Solving the equation of the form $a - \frac{1}{x}\ln{(\frac{x}{1-x})} + \frac{1}{x^2}$

Question

I got an equation of the form $a - \frac{1}{x}\ln{\frac{x}{1-x}} + \frac{1}{x^2} = 0$. $a$ is a constant and the values of $x \in [0,1]$.

Does a closed-form solution exist for this? If not, is there any approximate solution?

Answer 1

Not a complete answer:

$$a-\frac{1}{x}\ln\left(\frac{x}{1-x}\right)+\frac{1}{x^2}=0$$ $$x^2a-x\ln\left(\frac{x}{1-x}\right)+1=0$$

We see, this equation is a polynomial equation of more than one algebraically independent monomials ($x,\ln\left(\frac{x}{1-x}\right)$) and with no univariate factor. We therefore don't know how to rearrange the equation for $x$ by applying only finite numbers of elementary functions (elementary operations) we can read from the equation.

$x\to \frac{e^t}{1+e^t}$: $$\frac{ae^{2t}-te^t-te^{2t}+2e^t+e^{2t}+1}{(e^t+1)^2}=0$$ $$a(e^t)^2-t(e^t)^2+(e^t)^2-te^t+2e^t+1=0$$

We see, the equation is an algebraic equation of both $t$ and $e^t$ with no univariate factor. The equation cannot have solutions except $0$ that are elementary numbers therefore.

And we see, we cannot solve this equation in terms of Lambert W.