Is there a way to solve this equation? (maybe with Lambert's W Function?) I'd like to know if there is a way to solve the equation 
$$x\ln x=\alpha+\beta x$$
for known constants $\alpha,\beta\in\mathbb{R}$.
I know that Lambert's W Function $W$ can be used to solve
$$x\ln x=\alpha$$
because then $x=e^{W(\alpha )}$, but in the upper problem I can't reformulate the equation in a way that lets me use Lambert's W function. 
Does anyone know how to do this? Or any way to find a solution for $x$?
Thanks.
 A: Note that $x\ln(x) - x\beta = x\ln(xe^{-\beta})$, so we substitute $y=xe^{-\beta}$ and get
$$y \ln(y) = \alpha e^{-\beta}.$$
Thus $y = \exp(W(\alpha e^{-\beta}))$ and $x = \exp(\beta+W(\alpha e^{-\beta}))$.
A: We have
\begin{align*}
  x\log x &= \alpha x +\beta\\
 \iff \beta &= x(\log x - \alpha)\\
             &= x\log\bigl(x\exp(-\alpha)\bigr)\\
 \iff \exp(-\alpha)\beta &= \exp(-\alpha)x \log\bigl(x\exp(-\alpha)\bigr)\\
  \iff \exp(-\alpha)x &= \exp\bigl(W(\exp(-\alpha)\beta)\bigr)
\end{align*}
A: I will give you a step by step solution.


*

*First we have $x \ln x=\alpha +\beta x$.

*We rearrange the equation to get $x \ln x - \beta x = \alpha$.

*Factorising gives $x( \ln x - \beta ) = \alpha$.

*Now we substitute $x= \exp ( \ln x)$ and multiply both sides by $ \exp(-\beta)$.

*Up to now we have
$$( \ln x - \beta)e^{\ln x - \beta}= \alpha e^{-\beta}$$

*Now take $W$ of both sides and remember that $W(xe^{x})=x$, so
$$\ln x - \beta= W(\alpha e^{-\beta})$$.

*Rearranging and taking exponents of both sides gives you the required result.
$$x =e^{ W(\alpha e^{-\beta})+\beta}$$.

