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

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.

• 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". – user3002473 Apr 16 at 5:34
• @user3002473. Make an answer of that. It is smart and right. – Claude Leibovici Apr 16 at 6:00

## 1 Answer

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

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,

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