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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.

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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 $x = \exp(\beta+W(\alpha e^{-\beta}))$.

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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*}

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  • $\begingroup$ you swapped α and β $\endgroup$ – endolith Feb 8 '16 at 4:29
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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. $$x =e^{ W(\alpha e^{-\beta})+\beta}$$.
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