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Variable Separable Method

f a = log 2...

Question

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

A

2 a c

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B

2 b c

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C

2 a b

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D

None of these

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Solution

The correct option is C $2 b c$
Given
$a = l o g_{24} 12$ ; $b = l o g_{36} 24$ ; $c = l o g_{48} 36$

$1 + a b c = 1 + l o g_{24} 12 \times l o g_{36} 24 \times l o g_{48} 36$

$= 1 + \frac{l o g 12}{l o g 24} \times \frac{l o g 24}{l o g 36} \times \frac{l o g 36}{l o g 48}$

$= \frac{(l o g 48 + l o g 12)}{(l o g 48)}$

$= \frac{l o g (576)}{l o g 48}$

$= \frac{(l o g 24)^{2}}{l o g 48}$

$= \frac{2 l o g 24}{l o g 48}$

$= 2 \times \frac{(l o g 24)}{l o g 36} \frac{l o g 36}{l o g 48}$

$= 2 \times (l o g_{36} 24) (l o g_{48} 36)$

$= 2 b c$

So $1 + a b c = 2 b c$

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Similar questions

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

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

1 + a b c

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

then 1+abc is equal to

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

1 + a b c

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