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

$$
\left\{\begin{aligned}
& \log _{2}(y-x)=\log _{8}(19 y-13 x) \\
& x^{2}+y^{2}=13
\end{aligned}\right.
$$
# Solution:

The following equivalent transformations are valid:

$$
\left\{\begin{array} { c } 
{ y - x > 0 } \\
{ 19 y - 13 x > 0 } \\
{ ( y - x ) ^ { 3 } = 19 y - 13 x } \\
{ x ^ { 2 } + y ^ { 2 } = 13 }
\end{array} \Leftrightarrow \left\{\begin{array} { c } 
{ y - x > 0 } \\
{ ( y - x ) ^ { 3 } = 19 y - 13 x } \\
{ ( y - x ) ^ { 2 } = 13 - 2 x y }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
y-x>0 \\
(y-x)^{3}=19 y-13 x \\
(y-x)^{3}=(13-2 x y)(y-x)
\end{array} \Leftrightarrow\right.\right.\right.
$$

$$
\left\{\begin{array} { c } 
{ y - x > 0 } \\
{ ( y - x ) ^ { 3 } = 19 y - 13 x } \\
{ 19 y - 13 x = 13 y - 13 x - 2 x y ^ { 2 } + 2 x ^ { 2 } y }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
y-x>0 \\
(y-x)^{3}=19 y-13 x \Leftrightarrow \\
2 x y(y-x)=3 y
\end{array} \Leftrightarrow\right.\right.
$$

$$
\left[\begin{array}{c}
x 0 } \\
{ ( y - x ) ^ { 3 } = 19 ( y - x ) + 6 x } \\
{ 2 x ( y - x ) = 3 }
\end{array} \Leftrightarrow \left[\begin{array}{c}
y-x>0 \\
(y-x)^{3}=9(y-x)+\frac{3}{y-x}
\end{array} \Leftrightarrow\right.\right.
$$

$$
\left[\begin{array}{c}
\left\{\begin{array}{c}
x=-\sqrt{13} \\
y=0
\end{array}\right. \\
\left\{\begin{array}{c}
y-x>0
\end{array}\right. \\
(y-x)^{4}-9(y-x)^{2}+3=0 \\
x=\frac{3}{2(y-x)}
\end{array}\right.
$$

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

Answer: $\left\{(-\sqrt{13} ; 0) ;(-3 ;-2) ;\left(-\frac{1}{\sqrt{2}} ; \frac{5}{\sqrt{2}}\right)\right\}$.