Question

The value of $\frac{{\log }_{2}24}{{\log }_{96}2}-\frac{{\log }_{2}192}{{\log }_{12}2}$ is equal to

Answer 1

let $y=\frac{{\log }_{2}24}{{\log }_{96}2}-\frac{{\log }_{2}192}{{\log }_{12}2}$
$\Rightarrow y=\left({\log }_{2}24\right)\left({\log }_{2}96\right)-\left({\log }_{2}192\right)\left({\log }_{2}12\right)$
$\Rightarrow y=({\log }_{2}2+{\log }_{2}12)({\log }_{2}8+{\log }_{2}12)-\left({\log }_{2}12\right)({\log }_{2}16+{\log }_{2}12)$
$\Rightarrow y=(1+{\log }_{2}12)(3+{\log }_{2}12)-\left({\log }_{2}12\right)(4+{\log }_{2}12)$
$=3+{\log }_{2}12+3{\log }_{2}12+{\log }_{2}12{\log }_{2}12-(4{\log }_{2}12+{\log }_{2}12{\log }_{2}12)=3$

$\Rightarrow y=3$
Ans: 3