3 log 2
4 log 2
-3 log 2
-4 log 2
Correct answer is D
\(\log \frac{1}{8} + \log \frac{1}{2} = \log 8^{-1} + \log 2^{-1}\)
= \(\log 2^{-3} + \log 2^{-1}\)
= \(-3 \log 2 - 1 \log 2 = -4 \log 2\)
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