MCQ
If ${\log _{12}}27 = a,$ then ${\log _6}16 = $
- A$2.{{3 - a} \over {3 + a}}$
- B$3.{{3 - a} \over {3 + a}}$
- ✓$4.{{3 - a} \over {3 + a}}$
- DNone of these
${\log _6}16 = {{\log 16} \over {\log 6}} = {{4\log 2} \over {\log 2 + \log 3}}$
$ = {{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}}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.