1. Solve the equation

$$
\log _{x} a+\log _{\sqrt{x}} a+\ldots .+\log _{\sqrt[8]{x}} a=\log _{a} x, a>0
$$
Solution. We have

$$
\begin{aligned}
& \log _{x} a+\log _{\sqrt{x}} a+\ldots .+\log _{\sqrt[8]{x}} a=\log _{a} x, a>0 \\
& \log _{x} a+2 \log _{x} a+\ldots .+8 \log _{x} a=\log _{a} x, x>0 \\
& 36 \log _{x} a=\log _{a} x \\
& 36 \log _{a}^{2} x=1 \\
& \log _{a} x= \pm \frac{1}{6}
\end{aligned}
$$

The solutions to the equation are: $x=\sqrt[6]{a}$ and $x=\frac{1}{\sqrt[6]{a}}$.