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)