Is it possible to determine which of these numbers is greater: $\sqrt{2}^{\sqrt{3}}$ or $\sqrt{3}^{\sqrt{2}}$ without approximating anything?

Question

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?

Answer 1

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.

Answer 2

Let $f(x)=(\ln x)/x$. Calculus tells us this increases for $x<e$ and decreases for $x>e$. Therefore $f(\sqrt2)<f(\sqrt3)$ and so $\sqrt3\ln\sqrt2<\sqrt2\ln\sqrt3$. Exponentiating: $\sqrt2^{\sqrt3}<\sqrt3^{\sqrt2}$.

Answer 3

$$(\sqrt3^\sqrt2)^\sqrt3>(\sqrt3^\sqrt2)^\sqrt2=\sqrt9>\sqrt8=(\sqrt2^\sqrt3)^\sqrt3$$ now using $a^x>b^x$ implies $a>b$ we conclude that $\sqrt3^\sqrt2$ is bigger.