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Maths-

General

Easy

Question

If a² + 4b² = 12ab, then log (a + 2b) =

$\frac{1}{2}$ (log a + log b – log 2)
log a/2 + log b/2 + log 2
$\frac{1}{2}$ (log a + log b + 4 log 2)
$\frac{1}{2}$ (log a – log b + 4log 2)

The correct answer is: $fraction numerator 1 over denominator 2 end fraction$ (log a + log b + 4 log 2)

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