Question

The value of $\frac{{\log }_{2}24}{{\log }_{96}2}-\frac{{\log }_{2}192}{{\log }_{12}2}$ will be

• A. $3$
• B. $0$
• C. $2$
• D. $1$

Answer 1

The correct option is A $3$

$\frac{{\log }_{2}24}{{\log }_{96}2}-\frac{{\log }_{2}192}{{\log }_{12}2}={\log }_{2}24\times {\log }_{2}96-{\log }_{2}192\times {\log }_{2}12$
$={\log }_{2}24({\log }_{2}24+{\log }_{2}4)-{\log }_2(3\times 64)\times {\log }_2(3\times 4)$
$=({\log }_{2}8+{\log }_{2}3)({\log }_{2}3+{\log }_{2}8+{\log }_{2}4)-({\log }_{2}64+{\log }_{2}3)\times ({\log }_{2}3+{\log }_{2}4)$
$=(3+y)(5+y)-(6+y)(2+y)$, where $y={\log }_{2}3$
$=3$