# Thread: Which is greater: e^π or π^e?

1. ## Which is greater: e^π or π^e?

My Math 324 professor said we should try this for fun. Can someone explain how to come up with the answer? I thought maybe do a Taylor approximation since e^x is ∑ x^n / n!, but then I got something divergent.

Please contribute.

2. Originally Posted by Jamin2112
My Math 324 professor said we should try this for fun. Can someone explain how to come up with the answer? I thought maybe do a Taylor approximation since e^x is ∑ x^n / n!, but then I got something divergent.

Please contribute.
Could you write the instructions exactly as they were given please? I'm having trouble understanding what the problem is.

And BTW, the taylor series for e^x converges for all x.

3. This is very interesting, do you have to provide some sort of proof?

4. $\log(e^n)=n\log(e)$

$\log(n^e)=e\log(n)$

What do you think?

5. the answer is $e^x$ is greater

let $f(x) = x - e\ln(x)$

we have $f(x) \to +\infty$ , when $x \to 0$

and

$\lim_{x\to+\infty} x - e\ln(x) = \lim_{x\to +\infty} x ( 1 - \frac{e\ln(x)}{x} ) = +\infty$

consider

$f'(x) = 1 - \frac{e}{x}$

we can see when $x = e$ , $f'(x) = 0$

its second derivative gives $f''(e) = \frac{1}{e} > 0$

means $f(x)$ has only one local min.

come back to $f(e) = e - e\ln(e) = e-e = 0$

from the above calculations .

we can see $f(x) \geq 0$

$\implies x \geq e \ln(x) \implies e^x \geq x^e$

equality holds when $x = e$ (obviously )

ps . add a constraint $x > 0$

6. consider the function $f(x)=\frac{\ln x}x,$ which is strictly decreasing for $x>e.$

since $\pi>e,$ we get $\frac{\ln\pi}\pi<\frac{\ln e}e\implies e\ln\pi<\pi\ln e,$ and that yields $\ln\pi^e<\ln e^\pi,$ and finally since $g(x)=e^x$ is a strictly increasing function, then $e^{\ln \pi^e} and we're done.

7. Originally Posted by Krizalid
consider the function $f(x)=\frac{\ln x}x,$ which is strictly decreasing for $x>e.$

since $\pi>e,$ we get $\frac{\ln\pi}\pi<\frac{\ln e}e\implies e\ln\pi<\pi\ln e,$ and that yields $\ln\pi^e<\ln e^\pi,$ and finally since $g(x)=e^x$ is a strictly increasing function, then $e^{\ln \pi^e} and we're done.
Thanks, dude! I went through all your steps and now I understand it!