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Math Help - Which one is bigger between 2^n and n^logn ?

  1. #1
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    Which one is bigger between 2^n and n^logn ?

    Hi

    Which one is bigger between 2^n and n^logn ?
    How can I know?
    I try to put a log in both sides.
    But, still cannot get an answer..
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  2. #2
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    Re: Which one is bigger between 2^n and n^logn ?

    Since 2^x = e^{x\log(2)} and x^{\log x}= e^{(\log x)^2}, then 2^x \ge x^{\log x} when x\log 2 \ge (\log x)^2.

    Let f(x) = x\log(2) - (\log x)^2. You can easily calculate f(1) = \log(2) > 0. Then, f'(x) = \log(2) - 2\dfrac{\log x}{x}. f'(1) = \log(2). You want to show that f'(x) \ge 0 for all x>1. Suppose f'(x) = 0. Then \log(2) = 2\dfrac{\log x}{x}. This happens when x\log 2 = 2\log x. Since 2^x > x^2 for all x>1, f'(x)>0 for all x>1. So, 2^x>x^{\log x} for all x\ge 1.
    Thanks from yanirose
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  3. #3
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    Re: Which one is bigger between 2^n and n^logn ?

    I deeply appreciate your helps.
    But, may I ask about 2^x > x^2 for all x>1?
    Because when x=1,2 and 3, 2^x <= x^2? For example, when x=3, 8 < 9.
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  4. #4
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    Re: Which one is bigger between 2^n and n^logn ?

    Oh, I wasn't thinking. You are correct that x^2>2^x for 2<x<4. So, f(x) has local extrema at x=2,x=4. Then f(2) = 2\log(2) - (\log 2)^2>0 and f(4) = 4\log(2) - (\log 4)^2>0, so since f(x) increases everywhere else, it must be greater than zero for all x>1.
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