Can $t(x)$ be found from: $$A \, t + B\ln\frac{1-t}{t}=x \; ?$$ Here, $A>0, \; B < 0$ and $0 \lt t \lt 1$.
The $t(x)$ should be given in analytical form (even if you use, say, Lambert's W - function).
PS: Such an expression (left hand side) appears as the complex phase in the integral representation of the confluent hypergeometric function of the first kind $_1F_1$, as used in the Coulomb wave function for positive energies.
You can start with $t(\frac{A}{2}):=\frac{1}{2}$.
Derivation by $x$:
$t'(x)=\frac{1}{A-B(\frac{1}{1-t(x)}+\frac{1}{t(x)})}$
$t'(\frac{A}{2})=\frac{1}{A-4B}$
And so on: Taylor series with the development point $\frac{A}{2}$.
EDIT:
If you set $t(x+\frac{A}{2})-\frac{1}{2}:=\sum\limits_{k=0}^\infty x^k a_k$ (=> $a_0=0$) then you’ll get a recurrence formula for $a_{k+1}$ by using the equation $(t-\frac{1}{2})’(-A(t-\frac{1}{2})^2+\frac{A}{4}-B)=\frac{1}{4}-(t-\frac{1}{2})^2$.
It's more convenient to use $a_1$ instead of $\frac{1}{A-4B}$.