What are solutions to $2^x=x$? Are there any solutions (real, complex , matrix etc.) to $2^x=x$? The best I can come up with is $\ln 2 = \frac{\ln x}{x}$ or $x^{\frac{1}{x}}=2$
 A: A simple explanation for the answer given by @ADG:
I.$$x\leq 0\implies x \leq 0<2^x;$$
II.$$0<x\leq1\implies x\leq1=2^0<2^x;$$
III.$$1<x\implies x<1+[x]<2^{[x]}\leq2^x. $$
($[x]$ is the whole part of $x$.)
A: There is no elementary method for solving $a^x=x$. However the solution exists and it is given by $$x=-\frac{W(-\log (a))}{\log (a)}$$ where appears Lambert function (just as anon suggested). The problem is that, in the real domain, this would require $\log(a) \leq \frac 1e$, that is to say $a \leq e^{\frac{1}{e}}\approx 1.44467$.
Now, if $a \gt e^{\frac{1}{e}}$, there is a complex solution.
A: For $x<0$ there is no solutions as $2^x>0,x<0$. For $x=0$, $2^x=1$ and $2^x$ increases faster than $x$, so there seems no solution in $x>0$ too.

A: If by elementary you mean real solutions then there are none. It is clear that for negative $x$ there can be no roots. so then we can confine our attention to positive $x$. An obvious way to deal with this would be to plot $y=2^x$ and $y=x$ on the same axes.
Alternatively we may observe that for $x>0$ that: $$2^x=e^{\ln(x)}>1+x\ln(x)+x^2 \frac{(\ln(x))^2}{2}=P_2(x)$$
Then observe that $P_2(x)-x>0$ for $x>0$.
By observe, I mean demonstrate that the minimum of $P_2(x)-x$ is positive.
More generally this has solutions in terms of the Lambert $W$ function, and is more or less example 1 on the Wikipedia page on Lambert's $W$, but such solutions do not qualify as elementary for most interpretations of the term.
