How to solve this tetration equation $\;^n 2 = \;^2 n $? How would one find all real solutions to the following equation:
$\qquad$ $n^n = 2^{2^{2^{2^{\dots^2}}}} $(where the number of $2$s is equal to $n$)
generalizing to $n$ being a real value. In tetration-notation this is    
$\qquad $ find a solution to  $\displaystyle \;^2 n = \; ^n 2$ for real $n \ne 2$.
I know one solution is $n = 2$, but I wonder if any other solutions exist.
Edit: Could there be any negative number solutions to this equation?
 A: I wanted to add some graphs to Gottfried's solution.  First definitions; $\text{sexp}_2(z)= \;^z 2$ which is extended to the complex plane by Kneser's solution. I wrote a program to calculate the slog; which is the inverse of sexp and has some nice uniqueness properties.  The fatou.gp program works for a wide range of real and complex sexp bases and is written in pari-gp and is available on this site http://math.eretrandre.org/.   Instead of graphing $\text{sexp}_2(x)\;$and $\;x^x$, I will take the $\log_2(x)$ of both equations which works when both are positive.  So I am graphing 
$$\text{sexp}_2(x-1)\;\;\text{vs}\;\;\log_2(x^2)=\frac{x\ln(x)}{\ln(2)}$$
Here is a graph for sexp base 2 which shows a solution at x=0, x=2, and at x=3.4141760984020147407016

What about other bases?  Here is a graph for sexp base=2.1150455841, which has a parabolic crossing near 2.5360 so there are only two solutions.  
For larger bases, there is only one solution.  Here is sexp base e which only has a solution at x=0.
