$2^\sqrt{10}$ 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 $10$...
 A: $$2^{\sqrt{10}}=8^{\sqrt{1+\frac{1}{9}}}<8^{1+\frac{1}{18}}=2^{\frac{19}{6}}<9.$$
Because $$2^{19}=2(500+12)^2=524288<531441=(700+29)^2=3^{12}.$$
There is another way to show a last inequality, but it's not so nice:
$$2^{19}<3^{12}$$ it's
$$\left(1+\frac{1}{8}\right)^6>2,$$ which is true because
$$\left(1+\frac{1}{8}\right)^6>1+\frac{6}{8}+\frac{15}{64}+\frac{20}{512}>2.$$
A: It follows from the maclaurian series of the function $\sqrt{k^2+h}$ for $|h|\leq k$ that ,
$$\sqrt{k^2+h}<k+\frac{h}{2k}$$
Put $k=3$ and $h=1$ we get ,
$$\sqrt{10}<3+\frac{1}{2\cdot 3}=3+\frac{1}{6}$$
Hence we must have
$$2^{\sqrt{10}}<2^3\cdot \sqrt[6]{2}=8\cdot \sqrt[6]{2}$$
Therefore we only need to show that
$$8\cdot \sqrt[6]{2}<3^2$$
Or in other words we need to show that
$$\sqrt[6]{2}<\frac{9}{8}=1+\frac{1}{8}$$
or in other words if we raise both sides to $6th$ power ,we need to show that
$$2<\left( \frac{9}{8}\right)^6=\left(1+\frac{1}{8}\right)^6$$
Expanding right hand side using binomial theorem we're left to prove that
$$2<\sum \limits_{k=0}^6 {}^6C_k \cdot \frac{1}{8^k}$$
As the first term in the expansion on right hand side is obviously 1 therefore subtracting 1 both sides leaves us to prove that
$$1<\sum \limits_{k=1}^6 {}^6C_k \cdot \frac{1}{8^k}$$
Now it's working time we don't need to know the full expansion. We'll just leave where the rhs exceeds 1.
Now ,$${}^6C_1\cdot \frac{1}{8}=\frac{6}{8}=\frac{3}{4}$$
$${}^6C_2\cdot \frac{1}{8^2}=\frac{30}{128}=\frac{15}{64}$$
$${}^6C_3 \cdot \frac{1}{8^3}=\frac{20}{512}=\frac{5}{128}$$
let's check now whether this exceeds 1 or not. Adding we get
$$\frac{3}{4}+\frac{15}{64}+\frac{5}{128}=\frac{63}{64}+\frac{5}{128}=\frac{131}{128}=\frac{128+3}{128}=1+\frac{3}{128}>1$$
(Yay!! ;) ) Therefore the inequality is proven. Hence we have
$$2^{\sqrt{10}}<3^2$$
A: A trick is to blow up the gap
$[2^\sqrt{10},\;3^2] → [2^{\sqrt{10}-3},\;9/8] → [2,\;(9/8)^{\sqrt{10}+3}]$
$1.26^3 = 2.000376$
$(9/8)^2 = 81/64 = 1+1/4+1/16 > 1.26$
$(9/8)^{\sqrt{10}+3} > (9/8)^6 > 1.26^3 > 2\qquad\qquad$ ⇒ RHS is bigger
A: 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 $2^{\sqrt{10}} < 2^{19/6} < 9$.
