How to solve $(1+x)^{y+1}=(1-x)^{y-1}$ for $x$? Suppose $y \in [0,1]$ is some constant, and $x \in [y,1]$. How to solve the following equation for $x$:
$\frac{1+y}{2}\log_2(1+x)+\frac{1-y}{2}\log_2(1-x)=0$ ?
Or equivalently $1+x = (1-x)^{\frac{y-1}{y+1}}$?
Thanks very much.

If we set $f = \frac{1+x}{1-x}$ and $\eta = \frac{y+1}{2}$ some manipulation yields the following:
$$2f^{\eta} - f - 1 = 0$$
For rational $\eta = \frac{p}{q}$, this can be converted to a polynomial in $f^{\frac{1}{q}}$, and is likely "unsolveable" exactly, and would require numerical methods (like Newton-Raphson etc, which will work for irrational $\eta$ too).
