Example 2. Solve the equation:
a) $0.2 \log _{x} \frac{1}{32}=-0.5$;
b) $\log _{x-1} 3=2$;
c) $\log _{\log _{3}} x=2$.
Solution. a) Since $\frac{1}{32}=\frac{1}{2^{5}}=\left(\frac{1}{2}\right)^{5}$, then $\log _{x} \frac{1}{32}=$ $=5 \log _{x} \frac{1}{2}$, and therefore, the original equation is equivalent to the equation

$$
\log _{x} \frac{1}{2}=-\frac{1}{2}
$$

from which $x=(1 / 2)^{-2}=4$. The number 4 is the only root of the original equation.

b) The original equation is equivalent to the equation

$$
x-1=3^{1 / 2}
$$

from which $x=1+\sqrt{3}$. The number $1+\sqrt{3}$ is the only root of the given equation.

c) The domain of the original equation is determined by the system

$$
\left\{\begin{array}{l}
\log _{3} x>0 \\
\log _{3} x \neq 1
\end{array}\right.
$$

Therefore, the original equation is equivalent to the system

$$
\left\{\begin{array}{l}
\log _{3} x>0 \\
\log _{3} x \neq 1, \Leftrightarrow x=3^{\sqrt{3}} \\
\log _{3} x=3^{1 / 2}
\end{array}\right.
$$

The number $3^{\sqrt{3}}$ is the only root of the original equation.