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