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

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

How can prove this
(x^1/2)>ln x for x>1

2. 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 $x>1$

3. You'll need to replace that $\implies$ by an $\iff$ in order your proof work.

4. thanks, but I want the prove