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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Originally Posted by ha11 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 $\displaystyle x>1$
You'll need to replace that $\displaystyle \implies$ by an $\displaystyle \iff$ in order your proof work.
thanks, but I want the prove
Last edited by ha11; Feb 5th 2010 at 10:44 AM.
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