I have done the following but I'not satisfied.
if $x^a=a$ then by substitution follows that $x^a=x^{x^a}=x^{x^{x^a}}$ etc.
So $x^{x^{x^{16}}}=16$ is equivalent to $x^{16}=16$, and $x=2^{\frac{1}{4}}$
Substituting $x$ to $x^{x^{x^{12}}}$ returns $2$.
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6$x^{16}=16$ means that $x^{x^{x^{16}}}=16$, but $x^{x^{x^{16}}}=16$ does not mean that $x^{16}=16$. – Kenny Lau Aug 04 '16 at 10:35
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That is my problem – Brani Aug 04 '16 at 10:37
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For $x,y>1$ $x^y>1$ and is increasing in both $x$ and $y$. For $0\leq x \leq 1$, $y\geq 0$ we always have $0\leq x^y\leq 1$. It follows that your solution is unique. – H. H. Rugh Aug 04 '16 at 11:08
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It's obvious that if $x=\sqrt[16]{16}$, then $x^{16} = 16$, and $x^{x^{16}} =16$ and $x^{x^{x^{16}}} = 16$, so $x=\sqrt[16]{16}$ solves your equation.
So then $x^{12} = {16^{0.75}} = 8$, and $x^{x^{12}} = x^8 = \sqrt{16} = 4$, and $x^{x^{x^{12}}} = x^4 = \sqrt[4]{16} = 2$. Hence $x^{x^{x^{12}}} = 2$
Ok hang on, I've got this beautiful approach!
Since $x^{x^{x^{16}}} = 16$, it follows that $x^{x^{x^{x^{x^{x^{16}}}}}} = 16$, and that $x^{x^{x^{x^{x^{x^{x^{x^{x^{16}}}}}}}}} = 16$, and so on , so forth, leading to the equation: $x^{x^{x^{x^{...}}}} = 16$. Now, if we look at $x$ exponentiated by both these equal exponents (I'm avoiding complex and negative values of $x$ then), then we get: $x^{x^{x^{x^{...}}}} = x^{16}$. From the previous two equations we get $x^{16}=16$, which give us $x=\sqrt[16]{16}$.
Sarvesh Ravichandran Iyer
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Don't really need that. For $0\leq x\leq 1$, $0\leq x^y\leq 1$ for any positive $y$ so you can never get a solution smaller than 1 – H. H. Rugh Aug 04 '16 at 11:11
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You are correct in your computations. As others have pointed out, a real solution of $x^{16}=16$ does solve your problem, but it is not immediate that your problem is equivalent to it.
However, you can argue that $f(x)= x^{x^{x^{16}}}$, as a function from $[1,\infty)$ to $[1,\infty)$ is strictly increasing. For instance, by considering the sign of its derivative fo $x\ge1$: $$x^{x^{x^{16}}+x^{16}-1}+16x^{x^{x^{16}}+x^{16}+15}\ln^2 x+x^{x^{x^{16}}+x^{16}+15}\ln x$$
So, any solution of the problem $\begin{cases}x^{x^{x^{16}}}=c\\ x\ge1\end{cases}$ is unique.
You should also observe that, if $0\le x\le 1$, then $f(x)\le 1$; $f$ is still increasing on $[0,1]$, but the $\ln x$ makes it not-as-immediate to prove.
These two observations prove that your solution is actually the only one. I wonder, though, if there is a flashier approach.