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

Answer

Correct option: C.
${1 \over 2}[\log a + \log b + 4\log 2]$
c
(c) ${a^2} + 4{b^2} = 12ab$$ \Rightarrow $${a^2} + 4{b^2} + 4ab = 16ab$

$ \Rightarrow $${(a + 2b)^2} = 16ab$

$ \Rightarrow $$2\log (a + 2b) = \log 16 + \log a + \log b$

$\therefore $ $\log (a + 2b) = {1 \over 2}[\log a + \log b + 4\log 2]$

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