Example 22. Solve the system

$$
\left\{\begin{array}{l}
\log _{y} x-\log _{x} y=8 / 3 \\
x y=16
\end{array}\right.
$$
Solution. The set of admissible values of $x$ and $y$ in the given system is defined by the system of inequalities: $x>0, x \neq 1, y>0$, $y \neq 1$. Letting $z=\log _{y} x$ and considering that for $x$ and $y$ from the domain of admissible values

$$
\log _{x} y=\frac{1}{\log _{y} x}
$$

we obtain the equation

$$
z-1 / z=8 / 3
$$

The set of all solutions to this equation consists of the numbers $z_{1}=3$ and $z_{2}=-1 / 3$.

Thus, the given system on its set of admissible values is equivalent to the combination of two systems:

$$
\left\{\begin{array} { l } 
{ \operatorname { l o g } _ { y } x = 3 , } \\
{ x y = 1 6 , }
\end{array} \quad \left\{\begin{array}{l}
\log _{y} x=-1 / 3 \\
x y=16
\end{array}\right.\right.
$$

From the equality $\log _{y} x=3$ it follows that $x=y^{3}$; therefore, the first system of this combination is equivalent on the domain of admissible values to the system

$$
\left\{\begin{aligned}
x & =y^{3} \\
x y & =16
\end{aligned}\right.
$$

From this we find $x=8, y=2$.

The second system of the combination is equivalent on the domain of admissible values to the system

$$
\left\{\begin{aligned}
x & =1 / \sqrt[3]{y} \\
x y & =16
\end{aligned}\right.
$$

From this we find $x=1 / 4, y=64$.

Thus, the set of all solutions to the original system consists of two ordered pairs: $(8 ; 2),\left(\frac{1}{4} ; 64\right)$.