If {log _{12}}27 = a, then {log _6}16 = | vedclass

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(c) $a = {{\log 27} \over {\log 12}} = {{3\log 3} \over {\log 3 + 2\log 2}} \Rightarrow \log 3 = {{2a\log 2} \over {3 - a}}$

${\log _6}16 = {{\log

Vedclass · Accepted Answer

$4.{{3 - a} \over {3 + a}}$

Vedclass · Answer

(c) $a = {{\log 27} \over {\log 12}} = {{3\log 3} \over {\log 3 + 2\log 2}} \Rightarrow \log 3 = {{2a\log 2} \over {3 - a}}$

${\log _6}16 = {{\log 16} \over {\log 6}} = {{4\log 2} \over {\log 2 + \log 3}}$

$ = {{4\log 2} \over {\log 2 + {{2a\log 2} \over {3 - a}}}} = {{4(3 - a)} \over {3 - a + 2a}} = 4.{{3 - a} \over {3 + a}}$.

If ${\log _{12}}27 = a,$ then ${\log _6}16 = $

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