$\begin{align}  & \frac{\log \sqrt{27}-\log \sqrt{8}}{\log 3-\log 2}=\frac{\log {{27}^{\tfrac{1}{2}}}-\log {{8}^{\tfrac{1}{2}}}}{\log 3-log2} \\ & \frac{\log \sqrt{27}-\log \sqrt{8}}{\log 3-\log 2}=\frac{\log {{3}^{\tfrac{3}{2}}}-\log {{2}^{\tfrac{3}{2}}}}{\log 3-log2} \\ & \frac{\log \sqrt{27}-\log \sqrt{8}}{\log 3-\log 2}=\frac{\tfrac{3}{2}\log 3-\tfrac{3}{2}\log 2}{\log 3-\log 2} \\ & \frac{\log \sqrt{27}-\log \sqrt{8}}{\log 3-\log 2}=\frac{\tfrac{3}{2}(\log 3-\log 2)}{\log 3-\log 2} \\ & \frac{\log \sqrt{27}-\log \sqrt{8}}{\log 3-\log 2}=\frac{3}{2} \\\end{align}$