2020 AMC 12B Problem 13

Let $a = \log_2 3,$ so $\log_3 2 = \tfrac1a.$ Then $$\begin{gathered} \log_2 6 + \log_3 6 \\ {}= (1 + \log_2 3) \\ \quad {}+ (1 + \log_3 2) \\ {}= 2 + a + \frac1a. \end{gathered}$$ Meanwhile $\left(\sqrt{\log_2 3} + \sqrt{\log_3 2}\right)^2$ $= a + \frac1a + 2\sqrt{a \cdot \tfrac1a}$ $= a + \frac1a + 2,$ which equals the expression above. Taking square roots, $\sqrt{\log_2 6 + \log_3 6}$ $= \sqrt{\log_2 3} + \sqrt{\log_3 2}.$ Thus, the correct answer is D.