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

TMM
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4 Answers4

26

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. ain't a mathematician 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. –  Nov 28 '10 at 21:33
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    @Perfecto: Not "impossible"; the reason the Lambert function was constructed so that it becomes possible! – J. M. ain't a mathematician Nov 28 '10 at 22:49
19

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.

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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.

Adrián Barquero
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7

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 $x=1$

Bryan Yocks
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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 $x=1$.

Henry B.
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