Question

The value of $N=2{\log }_6^{3}3\left(1+3{\log }_3^{2}2\right)-{\log }_6^3\left(\frac{3}{2}\right)$ is equal to

• A. 0
• B. 1
• C. 2
• D. -1

Answer 1

Answer: (B)

$\frac{2}{{\left({\log }_{3}6\right)}^3}\left(1+3{\log }_3^{2}2\right)-\frac{{\left(1-{\log }_{3}2\right)}^3}{{\left(1+{\log }_{3}2\right)}^3}$
Let ${\log }_{3}2=t$
$=\frac{2\left(1+3t^2\right)}{(1+t{)}^3}-\frac{(1-t{)}^3}{(1+t{)}^3}$
$=\frac{2+6t^2-1+t^3+3t-3t^2}{(1+t{)}^3}$
$=\frac{t^3+3t+3t^2+1}{(1+t{)}^3}=1$