Is it possible to determine which of these numbers is greater: $\sqrt{2}^{\sqrt{3}}$ or $\sqrt{3}^{\sqrt{2}}$ without approximating anything? Is it possible to determine which of these numbers is greater: $\sqrt{2}^{\sqrt{3}}$ or $\sqrt{3}^{\sqrt{2}}$?
I tried to express both in terms of powers of 3, but in the first number I got $log_32 $ as part of the product of the overall power of two, which - again - makes things hard. 
Is there a simple way to do this?

We compare
$$\sqrt{2}^\sqrt{3} \text{ vs. } \sqrt{3}^\sqrt{2}$$
or (taking natural logarithms)
$$(\sqrt{3}/2) \ln 2 \text{ vs. } (\sqrt{2}/2) \ln 3$$ or (multiplying by $2$) $$\sqrt{3} \ln 2 \text{ vs. } \sqrt{2} \ln 3$$ or (rearranging) $$\sqrt{3/2} \text{ vs. } (\ln3)/(\ln2) = \log_2 3.$$
Now $2^{\sqrt{3/2}} < 2^{3/2} = 2 \sqrt{2} < 3$, because $(2\sqrt{2})^2 = 8$ whereas $3^2 = 9$. Thus $\sqrt{3/2} < \log_2 3$ and we conclude $\sqrt{2}^{\sqrt{3}} < \sqrt{3}^{\sqrt{2}}$, chasing back the comparisons. 
