(5) The set $M=\left\{(x, y) \mid \log _{4} x+\log _{4} y \leqslant 1, x, y \in \mathbf{N}^{*}\right\}$ has ( ) subsets. (A) 4
(B) 16
(C) 64
(D) 256
5 From $\log _{4} x+\log _{4} y \leqslant 1$, we get $1 \leqslant x y \leqslant 4$. Therefore, $(x, y)=(1,1),(1,2)$, $(1,3),(1,4),(2,1),(2,2),(3,1),(4,1)$, a total of 8. Thus, the subsets of set $M$ are $2^{8}=256$ (in number). Hence, the answer is D.