The general solution of $x^a = a^x$ for real $a >0$ What are the roots of
$$f(x) = x^a - a^x$$
for real $a > 0$?


*

*Case 1: For $0 < a < 1$ there is 1 solution, $x=a$.

*Case 2: For $1\le a < e$ there are 2 solutions: $x=a$ and $[x>a]$.

*Case 3: For $a > e$ there are 2 solutions: $x=a$ and $[x < a]$.  


if we let b = a^(1/a). x = b^^ [use mathematica for tetration]
or using the basic operators, 
x = b^(b^(b^...))...)
for a > e, this solves for the non-simple root < a 
but for a < e it solves for  x = a
i am looking for a similar solution for a < e
 A: For $a\neq 1$ $x^a=a^x$ is equivalent to 
$$x=a^{\frac{x}{a}}\\
xa^{-\frac{x}{a}}=1\\
xe^{-\frac{\ln a}{a}\cdot x}=1\\
(-\frac{\ln a}{a}\cdot x)e^{-\frac{\ln a}{a}\cdot x}=-\frac{\ln a}{a}\\
-\frac{\ln a}{a}\cdot x=W(-\frac{\ln a}{a})\\
x=-\frac{aW(-\frac{\ln a}{a})}{\ln a}$$
using Lambert W function. Note that this function is actually multivalued for negative arguments, so it includes both the trivial solution $x=a$ and the other, nontrivial one. One shouldn't expect to be able to solve for $x$ using only elementary functions.
A: This is not a complete answer, but an example showing that two (in some circumstances exact) solutions do actually exist.
Let $a := 2$. Then
$$
\begin{array}{rcrcl}
  x := 2 &\Rightarrow& x^2 =& 4 &= 2^x, \\
  x := 4 &\Rightarrow& x^2 =& 16 &= 2^x. \tag{*}
\end{array}
$$
Let $a := 4$. Then
$$
\begin{array}{rcrcl}
  x := 2 &\Rightarrow& x^4 =& 16 &= 4^x, \tag{*} \\
  x := 4 &\Rightarrow& x^4 =& 256 &= 4^x.
\end{array}
$$
The solutions (*) arise from the symmetry in the equation. I conjecture that, if more such examples exist, powers of $2$ might play a role.
