Question

Prove that $\frac{{\log }_{2}24}{{\log }_{96}2}-\frac{{\log }_{2}192}{{\log }_{12}2}=3.$

Answer 1

On taking LCM:

$\frac{{\log }_{2}24\times {\log }_{12}2-{\log }_{2}192\times {\log }_{96}2}{{\log }_{96}2\times {\log }_{12}2}$

$\Rightarrow \frac{{\log }_{12}24-{\log }_{96}192}{{\log }_{96}2\times {\log }_{12}2}$

$\Rightarrow \frac{1+{\log }_{12}2-1-{\log }_{96}2}{{\log }_{96}2\times {\log }_{12}2}$

$\Rightarrow \frac{\frac{\log 2}{\log 12}-\frac{\log 2}{\log 96}}{\frac{\log 2}{\log 96}\times \frac{\log 2}{\log 12}}$

$\Rightarrow \frac{\log 96-\log 12}{\log 2}$

$\Rightarrow {\log }_2\left(\frac{96}{12}\right)={\log }_{2}8=3$