If $a = \log_{24} 12, b = \log_{36} 24$ and $c = \log_{48} 36,$ then $1 + abc$ is equal to

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$2 ac$

$2 bc$

$0$

$2 ab$

answer is C.

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Detailed Solution

$a = \log_{24} 12 = \frac{\log 12}{\log 24} = \frac{2 \log 2 + \log 3}{3 \log 2 + \log 3}$

$b = \log_{36} 24 = \frac{3 \log 2 + \log 3}{2 (\log 2 + \log 3)}$

$c = \log_{48} 36 = \frac{2 (\log 2 + \log 3)}{4 \log 2 + \log 3}$

$∴ abc = \frac{2 \log 2 + \log 3}{4 \log 2 + \log 3}$

$\Rightarrow 1 + abc = \frac{6 \log 2 + 2 \log 3}{4 \log 2 + \log 3} = 2 . \frac{3 \log 2 + \log 3}{4 \log 2 + \log 3} = 2 bc .$

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