[Algebra]
Problem: Let $T=-10$. Given that $\log _{2} 4^{8 !}+\log _{4} 2^{8 !}=6 ! \cdot T \cdot x$, compute $x$.

Solution: Note that $4^{8 !}=2^{2 \cdot 8 !}$, thus $\log _{2} 4^{8 !}=2 \cdot 8$ !. Similarly, $\log _{4} 2^{8 !}=\frac{8 !}{2}$. Thus $2 \cdot 8 !+\frac{8 !}{2}=$ $6 !\left(2 \cdot 7 \cdot 8+7 \cdot \frac{8}{2}\right)=6 ! \cdot 140$. Thus $140=T x$, and with $T=-10, x=\mathbf{- 1 4}$.

Final answer: -14