If $a^{2} + 4 b^{2} = 12 a b$ , then log (a + 2b) is

$\frac{1}{2} [l o g a + l o g b - l o g 2]$

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$l o g \frac{a}{2} + l o g \frac{b}{2} + l o g 2$

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$\frac{1}{2} [l o g a + l o g b + 4 l o g 2]$

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$\frac{1}{2} [l o g a - l o g b + 4 l o g 2]$

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Solution

The correct option is C
$\frac{1}{2} [l o g a + l o g b + 4 l o g 2]$

$a^{2} + 4 b^{2} + 4 a b = 16 a b$
$∴ (a + 2 b)^{2} = 16 a b$
$∴ 2 l o g (a + 2 b) = 4 l o g 2 + l o g a + l o g b$

Therefore log (a +2b) = $\frac{1}{2} (4 l o g 2 + l o g a + l o g b)$

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Similar questions

a^{2} + 4 b^{2} = 12 a b

log (a + 2 b)

a^{2} - 12 a b + 4 b^{2} = 0

- log (a + 2 b) + \frac{1}{2} (log a + log b) + 2 log 2

a^{2} - 12 a b + {4 b}^{2} = 0,

log (a + 2 b) = \frac{1}{2} (log a + log b) + 2 log 2

If a-2b = 15 and find a²+ 4b² .

\frac{a^{2} - 4 b^{2}}{a^{2} - 9 b^{2}} \times \frac{a - 3 b}{a + 2 b}

\frac{a - 2 b}{a + 3 b}

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