The general solution of $x^a = a^x$ for real $a >0$

Question

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

Answer 1

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.

Answer 2

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.