What is the minimum value of $a$ such that $x^a \geq \ln(x)$ for all $x > 0$?

This is probably just elementary, but I don't know how to do it. I would like to find the minimum value of $a$ such that $x^a \geq \ln(x)$ for all $x > 0$. Numerically, I have found that this minimum value lies between 0.365 and 0.37 (i.e., $x^{0.37} > \ln(x)$ for all $x > 0$, but $x^{0.365}$ is not). Is there any analytical way to find out exactly this minimum value?

EDIT: Based on the received answers, I finally came up with my own one as follows.

Consider the function $f(x) = x^a - \ln(x).$ This function is convex in $x$, and hence, achieves the unique minimum as $x^*$ such that $f'(x^*) = 0.$ Solving that equation yields $$f_{\mathrm{min}} = \min\limits_{x>0} f(x) = \frac{\ln(a)+1}{a}.$$

Now, by letting $f_{min} = 0$, we get the desired value $a^* = 1/e.$

-
$1/e$ is one of those values that you recognize numerically after a while. –  asmeurer Dec 29 '10 at 6:09

$a = \max_{x>0} \frac{\ln(\ln(x))}{\ln(x)} = 1/e$.

-
Got it! Thanks, lhf. –  cresmoon Dec 29 '10 at 1:17
Consider the minimum of the function $f(x)=x^{1/e}-\ln(x)$.
EDIT: As a second step, verify that $x^a < \ln(x)$ at $x=(1/a)^{1/a}$, for any $0<a<1/e$.
This minimum occurs at $e^e \approx 15.1543$ and is 0. –  Ross Millikan Dec 28 '10 at 23:52
Since the minimum is $0$, $x^{1/e} \geq \ln(x)$ for all $x>0$. –  Shai Covo Dec 29 '10 at 0:01