$2^\sqrt{10}$ vs $3^2$ Is there a neat way to show that $2^\sqrt{10} < 3^2$?
I have tried raising to larger powers, like $(2^\sqrt{10})^{100}$ vs $3^{200}$ but the problem is the two functions $2^{x\sqrt{10}}$ and $3^{2x}$ are almost equivalent, and there is no point (that I can find) where one function is "obviously" larger than the other.
Looking at $2^{2\sqrt{10}}$ vs $3^4$  I tried to find a way of showing that $2^{2\sqrt{10}} < 2^6+2^4 = 2^4(2^2+1)$ but couldn't see any neat solution.
Any help or hints are appreciated.
edit: I should have specified when I say "neat solution" I was looking for a method readily done by hand. I realise this might be an unrealistic limitation, but it was why I'm interested.
Final thoughts before going to bed: I was looking at the functions $f(x)=2^{\sqrt{1+x^2}}$ and $g(x)=x^2$. At $x=0$, clearly $f>g$. They are equal at $x=2\sqrt{2}$. At $x=4$, again $f>g$. This mean that somewhere in the interval $2<x<4$, $g>f$ (as they are both convex). The task then is to try to find a point such that $x>3$ and $g>f$. But then another fun inequality pops out... $2^{\sqrt{11}}$ vs $10$...

By AM-GM, $\frac{3 + 10/3}{2} > \sqrt{10} \implies \frac{19}{6} > \sqrt{10}$. Thus:
$$2^{19/6} \ ? \ 9 \iff 8 \cdot 2^{1/6} \ ? \ 9 \iff 2 \ ?  \ (1 + 1/8)^6$$
$$\iff 2 \ ? \ 1 + 6/8 + 15/64 + 20/512 + \text{other positive terms}$$
$$\iff 2 < 1 + 96/128 + 30/128 + 5/128$$
hence $2^{\sqrt{10}} < 2^{19/6} < 9$.
