To evaluate the expression \( \log_{\sqrt{2}} 4 + \log_{\frac{1}{2}} 16 - \log_{4} 32 \), we can simplify each term separately. 1. Calculate \( \log_{\sqrt{2}} 4 \) \(\log_{\sqrt{2}} 4 = \frac{\log_2 4}{\log_2 \sqrt{2}}\) We know that \( \log_2 4 = 2 \) (since \( 2^2 = 4 \)) and \( \log_2 \sqrt{2} = \log_2 2^{1/2} = \frac{1}{2} \). Thus, \(log_{\sqrt{2}} 4 = \frac{2}{\frac{1}{2}} = 4\) 2. Calculate \( \log_{\frac{1}{2}} 16 \) \(\log_{\frac{1}{2}} 16 = \frac{\log_2 16}{\log_2 \frac{1}{2}}\) We know that \( \log_2 16 = 4 \) (since \( 2^4 = 16 \)) and \( \log_2 \frac{1}{2} = \log_2 2^{-1} = -1 \). Thus, \(\log_{\frac{1}{2}} 16 = \frac{4}{-1} = -4\) 3. Calculate \( \log_{4} 32 \) \(\log_{4} 32 = \frac{\log_2 32}{\log_2 4}\) We know that \( \log_2 32 = 5 \) (since \( 2^5 = 32 \)) and \( \log_2 4 = 2 \). Thus, \(\log_{4} 32 = \frac{5}{2}\) Now, we can substitute these results back into the original expression: \(\log_{\sqrt{2}} 4 + \log_{\frac{1}{2}} 16 - \log_{4} 32 = 4 - 4 - \frac{5}{2}\) Calculating this gives: \(4 - 4 - \frac{5}{2} = 0 - \frac{5}{2} = -\frac{5}{2}\) Thus, the final result is: \(-\frac{5}{2}\) = - 2.5