## Task B-3.6.

Determine all pairs of real numbers $(x, y)$ that are solutions to the system of equations

$$
\left\{\begin{array}{l}
y^{5 x^{2}-51 x+10}=1 \\
2 \log x+2 \log y=\log 9+\log 25
\end{array}\right.
$$
## Solution.

The conditions on the solutions are $x>0, y>0$.

The given system is equivalent to the following sequence of systems:

$$
\begin{aligned}
\left\{\begin{array}{l}
y^{5 x^{2}-51 x+10}=1, \\
\log x^{2}+\log y^{2}=\log 225 .
\end{array}\right. & \Longrightarrow\left\{\begin{array} { l } 
{ y ^ { 5 x ^ { 2 } - 5 1 x + 1 0 } = 1 } \\
{ \operatorname { log } ( x y ) ^ { 2 } = \operatorname { log } 1 5 ^ { 2 } }
\end{array} \Longrightarrow \left\{\begin{array}{l}
y^{5 x^{2}-51 x+10}=1 \\
(x y)^{2}=15^{2}
\end{array}\right.\right. \\
& \Longrightarrow\left\{\begin{array}{l}
y^{5 x^{2}-51 x+10}=1 \\
x y=15
\end{array}\right.
\end{aligned}
$$

From the equation $y^{5 x^{2}-51 x+10}=1$ it follows that $y=1$ or $5 x^{2}-51 x+10=0$.

If $y=1$, then $x=15$, so one solution is the ordered pair $(15,1)$.

If $5 x^{2}-51 x+10=0$, then $x=10$ or $x=\frac{1}{5}$.

From $x=10$ it follows that $y=\frac{3}{2}$, and from $x=\frac{1}{5}$ it follows that $y=75$.

Therefore, the solutions of the given system are the ordered pairs $(15,1),\left(10, \frac{3}{2}\right)$ and $\left(\frac{1}{5}, 75\right)$.