$\begin{align}  & {{\log }_{\sqrt{2}}}4+{{\log }_{\tfrac{1}{2}}}16-{{\log }_{4}}32 \\ & {{\log }_{{{2}^{\tfrac{1}{2}}}}}4+{{\log }_{{{2}^{-1}}}}16-{{\log }_{{{2}^{2}}}}32 \\ & \text{Note:}{{\log }_{{{a}^{n}}}}b=\tfrac{1}{n}{{\log }_{a}}b \\ & =\frac{1}{\tfrac{1}{2}}{{\log }_{2}}4+\frac{1}{-1}{{\log }_{2}}16-\frac{1}{2}{{\log }_{2}}32 \\ & =2{{\log }_{2}}4-{{\log }_{2}}16-\tfrac{1}{2}{{\log }_{2}}32 \\ & =2{{\log }_{2}}{{2}^{2}}-{{\log }_{2}}{{2}^{4}}-\tfrac{1}{2}{{\log }_{2}}{{2}^{5}} \\ & =2(2){{\log }_{2}}2-4{{\log }_{2}}2-\tfrac{5}{2}{{\log }_{2}}2\text{    (Note: lo}{{\text{g}}_{2}}2=1) \\ & =4-4-\tfrac{5}{2}=-\tfrac{5}{2}=-2.5 \\\end{align}$