# which is bigger $\log\log x$, $\sqrt x$ , $(\log x)^2$ [closed]

I need to compare these 3 and rank them based on which is bigger:

• $\log \log x$
• $\sqrt x$
• $(\log x)^2$

As $x\to\infty$

• For which $x$? As $x \to \infty$? Oct 18, 2013 at 11:17
• How stuck are you? These functions may look complicated, but they're actually pretty easy to visualize what order they go in, though actually proving it takes a little more effort. Do you have a guess? Oct 18, 2013 at 11:20
• I think square root is bigger, than log x squared, than log logx ... but I need to show it mathematically. Oct 18, 2013 at 11:27
• Go to wolframalpha.com and type in "please Plot[{Log[Log[x]],Sqrt[x],Log[x]^2},{x,0,6}]" Oct 18, 2013 at 11:27
• thanks #Nick kidman but I need to show it mathematically. Oct 18, 2013 at 11:28

Say $$\log { \log { x } } =y,$$
Take the $\log$ of the equations:$$\log { y } ,\quad (1)\\ \frac { 1 }{ 2 } { e }^{ y },\quad (2)\\ 2y,\quad (3).$$
Now it is clear that $(2)>(3)>(1)$ for $y\rightarrow \infty$
• $\log \log x$ is always less than the other 2.
• $\sqrt x$ is only smaller than $(\log x)^2$ for the small interval between the 2 points where they meet, when $\sqrt x = (\log x)^2$
For every $x>1$, you can find a $t$ such that $x:=\mathrm {exp}(\mathrm {exp}(t)))^2$.
Plug this reparameterization into you functions and you find a clear hierarchy: The first grows linearly in $t$, the second one doubly exponentially and the third only exponentially. As $t$ grows bigger (resp. $x$, because the functionality is monotonally increasing), at one point the doubly exponential function only even takes values which are bigger than that of the simple exponential function and the same goes for the even slower linear one.