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If $\displaystyle x^{x^9}=\sqrt{3^{\sqrt{3}}}$ and $\displaystyle y=x^{\left(\frac{1}{y^{y^x}}\right)}$, determinate the value of $y^{3x}$.

My try

It is easy to see that if we raise the first equation to $9$ and then we comparate terms, it is possible compute the value of $x$. In fact, I found that $x=3^{1/6}$.

However, I can't manipulate the second equation to find the value of $y$, so I can't proceed further. Any hints are appreciated.

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    $\begingroup$ This is a bit of a silly solution, but the second equation rearranged is $y^{y^{y^x}} = x$. Substituting $x$ back into this infinitely gives $y^{y^{y^{\cdots}}} = x$, or simply $y^x = x$, from which we get $y = \sqrt[x]{x}$, which we may check is indeed a solution. I'm sure someone can come up with a better solution that doesn't use any "hacks". $\endgroup$ – user3002473 Apr 16 at 5:34
  • $\begingroup$ @user3002473. Make an answer of that. It is smart and right. $\endgroup$ – Claude Leibovici Apr 16 at 6:00
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We are going to prove that $\;y^x=x\;.$

Since $\;x=\sqrt[6]3>1\;,\;$ then $\;y=x^{\left(\dfrac{1}{y^{y^x}}\right)}>1\;.$

If $\;y^x\;$ were greater than $\;x\;$, it would follow that

$y^{y^x}>y^x>x\;,\;$ consequently,

$\dfrac x{y^{y^x}}<1\;,\;$ hence,

$x<y^x=x^{\left(\dfrac x{y^{y^x}}\right)}<x\;,$

but it is a contradiction.

Analogously, if $\;y^x\;$ were less than $\;x\;$, we would get another contradiction.

So it proves that $\;y^x=x\;.$

Moreover,

$y^{3x}=x^3=\sqrt3\;.$

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