Solutions of $a^{a^x}=x$ for fixed $a>0$ I am interested in the equation $a^{a^x}=x$ for some fixed $a>0$. Is there some way to rearrange for $x$ or solve otherwise? What about the nature of the solutions? For which fixed $a>0$ are there any real solutions $x>0$, and how many?
I already worked with the equation $a^x=x$ and I can deal with it. I learned from the comments that a real solution exists for $a\le e^{1/e}$ and is given by 
$$x=-\frac{W(-\ln(a))}{\ln(a)}$$ 
with the Lambert W function. But the equation above is out of reach for me.
 A: Remark: I have not yet a closed-form formula for the occuring fixpoints, but to put the earlier comments together and to give at least a simple computational procedure to determine the additional fixpoints.                 


*

*) To "symmetrize" the formula we observe, that we can as well write 
$$ (a^x)^{(a^x)} = x^x \tag 1$$ 
This makes even more obvious, that $f(x)=a^{a^x}$ has the same fixpoints as $g(x)=a^x$  (We have $f(x)=g(g(x))$ so this is of course basically obvious).
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Let $t= \exp(-W_0(-\log(a))) $ then $a=t^{1/t}$ and $(t^{1/t})^t=t^1=t$ and $t$ is a fixpoint (may be complex).         

*) By a plot we find in four ranges for $a$ different behaviour.
$$ \begin{array} {} 
(1)& e^{1/e}& <a &<\infty  & 0 \text{ real fixpoint for $g(x)$ and $f(x)$ } \\
(2)& e^{1/e} &= a &        & 1 \text{ real fixpoint for $g(x)$ and $f(x)$ } \\
(3)& 1 &< a &< e^{1/e} &  2 \text{ real fixpoints for $g(x)$ and $f(x)$ } \\
(4)& e^{-e} &\le a & \le 1  & 1 \text{ real fixpoint for $g(x)$ and $f(x)$ } \\
(5)& 0 &< a & < e^{-e}  & 1 \text{ real fixpoint for $g(x)$ and $f(x)$ } \\
& & & & \text {and $2$ more real fixpoints for $f(x)$ }
\end{array} \tag 2$$

*) Where possible, the "standard" real fixpoint $t$ is computed using the LambertW using $W_0(x)$ and in case (3) we find another one using $W_{-1}(x)$. For case 5 we can find the two other fixpoints $u$ (lower),$v$ (upper) by the following routine             
Pseudocode:
let a=0.01
let u=0, v=1 
for k=1 to 10: u, v = a^v, a^u : end  \\ to approximate initially 
\\ use Newton-algorithm to approximate second fixpoint to high precision
err=1 
while err>1e-200 
  err = (a^a^u - u )/(a^a^u * a^u * log(a)^2 - 1)
  u = u-err
wend
\\ find the third fixpoint v
v = a^u

Example: For base $a=0.01$ I find the first fixpoint for $f(x)$ and $g(x)$ using the Lambert $W_0()$-function as $t=0.27798742481...$ .
The two additional fixpoints for $f(x)$ are $u=0.013092520508... $ and $v= 0.941488368575...$ .           


*) We have for bases of the case (5)
$$ \large \begin{array} {}
a^{a^t}=t & a^{a^u}=u &a^{a^v}=v & \small \text{all are fixpoints of $f(x)$ }\\
a^t = t &a^u = v & a^v = u \\
a=t^{1/t} &a = u^{1/v} & a = v^{1/u}
\end{array} $$         


Perhaps from the last equalitites one can derive more closed-form expressions using Lambert W...
Picture for the three fixpoints for bases $0<b<1/e^e$  (sorry that I used "b" instead of your "a" for the base, it is my long-trained notation)  

A: I found a way to rearrange for $x$ which works for some $a$ and yields some solution! Some rigorous analysis is necessary to completely understand this procedure and to find similar forms of the other (real) solutions (there are zero to three). I still hope this might help you.

So let's start from $a^{a^x}=x$ for $a,x>0$, and state it like Gottfried did as $(a^x)^{a^x}=x^x$. There is the well known way to parametrize some solutions of $x^x=y^y$, which is
$$x=t^{\frac{1}{1-t}},\qquad y=t^{\frac t{1-t}}$$
for $t>0$. So, to solve the above problem, we are looking for a value $t$ for which we have
$$x=t^{\frac{^1}{1-t}},\qquad a^x=t^{\frac t{1-t}}.$$
The left equation can be rearranged for $t$ and we get $t=\frac{W(x\log x)}{\log x}$. When we plug this into the right side we find
$$a^x = \left(t^{\frac 1{1-t}}\right)^t=x^t=x^{\frac{W(x\log x)}{\log x}}\quad\Rightarrow\quad a=x^{\frac{W(x\log x)}{x\log x}}=x^{\frac{W(u)}u}$$
with $u=x\log x$. This gives $x=u/W(u)$ and 
$$a=\left(\frac{W(u)}u\right)^{-\frac{W(u)}u}=z^{-z}$$
with $z=W(u)/u$. We can solve for $z$ and finally find
$$z=-\frac{\log a}{W(-\log a)}.$$
which can be used to find $u$ via $u=-\log (z)/z$. The final solution might look something like this:

Solution. $$
x
=\frac{u}{W(u)}
=\frac{-\frac{\log z}{z}}{W(-\frac{\log z}{z})}
=\frac{\frac{\log \left(-\frac{\log a}{W(-\log a)}\right)}{\frac{\log a}{W(-\log a)}}}{W\left(\frac{\log \left(-\frac{\log a}{W(-\log a)}\right)}{\frac{\log a}{W(-\log a)}}\right)}
$$

Of course, this monstrous formula should never be used. Instead use the resubstituation like this:
$$z=-\frac{\log a}{W(-\log a)} \quad\to\quad u=-\frac{\log (z)}{z} \quad\to\quad x=\frac{u}{W(u)}.$$
Example. Choosing $a=1/2$ and above formula gave me the solution $x\approx 0.641186$ which indeed solves $a^{a^{x}}=x$.
I also tested it with $a=2$, which yielded a complex solution $x\approx 0.824679 - 1.56743i$ which worked, but shed no light on whether there are any other real ones.
Gottfried mentioned in the comments that it seems not to work for e.g. $a=0.01$. This is why more investigations are necessary.

Gottfried hinted me to the fact that this can be simplified (at least for some $a$) to the function 
$$x(a)=\exp(-W(-\log(a))).$$
This seems to work for all $a$ (in contrast to my resubstituation formula above), but still gives only a single solution. Maybe the other branches of $W$ can give other real solutions, but not sure.
