Question

If ${\log }_{12}27=a$, then the value of ${\log }_{6}16$ in terms of $a$ will be

• A. $2\left(\frac{3-a}{3+a}\right)$
• B. $3\left(\frac{3-a}{3+a}\right)$
• C. $4\left(\frac{3-a}{3+a}\right)$
• D. $5\left(\frac{3-a}{3+a}\right)$

Answer 1

The correct option is C $4\left(\frac{3-a}{3+a}\right)$

Given that, $a={\log }_{12}27$
$\Rightarrow a={\log }_{12}{\left(3\right)}^3=3{\log }_{12}3[\because \log a^m=m\log a]$
$\Rightarrow a=\frac{3}{{\log }_{3}12}=\frac{3}{{\log }_3(3\cdot 4)}$
$=\frac{3}{1+{\log }_{3}4}[\because \log (ab)=\log a+\log b]=\frac{3}{1+2{\log }_{3}2}$
$\Rightarrow {\log }_{3}2=\frac{3-a}{2a}$
Then,
${\log }_{6}16={\log }_6{2}^4=4{\log }_{6}2=\frac{4}{{\log }_{2}6}$
$=\frac{4}{1+{\log }_{2}3}=\frac{4}{1+\frac{2a}{3-a}}=4\left(\frac{3-a}{3+a}\right)$
Ans: C