How can the following equation solved for x?
$\ln(x) - \ln(1-x) = \ln(1-y) - \ln(1-z) - \ln(v)$
I assume it simplifies to:
$\frac{x}{1-x} = \frac{1-y}{(1-z)*v} $
I have tried to solve this by factorizing the terms, but it does not lead anywhere. Any hints?
\begin{align*} \frac{x}{1-x} &= \frac{x\overbrace{-1+1}^{0}}{1-x} \\ &= \frac{x-1}{1-x} + \frac{1}{1-x} \\ &= -1 + \frac{1}{1-x} \end{align*} So you have \begin{align*} -1 + \frac{1}{1-x} &= \frac{1-y}{(1-z)v} \\ \frac{1}{1-x} &= 1 + \frac{1-y}{(1-z)v} \\ \frac{1-x}{1} &= \frac{1}{1 + \frac{1-y}{(1-z)v} } \\ x-1 &= \frac{-1}{1 + \frac{1-y}{(1-z)v} } \\ x &= 1 + \frac{-1}{1 + \frac{1-y}{(1-z)v} } \\ \end{align*}
Notice that your original equation has the constraints $x > 0$, $1-x > 0$, $1-y > 0$, $1-z > 0$, and $v > 0$. The above introduces the constraint $1 + \frac{1-y}{(1-z)v} \neq 0$ (although this follows as a consequence of the others).
$$\frac{x}{1-x} = \frac{1-y}{(1-z)v} \implies \frac{x}{1-y}=\frac{1-x}{(1-z)v} = \frac{x+1-x}{1-y+(1-z)v}=\frac{1}{1-y+(1-z)v}\\ \implies x = \frac{1-y}{1-y+(1-z)v}$$
where we used the fact that if $\frac ab = \frac cd$ then $$\frac ab = \frac cd = \frac{a+c}{b+d}\tag1$$
