Solving $e^\frac1x = x$ non-graphically? This question has come up twice in different tests and the instructions always point out that it should be solved using a graphic calculator. Fair enough, the answer is ≈ 1.76322...(goes on forever?). 
But how do you approach $e^\frac1x = x$ analytically for that solution? Is there a way?
 A: The equation is: 
$$e^{\frac 1x}=x$$
Raise everything to the $x$ power:
$$e=x^x$$
Now using super-square root in terms of Lambert's function leads us to:
$$x=\sqrt e_s =e^{W(1)}=\frac 1{W(1)}$$
A: Let $x=\frac{1}{u}$. Then
\begin{eqnarray}
e^{u} &=& \ln \left( \frac{1}{u} \right) \\
\implies e^{u} &=& \ln1 - \ln u \\
&=& - \ln u \\
\implies \ln u + e^{u} &=& 0 \\
\iff e^{\ln u + e^{u}} &=& 1 \\
\iff e^{\ln u}.e^{e^{u}} &=& 1 \\
\iff ue^{e^{u}}=1
\end{eqnarray}
I suspect the analytic solutions can be found through the Lambert W Function
A: That's $f(\frac{1}{x})=1$, with Lambert W, and not extremely nice. 
But it also formulates itself as a 
fixed point problem:

(computed here.)
Above we have $(g\circ g \circ \dots\circ g)(1)$, i.e. the value after 11 iterations of $g(x):=e^{\frac{1}{x}}$ applied to the solution guess $x=1$.
A: Now note that $e^{\frac{1}{x}}>0$ for $x<0$ and note that $x<0$ for $x<0$. Hence we only have to consider $x>0$.
If $e^{\frac{1}{x}}=x$, then $1/x=\ln(x)$ for $x>0$ , which means that $x\ln(x)=1$.
Now write $x\ln(x)=\ln(x)e^{\ln(x)}=1$ and note that Lambert's W function says then that $\ln(x)=W(1)$, hence $x=e^{W(1)}$
A: The following PARI/GP-program finds the solution using newton's method :
? x=1.5;for(j=1,6,x=x-(exp(1/x)-x)/(-1/x^2*exp(1/x)-1);print(j,"   ",x))
1   1.739986998347306487376609711
2   1.763077568142750917999087271
3   1.763222828791787414311562820
4   1.763222834351896702080716104
5   1.763222834351896710225201777
6   1.763222834351896710225201777

Since the derivate of $f(x)=e^{\frac{1}{x}}-x$ is $f'(x)=-\frac{1}{x^2}e^{\frac{1}{x}}-1$ , which is obviously negative for all $x>0$, the
 soltuion is moreover unique because for $x<0$, $f(x)$ is obviously positive.
A: Here are the steps
$$ \exp\left(\frac1x\right)=x $$
$$ \frac1x \exp\left(\frac1x\right)=1 $$
$$ \frac1x=W\left(1\right) =\Omega$$
$$ x=\frac1{\Omega}\approx\frac1{0.56714329}=1.76322$$
Where $W(z)$ is the Lambert W Function and $\Omega$ is the Omega Constant.
