Solving $\frac{\ln(x)\ln(y)}{\ln(1-x)\ln(1-y)}=1$ for $y$ 
I'm trying to solve for $y$ in terms of $x$ for the expression below.
$$\frac{\ln(x)\ln(y)}{\ln(1-x)\ln(1-y)}=1$$

First I multiplied both sides by $$ \frac{\ln(1-x)}{\ln(x)} $$
to get
$$ \frac{\ln(y)}{\ln(1-y)}=\frac{\ln(1-x)}{\ln(x)} $$
but I don't see how to isolate $y.$ I tried using every technique I know including logarithm properties.

The function $f(x)=\dfrac{\ln(1-x)}{\ln(x)}$ is monotonic in its domain $(0,1)$, hence it is invertible. So the relation between $x$ and $y$ is a bijection, and…
$$y=1-x.$$

Interestingly, the function is well approximated by $\left(\dfrac1x-1\right)^{-3/2}$, and a solution with $a$ in the RHS is approximately
$$\left(\dfrac1x-1\right)^{-3/2}=a\left(\dfrac1y-1\right)^{3/2},$$ or
$$y=\frac{1-x}{1+(a^{2/3}-1)x}.$$
