If a -log 24 12- b -log 36 24 and c -log 48 36 then 1- abc is equal toA- 2 bcB- 0C- 2 acD- 2 ab

The correct option is A 2bca=log_24 12=log 12/log 24=2`log 2+log 3/3 log 2+log 3 b=log_36 24 =3 log 2+log 3/2(log 2+log 3) c= log_48 36=2(log 2+log 3)/4 log 2+ ...

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Byju's Answer

Standard XI

Mathematics

Fundamental Laws of Logarithms

If a =log 24 ...

Question

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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If a=log24 12, b=log36 24 and c=log48 36 then 1+abc is equal to

The correct option is C 2bca=log24 12=log 12log 24=2‘log 2+log 33 log 2+log 3 b=log36 24=3 log 2+log 32(log 2+log 3) c=log48 36=2(log 2+log 3)4 log 2+log 3 ∴ abc=2 log 2+log 34 log 2+log 3 ⇒1+abc=6 log 2+2 log 34 log 2+log 3=2.3 log 2+log 34 log 2+log 3=2bc

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