5. When $a \neq b, c \neq d$, $P=3^{a}+7^{c}, Q=3^{b}+7^{d}, a, b, c, d$ are all natural numbers. Then ().
(A) $P$ and $Q$ must not be equal
(B) Only one set of values for $a, b, c, d$ makes $P=Q$
(C) There are a finite number but more than one set of values for $a, b, c, d$ that make $P=Q$.
(D) There are infinitely many sets of values for $a, b, c, d$ that make $P=Q$
5. If $3^{c}+7^{c}=3^{s}+7^{d}$, assume $a>\bar{n}$, then it must be that $c<d$.
Thus, $3^{h}\left(3^{\circ-\omega}-1\right)=7^{c}\left(7^{d-\epsilon}-1\right)$.
Since $(3,7)=1,3^{b} \neq 3^{c}$, it must be that $3^{b}=7^{d-c}-1$.
Since the units digit of $3^{b}$ is $3,9,7,1$; and the units digit of $7^{d-c}-1$ is $6,8,2,0$, it follows that $3^{b} \neq 7^{d-6}-1$. Therefore, $3^{a}+7^{b} \neq 3^{c}+7^{d}$.
Hence, the answer is A.