Mathematics Asked by Hongji Zhu on November 27, 2020
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…
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:
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.
Correct answer by Äres on November 27, 2020
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 $3$.
Answered by Ali Ashja' on November 27, 2020
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 $$x+y+zneq9.$$
Answered by Michael Rozenberg on November 27, 2020
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