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.
 A: If it helps:
$$\frac{\ln{(y)}}{\ln{(1-y)}} = \frac{\ln{(1-x)}}{\ln{(x)}}$$
$$e^{\frac{\ln{(y)}}{\ln{(1-y)}}} = e^\frac{\ln{(1-x)}}{\ln{(x)}}$$
$$ {{e^{\ln{(y)}}}^{\frac{1}{\ln{(1-y)}}}} =  {{e^{\ln{(1-x)}}}^{\frac{1}{\ln{(x)}}}}$$
$$y^{\frac{1}{\ln{(1-y)}}} = (1-x)^{\frac{1}{\ln{(x)}}}$$
$$y^{\ln{(x)}} = (1-x)^{\ln{(1-y)}}$$

From here you can make a substitution:
$1-y = t$.

And you get:
$$(1-t)^{\ln{x}} = (1-x)^{\ln{(t)}}$$
Since both $\ln{x}$ and $e^x$ are injective it follows:
$$x = t$$
And since $t = 1-y$, we conclude:
$$y = 1-x $$
A: 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
$$y=\frac{1-x}{1+(a^{2/3}-1)x}.$$
A: Because, in the general case, the equation depends on two algebraically independent monomials ($\ln(y),\ln(1-y))$, the equation cannot be solved for $y$ by only rearranging it by applying only finite numbers of elementary functions/operations.
Other tricks, Special functions, numerical or series solutions could help.
