If $a = l o g_{24} 12, b = l o g_{36} 24 a n d c = l o g_{48} 36$ then 1+abc is equal to

2ab

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2ac

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2bc

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Solution

The correct option is C 2bc
$a = l o g_{24} 12 = \frac{l o g 12}{l o g 24} = \frac{2 ‘ l o g 2 + l o g 3}{3 l o g 2 + l o g 3}$
$b = l o g_{36} 24 = \frac{3 l o g 2 + l o g 3}{2 (l o g 2 + l o g 3)}$
$c = l o g_{48} 36 = \frac{2 (l o g 2 + l o g 3)}{4 l o g 2 + l o g 3}$
$∴ a b c = \frac{2 l o g 2 + l o g 3}{4 l o g 2 + l o g 3}$
$\Rightarrow 1 + a b c = \frac{6 l o g 2 + 2 l o g 3}{4 l o g 2 + l o g 3} = 2. \frac{3 l o g 2 + l o g 3}{4 l o g 2 + l o g 3} = 2 b c$

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f $a = {log}_{24} 12, b = {log}_{36} 24, c = {log}_{48} 36$ , then 1+abc is equal to ______________.

a = l o g_{24} 12, b = l o g_{36} 24 a n d c = l o g_{48} 36

a = l o g_{24} 12, b = l o g_{36} 24 a n d c = l o g_{48} 36

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