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Basic of Logarithms
medium
If ${\log _{12}}27 = a,$ then ${\log _6}16 = $
A
$2.{{3 - a} \over {3 + a}}$
B
$3.{{3 - a} \over {3 + a}}$
C
$4.{{3 - a} \over {3 + a}}$
D
None of these
Solution
(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}}$.
Standard 11
Mathematics
