Question

If ${\log }_{12}27=a.$, then prove that : ${\log }_{6}16=\frac{4(3-a)}{3+a}.$

Answer 1

${\log }_{6}16={\log }_6{2}^4=4{\log }_{6}2=\frac{4}{{\log }_{2}6}$
$=\frac{4}{{\log }_{2}2+{\log }_{2}3}=\frac{4}{1+{\log }_{2}3}$ ...(1)

Given $a={\log }_{12}27={\log }_{12}{3}^3$

$=3{\log }_{12}3$

$=\frac{3}{{\log }_{3}12}$
$=\frac{3}{{\log }_{3}3+{\log }_{3}4}$

$\Rightarrow a=\frac{3}{1+2{\log }_{3}2}$
$\therefore a+2a{\log }_{3}2=3$

${\log }_{3}2=\frac{3-a}{2a}$
Hence ${\log }_{2}3=\frac{2a}{3-a}.$
Substituting this value of ${\log }_{2}3$ in (1), we get
${\log }_{6}16=\frac{4}{1+\frac{2a}{(3-a)}}=\frac{4(3-a)}{3+a}.$