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I know it it seems trivial that $x = 1$, but I would like to know a more rigorous solution involving algebra. I tried solving for it, but could not come up with a proper solution.

My attempt:

$e = xe^x \implies e^{1-x} = x \implies (1-x)\ln e = \ln x \implies 1-x = \ln x \implies 1 = \ln x + x$

And at this point I got stuck.

A is for Ambition
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    $f(x) = \ln x + x$ is a strictly monotonically increasing function for $x > 0$. Hence $x = 1$ is the unique solution of $f(x) = 1$ for positive $x$. It's not hard to convince yourself there are no solutions to the original equation $e = xe^x$ for $x \leq 0$. – Simon S Jan 27 '15 at 21:08
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    $1 = \ln x + x $ ... recall that for $x>1, \ln x > 0 \Rightarrow \ln x+x > 1$ and for $x<1, \ln x<0 \Rightarrow \ln x+x < 1 $ – Joffan Jan 27 '15 at 21:16

2 Answers2

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The function $f(x) = xe^x$ does not have an inverse that can be expressed in terms of any finite algebraic combination of the usual functions. The two branches that form its inverse are known as the Lambert W function(s).

The equation $$ y = xe^x $$ can be solved for $y$ using a numerical approach, if an approximate solution is desired.

Ben Grossmann
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    this does not even answer the question and gets accepted as an answer? – abel Jan 28 '15 at 20:30
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    @abel the question was how to solve the question algebraically. The only reasonable answer is to state that it is impossible to do so. – Ben Grossmann Jan 28 '15 at 20:33
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Consider $f(x) = xe^x-e$. You have that $x = 1$ is a root. And: $$f'(x) = e^x + xe^x > 0,$$ for $x > 0$, so we don't have any positive roots other than $1$. If $x \leq 0$, you would have $e = \text{something negative}$, which can't happen.

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Ivo Terek
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