11.2. Which of the numbers is greater, $2^{\sqrt{\log _{3}}}$ or $3^{\sqrt{\log _{2}}}$?
Answer. $3^{\sqrt{\log _{2} 3}}$ is greater than $2^{\sqrt{\log _{3} 2}}$.

Solution. Two is less than three, so $\log _{3} 2 < 1$ and $\sqrt{\log _{2} 3} > 1$, therefore, $3^{\sqrt{\log _{2} 3}} > 3^{1} = 3 > 2 > 2^{\sqrt{\log _{3} 2}}$, which is what we needed to prove.

Grading criteria. Approximate calculations on a calculator: 0 points.