The value of,log2⁡24log96⁡2−log2⁡192log12⁡2 is

Let log2⁡12=a, then 1log96⁡2=log2⁡96=log2⁡23×12=3+alog2⁡24=1+a⇒ log2⁡192=log2⁡(16×12)=4+a and 1log12⁡2=log2⁡12=a. Therefore, the given expression is: (1+a)(3+a)−(4+a)a=3