(c) $a = {\log _{24}}12 = {{\log 12} \over {\log 24}} = {{2\log 2 + \log 3} \over {3\log 2 + \log 3}}$

$b = {\log _{36}}24 = {{3\log 2 + \log 3} \over {2(\log 2 + \log 3)}}$

$c = {\log _{48}}36 = {{2(\log 2 + \log 3)} \over {4\log 2 + \log 3}}$

$\therefore abc = {{2\,\,\log 2 + \log 3} \over {4\log 2 + \log 3}}$

==> $1 + abc = {{6\log 2 + 2\log 3} \over {4\log 2 + \log 3}} = 2.{{3\log 2 + \log 3} \over {4\log 2 + \log 3}} = 2bc$.