(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}}$.