4. 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 $\log _{\sqrt{6}} x=\log _{6}(4 x-4)$. Then $\triangle A B C$
(A) is an isosceles triangle, but not a right triangle;
(B) is a right triangle, but not an isosceles triangle;
(C) is an isosceles right triangle;
(D) is neither an isosceles triangle nor a right triangle.
4. (B)

From the given equation, we have
$$
\log _{b} x^{2}=\log _{6}(4 x-4),
$$

which implies
$$
x^{2}-4 x+4=0,
$$

The roots obtained are
$$
\text { - } x_{1}=x_{2}=2 \text {. }
$$

Thus, $C=2A$ and $\sin B=2 \sin A$,
Since $A+B+C=180^{\circ}$, we have $3A+B=180^{\circ}$,
Therefore,
$$
\sin B=\sin 3A,
$$

So,
$$
2 \sin A-4 \sin ^{3} A=2 \sin A,
$$

which simplifies to $\sin A\left(1-4 \sin ^{2} A\right)=0$, but $\sin A \neq 0$,
Thus,
$$
\sin ^{2} A=\frac{1}{4},
$$

Since $\sin A>0$, we have $\sin A=\frac{1}{2}$,
Hence,
$$
A=30^{\circ}, \quad C=60^{\circ}, \quad B=90^{\circ} .
$$

Therefore, the answer is (B).