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

• February 4th 2010, 10:18 PM
ha11
Prove (x^1/2)>ln x for x>1
How can prove this
(x^1/2)>ln x for x>1
• February 4th 2010, 10:58 PM
VonNemo19
Quote:

Originally Posted by ha11
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 $x>1$
• February 5th 2010, 04:37 AM
Krizalid
You'll need to replace that $\implies$ by an $\iff$ in order your proof work. :D
• February 5th 2010, 10:13 AM
ha11
thanks, but I want the prove