Given $\log_2(\log_3x)=\log_3(\log_4y)=\log_4(\log_2z)$, find $x+y+z$. It seems so messy. I have no idea where to start. Can anyone share any ideas?
I found $x=3^{2^m}$, $y=4^{3^m}$ and $z=2^{4^m}$, and then I stopped again...
Sorry guys I misread the question... It should have been $\log_2(\log_3x)=\log_3(\log_4y)=\log_4(\log_2z)=0$... I am sorry...
 A: You have 2 equations over 3 variables, So there is no unique solution.
Although you may think, anyway $x+y+z$ can be constant, but:
$$\log_2 (\log_3 x) = \log_3 (\log_4 y) = \log_4 (\log_2 z) = t$$
$$\Longrightarrow x+y+z = 3^{2^t} + 4^{3^t} + 2^{4^t}$$
It's easy to see the right hand side can get any value in $(3,+\infty)$.
Maybe you must find its minimum, that is $3$.
A: Remember $\log_{b}(x)=y$ means that $b^{y}=x$.
So let $\log_2(\log_3x)=\log_3(\log_4y)=\log_4(\log_2z)=a$ then:

*

*$\log_2(\log_3x)=a\implies\log_3(x)=2^{a}\implies x=3^{2^{a}}$

*$\log_3(\log_4y)=a\implies\log_{4}(y)=3^{2}\implies y=4^{3^{a}}$

*$\log_4(\log_2z)=a\implies\log_{2}(z)=4^{a}\implies z=2^{3^a}.$
Thus $x+y+z=3^{2^{a}}+4^{3^{a}}+2^{3^{a}}$ and there are infinitely many solutions depending on $a.$
Taking $a=0$ we have $x+y+z=3+4+2=9$, , for $a=1$ you get $x+y+z=81$ for $a=2$ we have $x+y+z=262737$, etc.
A: The hint:
Let $$\log_2\log_3x=\log_3\log_4y=\log_4\log_2z=0.$$
So $$x+y+z=9.$$
Let $$\log_2\log_3x=\log_3\log_4y=\log_4\log_2z=1.$$
So $$x+y+z\neq9.$$
