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I know there are nice ways to compare $x^y$, $y^x$ such as here. And for multiple exponential, I think the problem becomes hard. For instance, I want to determine which of $x^{y^z}$ and $z^{y^x}$ is greater without computation. Applying logarithms to both side, we get \begin{align*} z\ln y+\ln\ln x,\quad x\ln y+\ln\ln z \end{align*} and I can't obtain useful information. Is there any nice way to compare multiple exponential easily?

suww
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  • What do you mean by compare?Are you familiar with the rate of growth of functions and computing derivatives? – Bumblebee Jan 04 '22 at 06:11
  • @Ajay which of $x^{y^z}$, $z^{y^x}$ is greater in general setting. – suww Jan 04 '22 at 06:16
  • Related (not intended to answer this question): https://math.stackexchange.com/questions/7892/comparing-pie-and-e-pi-without-calculating-them – Peter O. Jan 04 '22 at 14:52
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    What do you mean with "without computation"? Whatever you are going to do to compare these quantities will require some form of computation one way or another. – Hyperplane Jan 05 '22 at 13:15

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I don't think this will really generalize but here is my attempt-

We have, $ \sqrt{2}^{e^\pi}$ and $ \pi^{e^\sqrt{2}}$. Since we can take the $(e^{\sqrt{2}})$th root on both, this is equivalent to comparing $\sqrt{2}^{e^{\pi-\sqrt{2}}}$ and $\pi$. Further, we can take the $(e^{\pi})$th root on both sides to give $\sqrt{2}^{e^{-\sqrt{2}}}$ and $\pi^{e^{-\pi}}$. (We could do this because $f(x)=x^{\frac{1}{e^\pi}}$ and $f(x)=x^{\frac{1}{e^{\sqrt{2}}}}$ are both increasing functions)

Now consider the function $f(x)=x^{e^{-x}}$. The two final expressions we obtained are just this evaluated at $\sqrt{2}$ and $\pi$. It would be easy to compare now even directly, just by the standard methods of calculus if computationally annoying (thus defeating the entire purpose), but we can alternatively compare $\log_π{f(x)}$ as well, since $g(z)=\log_π(z)$ is an increasing function.

Amadeus
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