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Thread: Prove (x^1/2)>ln x for x>1

  1. February 4th 2010, 10:18 PM #1
    ha11
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    Prove (x^1/2)>ln x for x>1

    How can prove this
    (x^1/2)>ln x for x>1
    Last edited by mr fantastic; February 5th 2010 at 12:35 PM. Reason: Changed post title
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  2. February 4th 2010, 10:58 PM #2
    VonNemo19
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    Quote Originally Posted by ha11 View Post
    How can prove this
    (x^1/2)>ln x for x>1
    x^{1/2}>\ln{x}\Rightarrow{e}^{x^{1/2}}>x. which is clearly true for all x>1
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  3. February 5th 2010, 04:37 AM #3
    Krizalid
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    You'll need to replace that \implies by an \iff in order your proof work.
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  4. February 5th 2010, 10:13 AM #4
    ha11
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    thanks, but I want the prove
    Last edited by ha11; February 5th 2010 at 10:44 AM.
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