solutions of $x^x=a^a$, $a>1, x<0$ What are the solutions of the equation $x^x=a^a$, where $a>1$ and $x<0$?
My work so far:
I write $x=\frac{-2t}{2m+1}$, ($t,m\in \mathbb{N}^{*}$). (In order to deal with the exponational function with negative base, I assume only when $x$ is rational with odd divisor). So I get:
$$\large({\frac{-2t}{2m+1}}\large)^{\frac{-2t}{2m+1}}=a^a \quad (1)$$
or $$(2m+1)^{2t}=a^{a(2m+1)}(2t)^{2t}\quad (2)$$
From (1) we can deduce that if $a^a$ is not algebraic ($a^a$ is a root of the rational polynomial $x^{2m+1}-(\frac{2m+1}{2t})^{2t}$) then there is no solution. From (2) we also deduce that if $a^a\in \mathbb{N}$ then there is no solution as  $(2m+1)^{2t}$ is odd and $a^{a(2m+1)}(2t)^{2t}$ is always even. But what does that mean for $a>1$? should be algebraic and not a naturall number? Can we narrow a bit more the possible values for $a$? Thanks.
 A: Here's what I can say about $a$: The complete set of solutions for $(a,x)$ is given by $$
\left\{\left(\exp(W(\left(\frac{2t}{2m+1} \right)^{2t/(2m+1)})),-\frac{2t}{2m+1} \right)\mid t,m\in\mathbb{N}^+\right\}
$$
where $W$ is the Lambert W function. Every solution for $a$ is a transcendental numbers in $(1,1.3211 ...)$, and they form a dense subset of this interval.
As you observe, in order to have $x^x$ be a non-negative real number we need to have that $x$ is a rational number with odd denominator. As a proof, we note that all possible branches for the multi-valued exponential function can be written as $$
(-1)^{x} x^{-x} = \exp(n \pi i x) x^{-x}
$$
for $n$ an odd integer. If the output is to be a positive real number, obviously $nx$ must be an even integer. Thus, $x = -\frac{2t}{2m+1}$ for some positive integers $t,m$, and we get as you claim $$
(2m+1)^{2t} = a^{a(2m+1)} (2t)^{2t}
$$
I'll assume also that $2t$ and $2m+1$ are relatively prime. Let's substitute $b=a^a$, and we obtain $$
(2m+1)^{2t} = b^{2m+1} (2t)^{2t}
$$
From here, we can see that $b$ cannot be an integer because the LHS is odd and the RHS is even. Obviously $b$ is algebraic, and it is furthermore irrational: Suppose $b$ is rational with reduced form $\frac n d$, then $$
\frac{(2m+1)^{2t}}{(2t)^{2t}} = \frac{n^{2m+1}}{d^{2m+1}}
$$
Both fractions are reduced, so we must have $n^{2m+1} = (2m+1)^{2t}$, which implies that $\sqrt[2m+1]{2m+1}$ is an integer, and therefore we would have to have $m=0$, but this implies $x^x < 1 < a^a$. Therefore $b$ is an irrational algebraic number.
This implies that $a$ is transcendental by the Gelfond-Schneider theorem.
To show some bounds on $a$, we observe that $x^x = a^a$ implies $|x|^x = a^a$, or equivalently $$
{(-x)}^{x} = a^a
$$
Since $a>1$, we conclude that $x\in(-1,0)$. Since $a^a$ is strictly increasing, we note that for each $x\in(-1,0)$, there is a unique $a$ that solves the equation. The LHS is maximized at $x=-\frac1e$, thus all the solutions for $a$ satisfy $
a^a \le e^{\frac1e}
$ or equivalently $$
a \le \exp(W(1/e)) \approx 1.321099762\dots
$$
