Question

If ${\log }_{12}27=a$, then $\frac{3-a}{3+a}=$

• A. ${\log }_{6}2$
• B. $1+{\log }_{3}2$
• C. $1-{\log }_{3}2$
• D. ${\log }_{6}3$

Answer 1

The correct option is A ${\log }_{6}2$

$\frac{3-a}{3+a}=\frac{3{\log }_{12}12-{\log }_{12}27}{3{\log }_{12}12+{\log }_{12}27}=\frac{{\log }_{12}\left(\frac{(12{)}^3}{27}\right)}{{\log }_{12}\left(\right(12{)}^3.27)}={\log }_{6}2$