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

  1. #1
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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; Feb 5th 2010 at 12:35 PM. Reason: Changed post title
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by ha11 View Post
    How can prove this
    (x^1/2)>ln x for x>1
    $\displaystyle x^{1/2}>\ln{x}\Rightarrow{e}^{x^{1/2}}>x$. which is clearly true for all $\displaystyle x>1$
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  3. #3
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    You'll need to replace that $\displaystyle \implies$ by an $\displaystyle \iff$ in order your proof work.
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  4. #4
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    thanks, but I want the prove
    Last edited by ha11; Feb 5th 2010 at 10:44 AM.
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