If $l o g_{12} 27 = a$ , then find the value of $l o g_{6} 16$ .

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Solution

$l o g_{6} 16 = l o g_{6} 2^{4} = 4 l o g_{6} 2 = \frac{4}{l o g_{2} 6}$
$= \frac{4}{l o g_{2} 2 + l o g_{2} 3} = \frac{4}{1 + l o g_{2} 3}$
and $a = l o g_{12} 27 = l o g_{12} 3^{3} = 3 l o g_{12} 3$
$= \frac{3}{l o g_{3} 12} = \frac{3}{l o g_{3} 3 + l o g_{3} 4}$
$= \frac{3}{1 + 2 l o g_{3} 2}$
$∴ a + 2 a l o g_{3} 2 = 3$
or $l o g_{3} 2 = \frac{3 - 1}{2 a}$
Hence $l o g_{2} 3 = \frac{2 a}{3 - a}$

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