What is the larger of the two numbers? What is the larger of the two numbers?
$$\sqrt{2}^{\sqrt{3}} \mbox{ or } \sqrt{3}^{\sqrt{2}}\, \, \; ?$$
I solved this, and I think that is an interesting elementary problem. I want different points of view and solutions. Thanks!
 A: $$\sqrt2^{\sqrt 3}<^?\sqrt3^{\sqrt 2}$$
Raise both sides to the power $2\sqrt 2$, and get an equivalent problem:
$$2^{\sqrt 6}<^?9$$
Since $\sqrt 6<3$, we have:
$$2^{\sqrt 6}< 2^3 = 8 <9$$
So ${\sqrt 2}^{\sqrt 3}$ is smaller than $\sqrt3^{\sqrt 2}$.
A: Hint: If $a$ and $b$ are positive numbers, $a^b < b^a$  if and only if $\dfrac{\ln a}{a} < \dfrac{\ln b}{b}$.  Find intervals on which $\dfrac{\ln x}{x}$ is increasing or decreasing.
A: We have $\sqrt{2}>1$ and $\sqrt{3}>1$, so raising either of these to powers $>1$ makes them larger.
Call $x=\sqrt{2}^\sqrt{3}$ and $y=\sqrt{3}^\sqrt{2}$.
We have $x^{2\sqrt{3}}$=8 and $y^{2\sqrt{2}}=9.$
Since $2\sqrt{2} < 2\sqrt{3}$, we conclude $y>x$.
A: Hint: Use the Logarithm function.
A: In general, we can state two pertinent results:  (1)  If $a$ and $b$ are positive real numbers such that $b > a \ge e,$, then $a ^ {b} > b ^ {a}$; (2)  If a and b satisfy $e \ge b > a > 0$, then $b ^ {a} > a ^ {b}.$
A: $$\sqrt{2}^{\sqrt{3}} \approx 1.414^{1.732} \approx 1.822$$
$$\sqrt{3}^{\sqrt{2}} \approx 1.732^{1.414} \approx 2.174$$
$$\text{The rest is clear.}$$
