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Q. If a = $\log_{24} (12)$ , b = $\log_{36} (24)$ , c = $\log_{48} (36)$ then (1 + abc) = ?
a) 2ab b) 2ac c) 2bc d) 0

$G i v e n t h a t : a = \log_{24} 12, b = \log_{36} 24, a n d c = \log_{48} 36 N o w, 1 + a b c = 1 + (\log_{24} 12) (\log_{36} 24) (\log_{48} 36) = 1 + (\frac{\log 12}{\log 24}) (\frac{\log 24}{\log 36}) (\frac{\log 36}{\log 48}) [b e c a u s e \log_{y} x = \frac{\log x}{\log y}] = 1 + (\frac{\log 12}{\log 48}) = \frac{\log 48 + \log 12}{\log 48} = \frac{\log (48 \times 12)}{\log 48} = \frac{\log 576}{\log 48} = \frac{\log (24^{2})}{\log 48} [\log m^{n} = n \log m] = \frac{2 \log 24}{\log 48} = 2 \times \frac{\log 24}{\log 36} \times \frac{\log 36}{\log 48} = 2 {\times \log}_{36} 24 \times {\times \log}_{48} 36 = 2 b c H e n c e, 1 + a b c = 2 b c$

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