$7 \cdot 57$ In $\triangle A B C$, the sides opposite to angles $A, B, C$ are denoted as $a, b, c (b \neq 1)$, respectively, and $\frac{C}{A}, \frac{\sin B}{\sin A}$ are both roots of the equation $\log _{\sqrt{b}} x=\log _{b}(4 x-4)$. Then $\triangle A B C$ is
(A) an isosceles triangle, but not a right triangle.
(B) a right triangle, but not an isosceles triangle.
(C) an isosceles right triangle.
(D) neither an isosceles triangle nor a right triangle.
(China High School Mathematics League, 1992)
[Solution] From the given equation, we have $\log _{b} x^{2}=\log _{b}(4 x-4)$, thus $x^{2}-4 x+4=0$, solving this gives $x_{1}=x_{2}=2$.
$\therefore C=2 A$, and $\sin B=2 \sin A$, which implies $b=2 a$.
Also, $\because \quad A+B+C=180^{\circ}$,
$\therefore 3 A+B=180^{\circ}$,
then
$$
\begin{array}{ll} 
& \sin B=\sin 3 A=2 \sin A, \\
\therefore \quad & 3 \sin A-4 \sin ^{3} A=2 \sin A, \quad \text { i.e., } \quad \sin A\left(1-4 \sin ^{2} A\right)=0 . \\
\because \quad & \sin A \neq 0, \\
\therefore \quad & 1-4 \sin ^{2} A=0, \quad \text { which gives } \quad \sin A=\frac{1}{2} \text { (noting that } \sin A>0 \text { ). } \\
\therefore \quad & A=30^{\circ}, \quad C=60^{\circ}, \quad B=90^{\circ} .
\end{array}
$$

Therefore, the answer is $(B)$.