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?

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.
