Example 7. Solve the equation:
a) $\log _{x^{2}-1}\left(x^{3}+6\right)=\log _{x^{2}-1}\left(4 x^{2}-x\right)$;

b) $\log _{x^{3}+x}\left(x^{2}-4\right)=\log _{4 x^{2}-6}\left(x^{2}-4\right)$.
Solution. a) The given equation is equivalent to the system

$$
\left\{\begin{array}{l}
x^{3}+6>0 \\
x^{2}-1>0 \\
x^{2}-1 \neq 1 \\
x^{3}+6=4 x^{2}-x
\end{array}\right.
$$

The equation of this system $x^{3}-4 x^{2}+x+6=0$ has three roots: $x_{1}=-1, x_{2}=2, x_{3}=3$. The number $x_{1}=-1$ does not satisfy the condition $x^{2}-1>0$. The numbers $x_{2}=2$ and $x_{3}=3$ are solutions to this system, and therefore, to the original equation.

b) The equation is equivalent to the system

$$
\left\{\begin{array}{l}
x^{2}-4>0 \\
x^{3}+x>0 \\
x^{3}+x \neq 1 \\
x^{3}+x=4 x^{2}-6
\end{array}\right.
$$

The equation

$$
x^{3}-4 x^{2}+x+6=0
$$

has three roots: $x_{1}=-1, x_{2}=2, x_{3}=3$. Of these, only the number 3 satisfies all the requirements of the system. Therefore, the given equation has one root, the number 3.

An equation of the form

$$
\log _{\alpha(x)}\left(\log _{\beta(x)} f(x)\right)=0
$$

is equivalent to the system

$$
\left\{\begin{array}{l}
\alpha(x)>0 \\
\alpha(x) \neq 1 \\
\log _{\beta(x)} f(x)=1
\end{array}\right.
$$

which in turn is equivalent to the system

$$
\left\{\begin{array}{l}
\alpha(x)>0 \\
\alpha(x) \neq 1 \\
\beta(x)>0 \\
\beta(x) \neq 1 \\
f(x)=\beta(x) .
\end{array}\right.
$$