# How to solve $x\cdot\mathrm e^x=1$? [duplicate]

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Inverse of $y=xe^x$

I would like to solve the equation $x \cdot\mathrm e^x=1$. I know it has an answer, I could find it with a calculator, but I don't remember how to solve it on paper.

Any help?

edit

I know the answer is $x \approx 0.567143$. I don't want the answer, I want a method to find it.

## marked as duplicate by Hans Lundmark, J. M. is a poor mathematician, Sasha, Asaf Karagila♦, t.b.Oct 28 '11 at 14:37

• en.wikipedia.org/wiki/Lambert_W_function – Hans Lundmark Oct 28 '11 at 12:42
• Thanks @Hans for the comparision, but I don't think it's the same question. The links you gave don't answer me. – SteeveDroz Oct 28 '11 at 12:53
• Put $y=1$ in the question I linked to. The answer is that $x=W(1)$ where $W$ is a certain nonelementary function. You won't be able to write down any explicit expression for $x$ in terms of elementary functions, if that's what you're after. – Hans Lundmark Oct 28 '11 at 13:28

You can easily verify that there is only one solution:

1. if $x\leq0$ then $x\cdot\mathrm e^x\leq0<1$;

2. if $x>0$ then $x\cdot\mathrm e^x=1$ iff $\mathrm e^x = \frac1x$ (see the graph below); indeed, $\mathrm e^x$ increases and $\frac1x$ decreases on the set $\{x>0\}$ so there is no more than one solution. The solution exists since $\mathrm e^{0.1}<10$ but on the other hand $\mathrm e^1>1$ and hence by Intermediate Value Theorem there is a point $x\in (0.1,1)$ such that $\mathrm e^x = \frac1x$. This point you can easily find numerically: $x\approx 0.567143$ • Alternatively, you can also try intersecting $x$ and $\exp(-x)$. :) – J. M. is a poor mathematician Oct 28 '11 at 12:44
• Thanks, that's all very comprehensible. My question was: is there a way to solve the equation on paper, without appromiation. I'm not looking for the answer, I'm looking for the method. – SteeveDroz Oct 28 '11 at 12:51
• @Oltarus: nice to be of help, though solve the equation on paper is a bit incorrect problem. If I tell you that the answer is $\sin {(16! - 2\sqrt{15})}$ is it a solution (quite close to the real one, btw)? As you can read at Wiki page which Hans has cited, $W$ is not expressible through elementary function, so any answer will be either $x = W(1)$ or $x \approx 0.567$ – Ilya Oct 28 '11 at 12:57
• @oltarus: "is there a way to solve the equation on paper, without approxmiation." - so for example, if I say the solution of $\exp(x)=2$ is $x=\log\,2$, you'll object? – J. M. is a poor mathematician Oct 28 '11 at 12:59
• Hehe, I don't object, I might laugh. Ok, I understand: the function is not expressible except by $x=W(1)$. Thanks guys! – SteeveDroz Oct 28 '11 at 13:07

Newton-Rapshon Theorem:

$f(x)=xe^x-1$ (your function).

$f'(x)=e^x x+e^x$ (the derivative of your function).

After this, put in the Newton-Rapson formula:

$x_{k}=x_{k-1}-\frac{f(k-1)}{f'(k-1)}$

and define a first attempt, for example, $x_0=1$.

The final formula, for your equation will look like this:

$x_{k}=\frac{x_{k-1}^2+e^{-x_{k-1}}}{x_{k-1}+1}$, $\quad$ $x_0=1$.

Now you need to find this $x_k$ values because, the limit of $x_k$ when $k$ tends to infinity solves the original equation.

See this table: and this plot of $f(x)$: • Thank you for this different approach. That was not at all what I was looking for, but +1 anyway, because I think it's cool and gives a good approximation of values. – SteeveDroz Oct 31 '11 at 7:35

There are infinitely many complex numbers that satisfy your equation. Put another way, the Lambert function, which is the inverse function of $x\cdot\exp\,x$, has many branches. The unique real solution is $W(1)\approx0.567143290409783873$; the other complex solutions include $W_{-1}(1)\approx -1.5339133197935745079 - 4.3751851530618983855\,\, i$ and $W_1(1)\approx -1.5339133197935745079 + 4.3751851530618983855\,\,i$, among others...