Cho

B

- Hướng dẫn giải

Đáp án B

Ta có

$\begin{array}{l}{\log }_{12}27=a\iff {\log }_{12}{3}^3=a\\ \iff 3{\log }_{12}3=a\iff \frac{3}{{\log }_{3}12}=a\\ \iff \frac{3}{{\log }_3\left({3.2}^2\right)}=a\iff \frac{3}{1+2lo{g}_{3}2}=a\end{array}$

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

$\begin{array}{l}\iff {\log }_{6}24={\log }_6\left(6.4\right)=1+{\log }_6{2}^2\\ =1+\frac{2}{{\log }_{2}6}\\ =1+\frac{2}{1+{\log }_{2}3}=1+\frac{2}{1+\frac{2a}{3-a}}=\frac{9-a}{a+3}\end{array}$