# Solving the equation $x e^x = e$

I know that $x e^x = e$ means $x = 1$, but how do you solve for it?

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What have you tried? Can you rule out some ranges for $x$? Have you tried playing with logarithms (the natural thing to do)? –  Alex B. Nov 28 '10 at 8:38
@Raskolnikov: the problem is to show that x = 1 is the only (say, real) solution. I do not think "duh" is a constructive comment. –  Qiaochu Yuan Nov 28 '10 at 17:00
So guys, correct me if I am wrong: In many ways we can prove that x = 1, but given "xexp(x) = e", it is impossible to solve for x, that is, write x = {some formula} = ... = 1. –  Perfecto Bautista Nov 28 '10 at 21:30
@Perfecto: essentially, yes. –  Qiaochu Yuan Nov 29 '10 at 0:44
Are we going to debate the term "closed form" again? –  Ｊ. Ｍ. Nov 29 '10 at 1:03

If you want to prove the proposition that if $x e^x = e$ then $x = 1$, one way to do it is to just show that (1) $x = 1$ is indeed a solution of $x e^x = e$ by plugging the value in, and that (2) $x = 1$ is the only solution, for which you can argue that $x e^x$ is an increasing function on $x \ge 0$, and the equation clearly has no solution for $x < 0$.

If that makes it sounds like you have to know the answer beforehand before you can solve for it, well, that's not far from the truth. In general, if the right-hand side is an arbitrary real number $y$ instead of $e$, then you cannot solve $x e^x = y$ for $x$ in terms of elementary functions. One in fact defines the Lambert W function as precisely the solution of this equation.

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1 is the only real solution, yes, but remember that the Lambert function is multivalued like the logarithm. (For those with Mathematica: N[ProductLog[Range[-5, 5], E]] and then Chop[E - % Exp[%]]) –  Ｊ. Ｍ. Nov 28 '10 at 12:44
So guys, correct me if I am wrong: In many ways we can prove that x = 1, but given "xexp(x) = e", it is impossible to solve for x, that is, derive and write x = {some formula} = ... = 1. –  Perfecto Bautista Nov 28 '10 at 21:33
@Perfecto: Not "impossible"; the reason the Lambert function was constructed so that it becomes possible! –  Ｊ. Ｍ. Nov 28 '10 at 22:49

Another way is thinking of this as asking where do the graphs of the functions given by $f(x) = e^{x-1}$ and $g(x) = \frac{1}{x}$ intersect. This is because $$xe^{x} = e \Leftrightarrow e^{x-1} = \frac{1}{x} \quad \text{for x \neq 0}$$ Then the answer you have would correspond to the only intersection point $(1, 1)$. So if you have no idea of where to start you could just draw the corresponding graphs and that could give you an idea of what is happening.

Then the graphs maybe would suggest to you the following line of reasoning. If $0 < x < 1$ then $$\frac{1}{x} > 1 \quad \text{but} \quad e^{x-1} < e^{0} = 1$$ so their graphs can't meet in the interval $]0, 1[$. Then if $x > 1$ we have $$\frac{1}{x} < 1 \quad \text{but} \quad e^{x-1} > e^{0} = 1$$ so again there's no intersection point in the interval $]1, \infty[$.

This rules out all the possible solutions apart from $x = 1$ which you already knew and which could be guessed by looking at the graphs.

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If someone knows how to include pictures in an answer I'd like to know. –  Adrián Barquero Nov 28 '10 at 16:29
án: Clicking one of the buttons in the top portion of the "Edit" box (fifth button from the left) should bring up a box for uploading pictures. –  Ｊ. Ｍ. Nov 28 '10 at 16:33
@J.M. Thanks a lot! –  Adrián Barquero Nov 28 '10 at 18:21
+1 from me . –  Bryan Yocks Nov 28 '10 at 20:14
+1, but I think those $<$ and $>$ should be $\leq$ and $\geq$. –  JMCF125 Apr 8 '14 at 21:26

With logarithms.

Firstly $x>0$, Then :

$xe^{x}=e \Rightarrow \ln (xe^{x})=\ln e \Rightarrow \ln x + x = 1$.

So:

If $x>1$ then $\ln x >0$, thus $\ln x +x > 1+0 =1$ (contradiction)

If $x<1$ then $\ln x < 0$, thus $\ln x +x < 1+0 =1$ (contradiction)

Thus $x=1$

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Assuming that we are only looking for real solutions,

$$xe^x = e.$$

Note that $e^x$ is always positive and so no solution for this equation is possible for $x<0$ or $x=0$. As x increases, both x and $e^x$ are strictly increasing. Therefore, there is only one value for which $xe^x = e$. We already know that solution to be $x=1$.

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