Correct Answer: Option B

Explanation

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

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