Example 4 Given $x_{1}=1, x_{n}=\sqrt[4]{a x_{n-1}}(a>0)$. Find the general term formula $x_{n}$ of the sequence.
Since $x_{1}=1>0$, we have
$$
x_{2}>0, x_{3}>0, \cdots \cdots x_{n}>0 \text {. }
$$

Taking the common logarithm of both sides of $x_{n}=\sqrt[4]{a x_{n-1}}$, we get
$$
\lg x_{n}=\frac{1}{4}\left(\lg x_{n-1}+\lg a\right) \text {. }
$$

Let $b_{n}=\lg x_{n}, b_{n-1}=\lg x_{n-1}$. Then
$$
b_{n}=\frac{1}{4}\left(b_{n-1}+\lg a\right) \text {, }
$$

which is equivalent to $4 b_{n}=b_{n-1}+\lg a$.
Rearranging, we get
$4\left(b_{n}-\frac{1}{3} \lg a\right)=b_{n-1}-\frac{1}{3} \lg a$,

or $\frac{b_{n}-\frac{1}{3} \lg a}{b_{n-1}-\frac{1}{3} \lg a}=\frac{1}{4}$.
Given $b_{1}=\lg x_{1}=0$, we have
$b_{1}-\frac{1}{3} \lg a=-\frac{1}{3} \lg a$.
Thus, $b_{n}-\frac{1}{3} \lg a=\left(-\frac{1}{3} \lg a\right)\left(\frac{1}{4}\right)^{n-1}$.
Therefore, $b_{n}=\frac{1}{3} \lg a-\frac{1}{3} \lg a \cdot\left(\frac{1}{4}\right)^{n-1}$
$=\frac{1}{3}\left[1-\left(\frac{1}{4}\right)^{n-1}\right] \lg a$
$=\lg a^{\frac{1}{3}\left[1-\left(\frac{1}{4}\right)^{n-1}\right]}$.
Since $b_{n}=\lg x_{n}$, we have
$\lg x_{n}=\lg a^{\frac{1}{3}\left[1-\left(\frac{1}{4}\right)^{n-1}\right]}$.