Example 12 (17th National "Hope Cup" Competition Question) Solve the inequality about $x$
$$
\left|\log _{a} x\right|+\left|\log _{a}^{2} x-1\right|>a(a>1) .
$$
Let $\log _{a} x=t$, then $|t|+\left|t^{2}-1\right|>a(a>1)$.
Let $y_{1}=\left|t^{2}-1\right|, y_{2}=a-|t|$,
The original inequality is transformed into $y_{1}>y_{2}$.
Plot on the coordinate system $t O y$, localize the overall problem:
When $-1<t<0$,
$y_{1}=1-t^{2}, y_{2}=a+t, y_{1}(t)$ and $y_{2}(t)$ graphs are tangent, i.e., the system of equations $\left\{\begin{array}{l}y_{1}=1-t^{2}, \\ y_{2}=a+t\end{array}\right.$ has a unique solution, at this time, by $\Delta=0$, solve to get $a=\frac{5}{4}$.
The following discussion is based on the graph, by the symmetry of the even function, only the left side needs to be considered.
(1) When $1<a<\frac{5}{4}$, if $t<-1, y_{1}=t^{2}-1, y_{2}=a+t$,
By $y_{1}=y_{2}$, we get $t_{\text {A }}=\frac{1-\sqrt{5+4 a}}{2}$.
If $-1 \leqslant t<0, y_{1}=1-t^{2}, y_{2}=a+t$, by $y_{1}=y_{2}$, we get
$$
t_{B}=\frac{-1-\sqrt{5-4 a}}{2}, t_{C}=\frac{-1+\sqrt{5-4 a}}{2} .
$$

From the graph, at this time $t \in\left(-\infty, \frac{1-\sqrt{5+4 a}}{2}\right) \cup\left(\frac{-1-\sqrt{5-4 a}}{2}, \frac{-1+\sqrt{5-4 a}}{2}\right)$.
(2) When $a \geqslant \frac{5}{4}$, if $t<-1$ (due to the range of $a$, no need to consider $-1 \leqslant t<0$ ),
By $y_{1}=y_{2}$, we get $t_{A}=\frac{1-\sqrt{5+4 a}}{2}$,
From the graph, at this time $t \in\left(-\infty, \frac{1-\sqrt{5-4 a}}{2}\right)$, considering the symmetry of the even function, and substituting back, we get: when $1<a<\frac{5}{4}$,
$$
x \in\left(0, a^{\frac{1-\sqrt{5+4 a}}{2}}\right) \cup\left(a^{\frac{-1-\sqrt{5-4 a}}{2}}, a^{\frac{-1+\sqrt{5-4 a}}{2}}\right) \cup\left(a^{\frac{1-\sqrt{5-4 a}}{2}}, a^{\frac{1+\sqrt{5-4 a}}{2}}\right) \cup\left(a^{\frac{1+\sqrt{5+4 a}}{2}},+\infty\right) .
$$
When $a \geqslant \frac{5}{4}$,
$$
x \in\left(0, a^{\frac{1-\sqrt{5+4 a}}{2}}\right) \cup\left(a^{\frac{-1+\sqrt{5+4 a}}{2}},+\infty\right) .
$$