smc

Recall that: 1. ${\log }_b(uv)={\log }_{b}u+{\log }_{b}v.$ 2. ${\log }_{b}u\cdot {\log }_{u}b=1.$ We use these properties of logarithms to rewrite the original expression: $$\begin{array}{rl}\sqrt{{\log }_{2}6+{\log }_{3}6}& =\sqrt{({\log }_{2}2+{\log }_{2}3)+({\log }_{3}2+{\log }_{3}3)}\\ & =\sqrt{2+{\log }_{2}3+{\log }_{3}2}\\ & =\sqrt{{\left(\sqrt{{\log }_{2}3}+\sqrt{{\log }_{3}2}\right)}^2}\\ & =\boxed{\sqrt{{\log }_{2}3}+\sqrt{{\log }_{3}2}}.\end{array}$$