Solve the following equation for $x$:

$$
\sqrt{\log _{a} \sqrt[4]{a x}+\log _{x} \sqrt[4]{a x}}+\sqrt{\log _{a} \sqrt[4]{\frac{x}{a}}+\log _{x} \sqrt[4]{\frac{a}{x}}}=a
$$
According to the equation, both $a$ and $x$ are suitable to be the base of a logarithm, meaning they are positive numbers different from 1. (This, of course, represents an assumption for the parameter $a$, and a requirement for the unknown $x$.) Thus, $\log _{a} a=1, \log _{x} x=1,\left(\log _{a} x\right) \cdot\left(\log _{x} a\right)=1$, using these and other known identities of the logarithmic operation, we can write an equivalent equation to (1) in which only $\log _{a} x$ appears as the unknown:

$$
\sqrt{\frac{1}{4}\left(1+\log _{a} x\right)+\frac{1}{4}\left(\frac{1}{\log _{a} x}+1\right)}+\sqrt{\frac{1}{4}\left(\log _{a} x-1\right)+\frac{1}{4}\left(\frac{1}{\log _{a} x}-1\right)}=a
$$

Thus, for the number $\log _{a} x=y \neq 0$ as the unknown, the equation equivalent to the previous one must be satisfied:

$$
\begin{gathered}
\sqrt{\frac{y^{2}+2 y+1}{4 y}}+\sqrt{\frac{y^{2}-2 y+1}{4 y}}=a, \quad \text { that is } \\
\frac{|y+1|}{2 \sqrt{y}}+\frac{|y-1|}{2 \sqrt{y}}=a
\end{gathered}
$$

Therefore, only positive numbers can be accepted as $y$.

I. We solve the last equation by case analysis, first looking for a solution where $0 < y < 1$, which implies $0 < \log _{a} x < 1$, and thus $0 < a < 1$. (It cannot be $a^{2}=1$, since $a>0$ and $a \neq 1$.

II. If only numbers greater than 1 are accepted as $y$, then from (2)

$$
\frac{2 y}{2 \sqrt{y}}=\sqrt{y}=a, \quad y=a^{2}
$$

provided $a^{2}>1$.

Both obtained values of $y=\log _{a} x$ are positive, and their validity condition is $a>1$. In this case, (1) has two solutions:

$$
x_{1}=a^{y}=a^{a^{-2}} \quad \text { and } \quad x_{2}=a^{a^{2}}
$$

indeed, given $y>0$ and $a>1$, it holds that $x_{1,2}>1$. If, however, $a \leqq 1$, then the equation has no solution.

There is no need for verification, as we performed equivalent transformations everywhere based on the introduced assumptions.