If log 12 27- a- then log 6 16-A- 2 - 3-2B- 4 -3-a-3-aC- None of theseD- 3 -3-a-3-a

The correct option is B 4. 3 a/3 + aa = log 27/log 12 = 3 log 3/log 3 + 2log 2⇒ log 3 = 2a log 2/3 a log_6 16 = log 16/log 6 = 4 log 2/log 2 + log 3 = 4 ...

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Byju's Answer

Standard XI

Mathematics

Fundamental Laws of Logarithms

If log 12 27=...

Question

If $l o g_{12} 27 = a, t h e n l o g_{6} 16 =$

2. \frac{3 - a}{3 + a}

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3. \frac{3 - a}{3 + a}

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4. \frac{3 - a}{3 + a}

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Solution

The correct option is C $4. \frac{3 - a}{3 + a}$
$a = \frac{l o g 27}{l o g 12} = \frac{3 l o g 3}{l o g 3 + 2 l o g 2} \Rightarrow l o g 3 = \frac{2 a l o g 2}{3 - a} l o g_{6} 16 = \frac{l o g 16}{l o g 6} = \frac{4 l o g 2}{l o g 2 + l o g 3} = \frac{4 l o g 2}{l o g 2 + \frac{2 a l o g 2}{3 - a}} = \frac{4 (3 - a)}{3 - a + 2 a} = 4. \frac{3 - a}{3 + a}$

Suggest Corrections

If log1227=a,then log616=

The correct option is C 4.3−a3+aa=log 27log 12=3 log 3log3+2log 2⇒log 3=2alog 23−alog616=log 16log 6=4 log 2log 2+log 3=4 log 2log 2+2alog 23−a=4(3−a)3−a+2a=4.3−a3+a

If $l o g_{12} 27 = a, t h e n l o g_{6} 16 =$