Medium
If $a^2 + 4b^2 = 12ab,$ then $\log(a + 2b)$ is
- A$\frac{1}{2}[\log a + \log b - \log 2]$
- B$\log \frac{a}{2} + \log \frac{b}{2} + \log 2$
- C$\frac{1}{2}[\log a + \log b + 4\log 2]$
- D$\frac{1}{2}[\log a - \log b + 4\log 2]$
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