On solution for the simple equation $e^{-x}=x$ I am wondering if it is possible to work out the solution for the equation $\large{e^{-x}=x}$ directly. 
The obvious way to get the answer is to plot the two figures and find the intersection, but is it possible to get the solution analytically without the help of plotting? Do you have any ideas on this?
Thanks a lot in advance.

 A: $e^x=x$ has no real root.
The root of $e^{-x}=x$ or $xe^x=1$ is $x=W(1)\simeq 0.567143$
$W(X)$ is the Lambert W function, which express the root(s) of $We^{W}=X$ where $W$ is the unknown. The function $W(X)$ is multivaluated in $-e^{-1}<X<0$ , as it appears on the figure below : 

From p.15 in : https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function
Example :
$$e^{-x}x^a=b$$
$$e^{-\frac{x}{a}}x=b^{\frac{1}{a}}$$
$$xe^{-\frac{x}{a}}=b^{\frac{1}{a}}$$
$$(-\frac{x}{a})e^{-\frac{x}{a}} =(-\frac{1}{a})b^{\frac{1}{a}}$$
With $y=-\frac{x}{a}$ and $X=-\frac{b^{\frac{1}{a}}}{a}$
$$ye^y =X $$
$$y=W(X)$$
$$-\frac{x}{a}=W\left(-\frac{b^{\frac{1}{a}}}{a}\right)$$
$$x=-aW\left(-\frac{b^{\frac{1}{a}}}{a}\right)$$
A: The solution cannot be expressed with elementary functions.
It can be rewritten as $-1 = (-x)e^{(-x)}$, and the solution of this equation is given/defined by the Lambert W function :
https://en.wikipedia.org/wiki/Lambert_W_function
A: The Newton iteration for the equation $e^{-x}=x$, rewritten in the form $e^{-x}-x=0$, is
$$x_{k+1}=x_k-\frac{e^{-x_k}-x_k}{-e^{-x_k}-1}.$$
Simplify a bit for convenience:
$$x_{k+1}=x_k+\frac{e^{-x_k}-x_k}{e^{-x_k}+1}=\frac{e^{-x_k}-x_k+x_ke^{-x_k}+x_k}{e^{-x_k}+1}=\frac{e^{-x_k}+x_k e^{-x_k}}{e^{-x_k}+1} = \frac{x_k+1}{e^{x_k}+1}.$$
Using the graph we make the naive guess $x_0=0$. Then $x_1=\frac{1}{2}$, $x_2=\frac{3}{2e^{1/2}+2} \approx 0.5663$. The error is already less than $10^{-3}$. The number of correct digits roughly doubles at each subsequent step from this point on, so you get floating point accuracy in three or four more steps.
