# E and π

1. Nov 9, 2012

### Bipolarity

1. The problem statement, all variables and given/known data
Determine analytically which is greater, $e^{π}$ or $π^{e}$

2. Relevant equations

3. 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

2. Nov 9, 2012

### Zondrina

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.

Last edited: Nov 9, 2012
3. Nov 9, 2012

### Bipolarity

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

BiP

4. Nov 9, 2012

### Zondrina

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.

5. Nov 9, 2012

### Bipolarity

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

BiP

6. Nov 9, 2012

### 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

7. Nov 9, 2012

### Zondrina

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?

8. Nov 9, 2012

### Bipolarity

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

BiP

9. Nov 9, 2012

### Zondrina

Take a derivative and start analyzing things.