MHB Comparing $e^\pi$ and $\pi ^e$ Without Computing

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The discussion centers on comparing the values of $e^\pi$ and $\pi^e$ without computational tools. A participant explains that by analyzing the function $x^{1/x}$, which reaches its maximum at $x = e$, it can be concluded that $e^{1/e} > \pi^{1/\pi}$. Consequently, raising both sides to the power of $\pi e$ leads to the conclusion that $e^\pi > \pi^e$. The thread highlights the challenge of solving such problems without immediate answers, emphasizing the enjoyment of mathematical exploration. Overall, the consensus is that $e^\pi$ is indeed greater than $\pi^e$.
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Without using any computing devices, show which number is larger: $e^\pi$ or $\pi ^e$.
 
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Hint:
Consider the function $y=\frac{x}{\ln(x)}$.
 
Last edited:
eddybob123 said:
So does no one know or no one bothers to post their solution?...

Typically, you should not expect a solution to be posted so quickly; we ask in our http://mathhelpboards.com/challenge-questions-puzzles-28/guidelines-posting-answering-challenging-problem-puzzle-3875.html that you give our members about a week to respond. Sometimes you will get a quick response, but sometimes not.

In my case, I have seen this problem before along with its solution, so I felt it would only be fair to leave it for the enjoyment of someone who has not seen it before.

The vast majority of problems posted as challenges here are solved, but most people are not online all the time, and so it may be a while before someone comes along who will solve the problem and post their solution. (Sun)
 
(Headbang)
 
Last edited:
eddybob123 said:
So does no one know or no one bothers to post their solution?

I haven't even had my breakfast yet. :p

[sp]Using your (much needed) hint.

$$\frac{dy}{dx}=\frac{\ln x -1}{(\ln x )^2}$$

So $$\frac{dy}{dx}=0 \text{ when } x=e \text{ and } \frac{dy}{dx}>0 \text{ when } x>e$$

$$y=e$$ when $$x=e$$ and $$y>e$$ when $$x>e$$.

So when $$x=\pi$$ we have $$\frac{\pi}{\ln \pi}>e \Rightarrow \pi>e \ln \pi \Rightarrow e^{\pi}> \pi^ e $$

[/sp]

Actually, I haven't even done breakfast for my children.
 
(Clapping)

(Dance)(Dance)(Dance)(Dance)(Dance)(Dance)(Dance)(Dance)(Dance)
 
I have shown in my blog at Fun with maths
that

x^(1/x) is maximum at x = e so e^(1/e) > π^(1/π)

and hence e^π > π^e after raising both sides to power πe
 

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