7. If the decimal part of $\log _{c} b$ is zero, and $\log _{a} \frac{1}{b}>\log _{a} \sqrt{b}>\log _{b} a^{2}$, then among the following four conclusions:
(1) $\frac{1}{b}>\sqrt{b}>a^{2}$;
(2) $\log b+\log _{a} a=0$;
(3) $0<a<b<1$;
(4) $a b-1=0$.

the number of correct conclusions is
(A) 1 ;
(B) 2 ;
(C) 3 ;
(D) 4
7. (A)

Solution: From $\log _{a} \frac{1}{b}>\log _{a} \sqrt{b}$ we get $-\log _{a} b>\frac{1}{2} \log _{a} b$.
Therefore, $\log _{a} b<0$. From $-\log _{a} b>\frac{1}{2} \log _{a} b$, we get $-\frac{3}{2} \log _{a} b>0$, which implies $\log _{a} b<0$. This means $0<a<1, b>1$ or $a>1, b<1$. From $\frac{1}{b}>\sqrt{b}$, we get $b<1$. Therefore, $0<a<1$ and $b>1$.
From $\log _{a} \frac{1}{b}>\log _{a} \sqrt{b}$, we get $\log _{a} b<0$. From $\log _{a} b<0$, we get $\log _{b} a>0$. From $\log _{a} b<0$, we get $\log _{b} a^{2}<0$. From $\log _{a} b<0$, we get $\frac{1}{2} \log _{3} b>2 \log _{b} a=\frac{1}{\log _{a} b}$.
Since $\log _{a} b<0$, we have $\left(\log _{a} b\right)^{2}<4$, which means $-2<\log _{a} b<0$.
Since $\log _{a} b$ is an integer, we have $\log _{a} b=-1$,
which means $b=\frac{1}{a}$, thus $a b=1$, conclusion (4) is correct.