How can prove this

(x^1/2)>ln x for x>1

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- Feb 4th 2010, 10:18 PMha11Prove (x^1/2)>ln x for x>1
How can prove this

(x^1/2)>ln x for x>1 - Feb 4th 2010, 10:58 PMVonNemo19
- Feb 5th 2010, 04:37 AMKrizalid
You'll need to replace that $\displaystyle \implies$ by an $\displaystyle \iff$ in order your proof work. :D

- Feb 5th 2010, 10:13 AMha11
thanks, but I want the prove