1. Find all values of $x$, for each of which one of the three given numbers $\log _{x}\left(x-\frac{1}{3}\right)$, $\log _{x-\frac{1}{3}}(x-3)$, and $\log _{x-3} x$ is equal to the product of the other two.
Answer: $x=\frac{10}{3}, x=\frac{3+\sqrt{13}}{2}$.

Solution. Note that on the domain of definition, the product of all three logarithms is

$$
\log _{x}\left(x-\frac{1}{3}\right) \cdot \log _{x-\frac{1}{3}}(x-3) \cdot \log _{x-3} x=1
$$

Let the number that is the product of the other two be denoted by $c$, and the two remaining numbers by $a$ and $b$. Then $c=a b$ and $a b c=1$, from which it follows that $c^{2}=1, c= \pm 1$, i.e., one of the three given logarithms is equal to $\pm 1$.

The converse is also true, namely, if one of the three given logarithms is equal to $\pm 1$, then since the product of all three logarithms is 1, the product of the other two is also $\pm 1$, i.e., equal to the first logarithm.

Note that for all three given logarithms, the base and the argument are never equal for any $x$, so they are different from 1.

Thus, the requirement of the problem is satisfied if and only if one of the logarithms is equal to -1 (and all logarithms exist). A logarithm is equal to -1 when the product of its base and argument is equal to 1. We obtain the system

$$
\left[\begin{array} { l } 
{ x ( x - \frac { 1 } { 3 } ) = 1 , } \\
{ x ( x - 3 ) = 1 , } \\
{ ( x - 3 ) ( x - \frac { 1 } { 3 } ) = 1 }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=\frac{1 \pm \sqrt{37}}{6} \\
x=\frac{3 \pm \sqrt{13}}{2} \\
x=\frac{10}{3} \\
x=0
\end{array}\right.\right.
$$

Of the found values of the variable, only $x=\frac{10}{3}$ and $x=\frac{3+\sqrt{13}}{2}$ satisfy the domain of definition.