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Which increases faster e^x or x^e ?

  1. Mar 26, 2012 #1
    1. The problem statement, all variables and given/known data
    which increases faster e^x or x^e ?


    2. Relevant equations



    3. The attempt at a solution

    My attempt was taking the log of both, assuming it doesnt change anything (is this assumption correct?)

    x*ln(e) ------------------------ e*ln(x)

    now I took the derivative

    1 ------------------------ e/x

    and I said that if
    0<x<e, x^e increases faster
    otherwise, e^x is faster

    Is my logic correct?

    Thanks
     
  2. jcsd
  3. Mar 26, 2012 #2
    No, you can't just take the log of both. You aren't dealing with an equality you are comparing two equations. For specific values, your procedure of deriving both and comparing when one is greater works fine, but you have to derive the original equations.

    In general though, exponential growth (blank^x) grows faster than x^blank (whatever that is called), and you can prove that with limits.
     
  4. Mar 26, 2012 #3
    Thanks for the reply.

    How can you prove it with limits?
    lim x-> inf
    and then lim x-> -inf
    ?
     
  5. Mar 26, 2012 #4
    Or simply consider derivatives:
    d/dx (e^x) = e^x and d/dx(x^e)=ex^(e-1)
    These tell you how fast each function is increasing at each x, hence you can work out which function is increasing faster at each x.
     
  6. Mar 26, 2012 #5
    But you are using the fact:

    "exponential growth (blank^x) grows faster than x^blank (whatever that is called)"

    I want to prove it.
    how do you see that
    e^x grows faster then ex^(e-1) ?
     
  7. Mar 26, 2012 #6
    You have to consider the limit of the ratio of the two equations as x-->inf. Using l'hopital's rule, you can see the difference in the rates of change.
     
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