$\newcommand{\bigxl}[1]{\mathopen{\displaystyle#1}} \newcommand{\bigxr}[1]{\mathclose{\displaystyle#1}} $ $$\large e^{\bigxl(\pi^{(e^\pi)}\bigxr)}\quad\text{or}\quad\pi^{\bigxl(e^{(\pi^e)}\bigxr)}$$ Which one is greater?

Effort. I know that $$e^\pi\ge \pi^e$$

Then $$\pi^{(e^\pi)}\ge e^{(\pi^e)}$$

But I can't say $$e^{\bigxl(\pi^{(e^\pi)}\bigxr)}\le \pi^{\bigxl(e^{(\pi^e)}\bigxr)}$$


$$e^{\bigxl(\pi^{(e^\pi)}\bigxr)}\ge \pi^{\bigxl(e^{(\pi^e)}\bigxr)}$$

  • $\begingroup$ Roughly speaking, you want big numbers as high in the stack as possible, so the first should be larger. Others have proven this. $\endgroup$ – Ross Millikan Jul 13 '16 at 21:15
  • $\begingroup$ You should probably change all $ \geq$ to $>$, because there is obviously no equality here $\endgroup$ – Yuriy S Jul 13 '16 at 22:05

We use the following fact in the proof:

Let $c > 0$. Then $\ln(x + c) < \ln(x) + c$ for $x \geq 1$.

For notational convenience, we use the notation $f(x) \rightarrow g(x)$ to denote that $g(x) = \ln f(x)$. We have

$$ e^{\pi^{e^\pi}} \to \pi^{e^\pi} \rightarrow e^\pi\ln \pi \to \pi + \ln\ln \pi $$ and $$ \pi^{e^{\color{red}{\pi^e}}} < \pi^{e^{\color{red}{e^\pi}}} \to e^{e^\pi}\ln \pi \to e^\pi + \ln\ln \pi \to \ln(e^\pi + \ln\ln\pi) < \ln(e^\pi) + \ln\ln\pi = \pi + \ln\ln\pi $$ Therefore, $$ e^{\pi^{e^\pi}} > \pi^{e^{\pi^e}} $$


Starting from $\pi^e\lt e^{\pi}$, we have, by taking the logarithm twice and doing a trivial bit of algebra,

$$\pi^e\lt e^{\pi}\implies e\ln\pi\lt\pi\implies1+\ln\ln\pi\lt\ln\pi\implies\ln\ln\pi\lt\ln\pi-1$$

We'll use the two ends of the above in the following, which begins by taking a logarithm, then does some trivial algebra, and ends by exponentiating twice:

$$\begin{align} e\lt\pi&\implies1\lt\ln\pi\\ &\implies e^{\pi}-1\lt(e^{\pi}-1)\ln\pi\\ &\implies e^{\pi}+\ln\pi-1\lt e^{\pi}\ln\pi\\ &\implies\pi^e+\ln\ln\pi\lt e^{\pi}\ln\pi\quad\text{(using }\pi^e\lt e^{\pi}\text{ and }\ln\ln\pi\lt\ln\pi-1)\\ &\implies e^{\pi^e}\ln\pi\lt\pi^{e^{\pi}}\\ &\implies\pi^{e^{\pi^e}}\lt e^{\pi^{e^{\pi}}} \end{align}$$


The function $x^{\frac{1}{x}}$ is strictly decreasing for $x>e$, the maximum is $e^{\frac{1}{e}}$.


$e\le a<b$ => $a^{\frac{1}{a}}>b^{\frac{1}{b}}$ => $a^b>b^a$ => $a^b-1>b^a-1$ => $(a^b-1)\ln b>(b^a-1)\ln a$

=> $b^{a^b-1}>a^{b^a-1}$ => $b^{a^b-1}\frac{\ln a}{a}>a^{b^a-1}\frac{\ln b}{b}$ => $a^{b^{a^b}}>b^{a^{b^a}}$

Here: $a:=e$ and $b:=\pi$

  • $\begingroup$ This is the answer I was looking for! Generalized version, without limiting oneself to $e$ and $\pi$. Thank you :-) $\endgroup$ – Kusavil Jan 13 '18 at 21:48
  • $\begingroup$ @Kusavil : I am glad that it helps you! :-) $\endgroup$ – user90369 Jan 14 '18 at 12:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.