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E and π

  1. Nov 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Determine analytically which is greater, [itex] e^{π} [/itex] or [itex] π^{e} [/itex]


    2. Relevant equations



    3. The attempt at a solution
    It is known that 2<e<3.
    It is known that 3<π<4.

    Thus, [itex] 2^{π} < e^{π} < 3^{π} [/itex].
    What from there?

    BiP
     
  2. jcsd
  3. Nov 9, 2012 #2

    Zondrina

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    You have : [itex]2^\pi < e^\pi < 3^\pi[/itex]

    Your other inequality gives : [itex]3^e < \pi^e < 4^e[/itex]

    Now, it is clear without much proof that [itex]3^e < 3^\pi[/itex], right?

    EDIT : Another easier way is to use logarithms.
     
    Last edited: Nov 9, 2012
  4. Nov 9, 2012 #3
    I just realized that there is not sufficient information in the problem to do the proof.

    BiP
     
  5. Nov 9, 2012 #4

    Zondrina

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

    Start by assuming that [itex]e^\pi > \pi^e[/itex] and then show its either true or false.

    EDIT : You could also assume [itex]e^\pi < \pi^e[/itex] it works either way.
     
  6. Nov 9, 2012 #5
    Ooo I solved it, I ended up with [itex] e < \frac{\pi}{ln(\pi)} [/itex] which I then proved.
    Thanks dude.

    BiP
     
  7. Nov 9, 2012 #6
    Actually Zondrina, now that I think about it, I made a mistake. I was not yet able to correctly prove that [itex] e < \frac{\pi}{ln(\pi)} [/itex]

    BiP
     
  8. Nov 9, 2012 #7

    Zondrina

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

    Start by assuming : [itex]e^\pi > \pi^e[/itex]

    Now let x = pi so : [itex]e^x > x^e[/itex]

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

    Now, suppose that [itex]f(x) = \frac{x}{ln(x)}[/itex] where x is still equal to π. How can you use this to show f(x) > e?
     
  9. Nov 9, 2012 #8
    I don't know, perhaps some type of MVDT?

    BiP
     
  10. Nov 9, 2012 #9

    Zondrina

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    Take a derivative and start analyzing things.
     
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