Given $log_2(log_3x)=log_3(log_4y)=log_4(log_2z)$, find $x+y+z$.

Question

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...

Äres · Accepted Answer

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:

$log_2(log_3x)=aimplieslog_3(x)=2^{a}implies x=3^{2^{a}}$
$log_3(log_4y)=aimplieslog_{4}(y)=3^{2}implies y=4^{3^{a}}$
$log_4(log_2z)=aimplieslog_{2}(z)=4^{a}implies z=2^{3^a}.$

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.

Ali Ashja' · Answer

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 $3$.

Michael Rozenberg · Answer

The hint:
Let $$log_2log_3x=log_3log_4y=log_4log_2z=0.$$
So $$x+y+z=9.$$
Let $$log_2log_3x=log_3log_4y=log_4log_2z=1.$$
So $$x+y+zneq9.$$

Given $log_2(log_3x)=log_3(log_4y)=log_4(log_2z)$, find $x+y+z$.

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