[Algebra]
Problem: Let $T=8640$. Compute $\left\lfloor\log _{4}\left(1+2+4+\cdots+2^{T}\right)\right\rfloor$.

Solution: Let $S=\log _{4}\left(1+2+4+\cdots+2^{T}\right)$. Because $1+2+4+\cdots+2^{T}=2^{T+1}-1$, the change-of-base formula yields

$$
S=\frac{\log _{2}\left(2^{T+1}-1\right)}{\log _{2} 4}
$$



Let $k=\log _{2}\left(2^{T+1}-1\right)$. Then $T<k<T+1$, so $T / 2<S<(T+1) / 2$. If $T$ is even, then $\lfloor S\rfloor=T / 2$; if $T$ is odd, then $\lfloor S\rfloor=(T-1) / 2$. With $T=8640$, the answer is 4320 .

Final answer: 4320