4. (4 points) Solve the system of equations:

$$
\left\{\begin{array}{c}
\log _{2}(y-x)=\log _{8}(9 y-15 x) \\
x^{2}+y^{2}=15
\end{array}\right.
$$
# Solution:

The following equivalent transformations are valid:

$$
\begin{aligned}
& \left\{\begin{array} { c } 
{ y - x > 0 } \\
{ 9 y - 1 5 x > 0 } \\
{ ( y - x ) ^ { 3 } = 9 y - 1 5 x } \\
{ x ^ { 2 } + y ^ { 2 } = 1 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { c } 
{ y - x > 0 } \\
{ ( y - x ) ^ { 3 } = 9 y - 1 5 x } \\
{ ( y - x ) ^ { 2 } = 1 5 - 2 x y }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
y-x>0 \\
(y-x)^{3}=9 y-15 x \\
(y-x)^{3}=(15-2 x y)(y-x)
\end{array} \Leftrightarrow\right.\right.\right. \\
& \left\{\begin{array} { c } 
{ y - x > 0 } \\
{ ( y - x ) ^ { 3 } = 9 y - 1 5 x } \\
{ 9 y - 1 5 x = 1 5 y - 1 5 x - 2 x y ^ { 2 } + 2 x ^ { 2 } y }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
y-x>0 \\
(y-x)^{3}=9 y-15 x \Leftrightarrow \\
2 x y(y-x)=6 y
\end{array}\right.\right. \\
& {\left[\begin{array} { c } 
{ x  0 } \\
{ ( y - x ) ^ { 3 } = 9 ( y - x ) - 6 x } \\
{ 2 x ( y - x ) = 6 }
\end{array} \Leftrightarrow \left[\begin{array}{c}
\left\{\begin{array}{c}
x=-\sqrt{15} \\
y=0
\end{array}\right. \\
\left\{\begin{array}{c}
y-x>0 \\
(y-x)^{3}=9(y-x)-\frac{18}{y-x} \\
x=\frac{3}{y-x}
\end{array}\right.
\end{array} \Leftrightarrow\right.\right.}
\end{aligned}
$$

$$
\left[\begin{array}{c}
\left\{\begin{array}{c}
x=-\sqrt{15} \\
y=0
\end{array}\right. \\
\left\{\begin{array}{c}
y-x>0
\end{array}\right. \\
(y-x)^{4}-9(y-x)^{2}+18=0 \\
x=\frac{3}{y-x}
\end{array}\right.
$$

By solving the resulting biquadratic equation in terms of $y-x$ after some transformations, the answer can be obtained.

Answer: $\left\{(-\sqrt{15} ; 0) ;\left(\sqrt{\frac{3}{2}} ; 3 \sqrt{\frac{3}{2}}\right) ;(\sqrt{3} ; 2 \sqrt{3})\right\}$.