If $K=\frac{{\log }_2^{24}}{{\log }_{96}^2}-\frac{{\log }_2^{192}}{{\log }_{12}^2},$ then the values of ${\log }_{81}^{K^{100}}$ is_____ 

$$\begin{gathered} K=\frac{{\log }_2^{24}}{{\log }_{96}^2}-\frac{{\log }_2^{192}}{{\log }_{12}^2} \\ =\left({\log }_{2}24\right)\left({\log }_{2}96\right)-\left({\log }_{2}192\right)\left({\log }_{2}12\right) \\ =\left({\log }_{2}2+{\log }_{2}12\right)\left({\log }_2{2}^3+{\log }_{2}12\right)-\left({\log }_{2}12\right)\left({\log }_2{2}^4+{\log }_{2}12\right) \\ =\left(1+{\log }_{2}12\right)\left(3+{\log }_{2}12\right)-\left({\log }_{2}12\right)\left(4+{\log }_{2}12\right) \\ =3+{\log }_{2}12+3{\log }_{2}12+{\log }_{2}12\cdot {\log }_{2}12-\left(4{\log }_{2}12+{\log }_{2}12\cdot {\log }_{2}12\right) \\ =3 \end{gathered}$$
$$\begin{gathered} {\log }_{81}^{K^{100}}=\log {}_{3^4}3^{100} \\ =\frac{100}{4}\log {}_{3}3 \\ =25 \end{gathered}$$