Express \(\log \frac{1}{8} + \log \frac{1}{2}\) in terms of \(\log 2\).
Answer Details
We know that:
\[\log a + \log b = \log ab\]
Using this property, we can rewrite the given expression as:
\[\log \left(\frac{1}{8} \cdot \frac{1}{2}\right) = \log \frac{1}{16}\]
Now, we use another property of logarithms:
\[\log a^b = b\log a\]
to rewrite \(\log \frac{1}{16}\) in terms of \(\log 2\):
\[\log \frac{1}{16} = \log 2^{-4} = -4\log 2\]
Therefore, \(\log \frac{1}{8} + \log \frac{1}{2} = -4\log 2\) and the answer is: \(-4 \log 2\).