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how to solve the following equation?

$x+\log(x)=n+c$,where $c$ is constant, and $n$ is positive number and allow to be infinity.

Intuitively, if let $n \to \infty$， the approximate solution will be $x=n+c-\log(n)$, can any one help with the exact solution?

Did
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cindy
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    Search up lambert W function. – Ahmed S. Attaalla Sep 13 '15 at 04:27
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    Exponentiating both sides gives $x e^x = e^c$, which can be solved (more or less by definition) by the Lambert W function @AhmedS.Attaalla mentions. See this answer to the question I suggested as a duplicate: http://math.stackexchange.com/a/417666/155629 – Travis Willse Sep 13 '15 at 04:29
  • OP: Please do not deface your question after people answered it and commented on it. – Did Sep 13 '15 at 06:04
  • @Did I think deface is a too strong a word for the change OP make. the change of $\log(x)$ to $3\log(x)$ can be a legitimate one. – achille hui Sep 13 '15 at 06:13
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    @achillehui AFAIAC, I can imagine no reason for the 3. But more importantly, the revision excluded Lambert W function, thus disqualifying nearly everything contributed to the page by others so far (and making the question unanswerable, but this is still another point). – Did Sep 13 '15 at 06:18

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you can write that as $x\log e+\log x=n+c$

$\log xe^x=n+c$

and now you x on one side and constants on another

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