Task B-3.7. Solve the system of equations

$$
\begin{aligned}
& x=2^{\frac{4}{1+\log _{2} z}} \\
& y=2^{\frac{4}{1+\log _{2} x}} \\
& z=2^{\frac{9}{1+\log _{2} y}}
\end{aligned}
$$
Solution. By taking the logarithm (base 2) of the given equations, the system transforms into

$$
\begin{aligned}
& \log _{2} x=\frac{4}{1+\log _{2} z} \\
& \log _{2} y=\frac{4}{1+\log _{2} x} \\
& \log _{2} z=\frac{9}{1+\log _{2} y}
\end{aligned}
$$

where $x, y, z$ are positive real numbers different from $\frac{1}{2}$.

After introducing the substitution

$$
\begin{aligned}
a & =\log _{2} x \\
b & =\log _{2} y \\
c & =\log _{2} z
\end{aligned}
$$

the system transforms into

$$
\begin{aligned}
a & =\frac{4}{1+c} \\
b & =\frac{4}{1+a} \\
c & =\frac{9}{1+b}
\end{aligned}
$$

If we substitute $a$ from the first equation into the second equation, the system reduces to two equations with two unknowns

$$
\begin{aligned}
b\left(1+\frac{4}{1+c}\right) & =4 \\
c & =\frac{9}{1+b}
\end{aligned}
$$

After simplifying, we get

$$
\left\{\begin{array} { l l } 
{ b ( \frac { 5 + c } { 1 + c } ) } & { = 4 } \\
{ 1 + b } & { = \frac { 9 } { c } }
\end{array} \quad \text { or } \quad \left\{\begin{array}{ll}
b & =4 \cdot \frac{1+c}{5+c} \\
b & =\frac{9}{c}-1
\end{array}\right.\right.
$$

This leads to the equation

$$
4 \cdot \frac{1+c}{5+c}=\frac{9}{c}-1
$$

From this, we get $c^{2}=9, c_{1}=3, c_{2}=-3$.

From the first solution $a_{1}=1, b_{1}=2, c_{1}=3$ it follows that $\log _{2} x=1, \log _{2} y=2, \log _{2} z=3$

From the second solution $a_{2}=-2, b_{2}=-4, c_{2}=-3$ it follows that $\log _{2} x=-2, \log _{2} y=-4, \log _{2} z=-3$

or $x_{2}=\frac{1}{4}, y_{2}=\frac{1}{16}, z_{2}=\frac{1}{8}$.