[4(3-a)/(3+a)] If log_{12} 27 = a, then log_{6} 16 is:

For Competitive Exams

10th CBSE Maths

If $log_{12} 27$ = a, then $log_{6} 16$ is:

(3-a)/4(3+a)

(3+a)/4(3-a)

4(3+a)/(3-a)

4(3-a)/(3+a)

Explanation:

$log_{12} 27$ = a

=> log 27/ log 12 = a

=> $log 3^{3}$ / $log (3 * 2^{2})$ =a

=> 3 log 3 / log 3 + 2 log 2 = a

=> (log 3 + 2 log 2)/ 3 log 3 = 1/a

=> (log 3/ 3 log 3) + (2 log 2/ 3 log 3) = 1/3

=> (2 log 2)/ (3 log 3) = 1/a – 1/3 = (3-a)/ 3a

=> log 2/ log 3= (3-a)/3a

=> log 3 = (2a/3-a)log2

$log_{16} 16 $= log 16/ log 6

= $log 2^{4}/ log (2*3) $

= 4 log 2/ (log 2 + log 3)

= 4(3-a)/ (3+a)

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