If a = log_{24} 12, b = log_{36} 24 and c = l | vedclass

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(C) Given $a = \log_{24} 12 = \frac{\log 12}{\log 24} = \frac{2\log 2 + \log 3}{3\log 2 + \log 3}$.$b = \log_{36} 24 = \frac{\log 24}{\log 36} = \frac

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$2bc$

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(C) Given $a = \log_{24} 12 = \frac{\log 12}{\log 24} = \frac{2\log 2 + \log 3}{3\log 2 + \log 3}$.$b = \log_{36} 24 = \frac{\log 24}{\log 36} = \frac{3\log 2 + \log 3}{2\log 2 + 2\log 3}$.$c = \log_{48} 36 = \frac{\log 36}{\log 48} = \frac{2\log 2 + 2\log 3}{4\log 2 + \log 3}$.Multiplying these,$abc = \left(\frac{2\log 2 + \log 3}{3\log 2 + \log 3}ight) 	imes \left(\frac{3\log 2 + \log 3}{2\log 2 + 2\log 3}ight) 	imes \left(\frac{2\log 2 + 2\log 3}{4\log 2 + \log 3}ight)$.Canceling common terms,$abc = \frac{2\log 2 + \log 3}{4\log 2 + \log 3}$.Then $1 + abc = 1 + \frac{2\log 2 + \log 3}{4\log 2 + \log 3} = \frac{4\log 2 + \log 3 + 2\log 2 + \log 3}{4\log 2 + \log 3} = \frac{6\log 2 + 2\log 3}{4\log 2 + \log 3} = 2 	imes \frac{3\log 2 + \log 3}{4\log 2 + \log 3}$.Since $b = \frac{3\log 2 + \log 3}{2\log 2 + 2\log 3}$ and $c = \frac{2\log 2 + 2\log 3}{4\log 2 + \log 3}$,we have $bc = \frac{3\log 2 + \log 3}{4\log 2 + \log 3}$.Thus,$1 + abc = 2bc$.

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

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