1. Solve the equation

$$
\sqrt{\log _{a} \sqrt[4]{a x}+\log _{x} \sqrt[4]{a x}}+\sqrt{\log _{a} \sqrt[4]{\frac{x}{a}+\log _{x} \sqrt[4]{\frac{a}{x}}}}=a
$$

where \( a \) is a positive real number different from 1.
Solution. Note that $x \neq 1$ and $x>0$ (for $\log _{x} \sqrt[4]{a x}$ to exist). By converting the logarithms to base $a$, the equation will transform into

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

and with further transformations, it takes the form:

$$
\sqrt{\frac{\left(\log _{a} x+1\right)^{2}}{4 \log _{a} x}}+\sqrt{\frac{\left(\log _{a} x-1\right)^{2}}{4 \log _{a} x}}=a
$$

From this, $\log _{a} x>0$ (due to the non-negativity of the expressions under the square roots). The last equation is equivalent (due to $x \neq 1$, i.e., $\log _{a} x \neq 0$) to

$$
\left|\log _{a} x+1\right|+\left|\log _{a} x-1\right|=2 a \sqrt{\log _{a} x}
$$

To solve equation (1), we will consider two cases:

i) Let $\log _{a} x>1$, then the equation transforms into $\log _{a} x=a \sqrt{\log _{a} x}$. From this, we obtain the solution $x_{1}=a^{a^{2}}$. Furthermore, due to $\log _{a} x>1$, we get that $a>1$.
ii) Now let $0<\log _{a} x \leq 1$, then the equation transforms into $2=2 a \sqrt{\log _{a} x}$. From this, we obtain the solution $x_{2}=a^{\frac{1}{a^{2}}}$, and due to $\log _{a} x \leq 1$, we get that $a>1$. In this way, we have exhausted all possibilities (since $\log _{a} x \leq 0$). Therefore, the initial equation for $a>1$ has two solutions $x_{1}=a^{a^{2}}, x_{2}=a^{\frac{1}{a^{2}}}$, while for $0<a<1$ the equation has no solution.