I'm trying to inverse the function $f:x\mapsto Ax+Bx\log x$.
I know from Wolfram Alpha that the result is: $f^{-1}(x)=\frac{A}{BW(\frac{Ae^{x/B}}{B})}$ where $W$ is the W-Lambert (or product logarithm) function but I can't find the necessary steps to confirm it.
Note: in my specific settings, $A>0$, $B<0$ and $x \gg 1$ so $\frac{Ae^{x/B}}{B} > 1/e$ and $W$ is defined.
Best Answer
My solution (notation: we are looking for $x$ expressed with $y$ which was the previous left hand side):
Write $x=e^t$:
$$y=Ae^t+Bt e^t=e^t(A+Bt)$$ write $A/B+t=u$ to get rid of addition (you want a standard form of $xe^x$ to use the Lambert's function): $$y=Bu e^{u-A/B}=Be^{-A/B}u e^u$$ Divide to get to the standard form: $$e^{A/B}\frac{y}{B}=ue^u$$ Invert:
$$u=W\left(e^{A/B}\frac{y}{B}\right)$$ Now just express $x=e^{t}=e^{u-A/B}$: $$x=e^{-A/B}\exp\left[W\left(e^{A/B}\frac{y}{B}\right)\right]$$ It's not exactly the same form as you expected, but you can use other identities, such as $$e^{W(x)}=\frac{x}{W(x)}$$ to transform it into $$x=\frac{y}{BW\left(e^{A/B}\frac{y}{B}\right)}$$
In your expression, $A$ and $y$ (or actually $x$) are reversed, I'm not sure about that.