Question

Let ${\log }_{12}27=a$ then find the value of ${\log }_{6}16$.

Answer 1

Given,

${\log }_{12}27=a$

or, ${\log }_{27}12=\frac{1}{a}$

or, ${\log }_{3^3}(3\times 4)=\frac{1}{a}$

or, $\frac{1}{3}\left({\log }_3\right(3\times 4\left)\right)=\frac{1}{a}$

or, ${\log }_{3}3+{\log }_{3}4=\frac{3}{a}$

or, $2{\log }_{3}2=\frac{3-a}{a}$

or, ${\log }_{3}2=\frac{3-a}{2a}$

or, ${\log }_{2}3=\frac{2a}{3-a}$.......(1).

Now,

${\log }_{16}6$

$={\log }_{2^4}6$

$=\frac{1}{4}{\log }_2(2\times 3)$

$=\frac{1}{4}({\log }_{2}3+1)$

$=\frac{3+a}{4(3-a)}$[ Using (1)].......(2).

From (2) we get,

${\log }_{6}16=\frac{4(3-a)}{3+a}$