a "reasonable" guess of the solution.
To some extent, this problem is a red herring and, to me, the worst thing to do would be to write $y=f(x)$.
For a given $y$, you know the value of $\left(\frac{y+1}{y-1}\right)$. So, to make the problem much nicer, define $k=\log \left(\frac{y+1}{y-1}\right)$ and, taking logarithms, consider that you look for the zero of function
$$g(x)=(1+x) \log (1+x)+(1-x) \log (1-x)-k$$ which is quite nice and very smooth (almost looking like a parabola).
Expanding the rhs as an infinite series
$$k=\sum_{n=1}^\infty \frac{x^{2 n}}{n (2 n-1)}$$ Using series reversion
$$x=\sqrt k \,\sum_{n=1}^\infty a_n k^n$$ The $a_n$'s make the sequence
$$\left\{1,-\frac{1}{12},-\frac{13}{1440},-\frac{97}{40320},-\frac{25307}{29030400},-\frac{56827}{153280512},\cdots\right\}$$ and this is almost the exact solution.
A: If you are just trying to solve for $y$ (and not have a particular value, which I think there are multiple valid values), then it is pretty straightforward.  Multiply both sides by $y - 1$.  Then, collect all of the $y$ terms to one side.  Then, pull out the $y$ term.  Finally, divide by the rest to isolate $y$ by itself.
