When does $(x^x)^x=x^{(x^x)}$ in Real numbers? I have tried to solve this equation:$(x^x)^x=x^{(x^x)}$ in real numbers I  
got only $x=1,x=-1,x=2$ , are there others solutions ?
Note: $x$ is real number .
Thank you for your help .
 A: Just a sketch for $x>0$:
Let $y=x^x$ and rewrite $x^y=y^x$ by taking $log$'s as $\log (x)/x = \log(y)/y$.
For each $x \in (1, e)$ there is a unique $y(x)$ such that $f(x) = f(y(x))$.
As a function of $y(x)$, you can show that it is increasing. So we are looking for solutions of $y(x) = x^x$. Because $g(x) = y(x)-x^x$ is increasing, it can have only one root. So there is only one positive $x$ that satisfies the equation with $x\neq 1$.
EDIT: For integer $x < 0$ non-integer, things get complicated because we will involve complex numbers and raising numbers to complex powers. 
A: Hint:
for $x>1$
$$x^{x^2} = x^{x^x}$$
$$x^2=x^x$$
$$2\log x=x\log x$$
$$x=2$$
A: First suppose $x>0$. Taking logs,
$$ x \log{x^x} = x^x \log{x}. $$
Applying the property $\log{a^b}=b\log{a}$ gives
$$ x^2 \log{x} = x^x \log{x}. $$
So at this poing either $\log{x}=0$ (i.e. $x=1$) or
$$ x^2=x^x. $$
Taking logs again,
$$ 2\log{x} = x\log{x}, $$
and we already have established that $\log{x} \neq 0$, so $x=2$.
Right, now let's think about $x<0$. We have
$$ (x^x)^x = x^{x^2}, $$
and $x^2>0$. Dividing gives
$$ 1 = x^{x^x-x^2} $$
Okay, now let's cheat and write $x=e^{i\pi} r$, $r>0$. Then we have
$$ 1 = (re^{i\pi})^{(re^{i\pi})^{re^{i\pi}}-r^2}. $$
Since $r>0$, we can at least pull it out of things...
$$ 1 = (re^{i\pi})^{r^{-r}e^{-i\pi r}-r^2} = r^{r^{-r}e^{-i\pi r}}r^{-r^2}e^{i\pi (r^{-r}e^{-i\pi r})}e^{-i\pi r^2} = e^{r^{-r}(\cos{\pi r}-i\sin{\pi r})\log{r}} r^{-r^2} e^{\pi r^{-r}(i\cos{\pi r}+\sin{\pi r})} e^{-i\pi r^2} $$
Okay, let's get an absolute value.
$$ 1 = e^{r^{-r}\cos{\pi r}\log{r}} r^{-r^2} e^{\pi r^{-r}\sin{\pi r}} $$
We can happily take the log of this:
$$ 0 = (r^{-r}\cos{\pi r}-r^2)\log{r}+\pi r^{-r}\sin{\pi r} = r^{-r}(\cos{\pi r}\log{r}+\pi\sin{\pi r})-r^2\log{r} $$
The dominant term on the right-hand side is clearly $-r^2\log{r}$, and therefore we can see that all the roots will have small $r$ (suffices that $r^2\log{r}>\log{r}+\pi$ for there to be no subsequent roots, for example—and $3$ is quite sufficient). $r=1$ is clearly a root. However, it is clear that the right-hand side tends to $-\infty$ for $r \to \infty$, and more careful examination shows that it also tends to $-\infty$ for $r \to 0$. The function is analytic, so there should only be finitely many roots; in particular, the behaviour suggests an even number. Okay, so that hasn't got us far: let's look at the argument.
$$ e^{-ir^{-r}\sin{\pi r}\log{r}} e^{i\pi r^{-r}\cos{\pi r}} e^{-i\pi r^2} = e^{i(\pi r^{-r}\cos{\pi r}-r^{-r}\sin{\pi r}\log{r}-\pi r^2)} $$
The factor in the exponential needs to be a multiple of $2\pi i$.
Okay, at this point I give up analysis and start drawing pictures.
Log of the absolute value. Obviously only two roots, one at $r=1$.

Argument, with lines of $2k\pi$. Obviously the other root in the former diagram does not have the right argument. Hence $r=1$ (i.e. $x=-1$) is the only other root.
A: You have $(x^x)^x = x^{x^2}$. Assuming $x \not \in \{-1,1,1\}$,
$$x^{x^2} = x^{x^x}$$
happens if and only if the exponents agree, i.e. $x^2 = x^x$ which under our assumptions is equivalent to $x=2$.
