4) Consider the two numbers $x=\left(\sqrt{3}^{\sqrt{2}}\right)^{\sqrt{2}}$ and $y=\left(\sqrt{2}^{\sqrt{3}}\right)^{\sqrt{3}}$. We have that

$\begin{array}{llll}\text { (A) } x=y & \text { (B) } x>y & \text { (C) } x1\end{array}$

(E) $x$ and $y$ cannot be compared.
4) The answer is (B).

Recall that $\left(a^{b}\right)^{c}=a^{b \cdot c}$; therefore, $x=\left(\sqrt{3}^{\sqrt{2}}\right)^{\sqrt{2}}=\sqrt{3}^{\sqrt{2} \cdot \sqrt{2}}=\sqrt{3}^{2}=3 \mathrm{e}$

$y=\left(\sqrt{2}^{\sqrt{3}}\right)^{\sqrt{3}}=\sqrt{2}^{\sqrt{3} \cdot \sqrt{3}}=\sqrt{2}^{3}=\sqrt{8}$. Since $3^{2}=9>8$, we have $x>y$, but $x^{2}-y^{2}=1$.