Inverse of function $y(x)=\frac1b(\ln(x+1)-x)$

Question

I'd like to ask a question about the function below : $$yb = ln ( x + 1 ) - x$$ I'm trying to find it's inverse. What I've tried so far : \begin{align} yb + x & = ln(x+1) \\ e^{yb + x} & = x+1 \\ e^{yb + x} - x & = \\ \end{align}

I'm stuck there and I don't know what to do. ${y,b}$ and ${x}$ are not whole numbers.

I've tried continuing to

\begin{align} e^{yb} . e^{x} - x & = 1 \\ e^{x} - {x\over e^{yb}}& = {1\over e^{yb} } \\ \end{align}

But it seems to make it worse. Any suggestion?

Thanks for your help :)

Answer 1

Subtract 1 from both sides of the equation $$yb-1 = \ln ( x + 1 ) - (x+1)$$ then set $X = x+1$ and exponentiate $$e^{yb-1} = Xe^{-X}$$ Now substitute $Z=-X$ and get $$-e^{yb-1} = Ze^{Z}$$ This is the standard situation for Lambert W and it gives $$Z=W\left(-e^{yb-1}\right)$$ Undo the substitutions to have the final result $$x=-W\left(-e^{yb-1}\right)-1$$