Debate: Which is Greater - e^π or π^e?

[Bipolarity]

Homework Statement

Determine analytically which is greater, $e^{π}$ or $π^{e}$

Homework Equations

The Attempt at a Solution

It is known that 2<e<3.
It is known that 3<π<4.

Thus, $2^{π} < e^{π} < 3^{π}$.
What from there?

BiP

 

[STEMucator]

You have : $2^\pi < e^\pi < 3^\pi$

Your other inequality gives : $3^e < \pi^e < 4^e$

Now, it is clear without much proof that $3^e < 3^\pi$, right?

EDIT : Another easier way is to use logarithms.

 

[Bipolarity]

You have : $2^\pi < e^\pi < 3^\pi$

Your other inequality gives : $3^e < \pi^e < 4^e$

Now, it is clear without much proof that $3^e < 3^\pi$, right?

EDIT : Another easier way is to use logarithms.

I just realized that there is not sufficient information in the problem to do the proof.

BiP

 

[STEMucator]

I just realized that there is not sufficient information in the problem to do the proof.

BiP

Of course there is, take the easy route by taking logarithms.

Start by assuming that $e^\pi > \pi^e$ and then show its either true or false.

EDIT : You could also assume $e^\pi < \pi^e$ it works either way.

 

[Bipolarity]

Of course there is, take the easy route by taking logarithms.

Start by assuming that $e^\pi > \pi^e$ and then show its either true or false.

Ooo I solved it, I ended up with $e < \frac{\pi}{ln(\pi)}$ which I then proved.
Thanks dude.

BiP

 

[Bipolarity]

Actually Zondrina, now that I think about it, I made a mistake. I was not yet able to correctly prove that $e < \frac{\pi}{ln(\pi)}$

BiP

 

[STEMucator]

Something obvious just hit me, I'm sorry I didn't mention this earlier.

Start by assuming : $e^\pi > \pi^e$

Now let x = pi so : $e^x > x^e$

Now after taking logarithms, our problem reduces to proving : $\frac{x}{ln(x)} > e$

Now, suppose that $f(x) = \frac{x}{ln(x)}$ where x is still equal to π. How can you use this to show f(x) > e?

 

[Bipolarity]

Something obvious just hit me, I'm sorry I didn't mention this earlier.

Start by assuming : $e^\pi > \pi^e$

Now let x = pi so : $e^x > x^e$

Now after taking logarithms, our problem reduces to proving : $\frac{x}{ln(x)} > e$

Now, suppose that $f(x) = \frac{x}{ln(x)}$ where x is still equal to π. How can you use this to show f(x) > e?

I don't know, perhaps some type of MVDT?

BiP

 

[STEMucator]

Take a derivative and start analyzing things.