\section*{Problem 3A - 201043A}

Determine all pairs \((x ; y)\) of real numbers with \(y>0\) and \(y \neq 1\), for which the following system of equations (1), (2) is satisfied!

\[
\begin{aligned}
x \log _{y}(\sqrt{3}-\sqrt{2})^{x} & =2 \log _{y}(5-\sqrt{24}) \\
y-x & =2
\end{aligned}
\]
}
Assume a pair \((x ; y)\) of real numbers with \(y>0\) and \(y \neq 1\) satisfies (1) and (2). Since

\[
(\sqrt{3}-\sqrt{2})^{2}=5-\sqrt{24} \quad \text { it follows that } \quad x^{2} \cdot \log _{y}(\sqrt{3}-\sqrt{2})=4 \cdot \log _{y}(\sqrt{3}-\sqrt{2})
\]

Since \(\sqrt{3} \neq \sqrt{2}\) and \(\sqrt{3}-\sqrt{2} \neq 1\), \(\log _{y}(\sqrt{3}-\sqrt{2})\) is defined and not equal to zero. This implies \(x^{2}=4\). Due to (2) and \(y>0\), \(x=y-2>-2\), so \(x\) must be 2. From (2), it follows that \(y=4\). Therefore, only the pair \((x ; y)=(2 ; 4)\) can satisfy the system of equations (1), (2). Since

\[
2 \cdot \log _{4}(\sqrt{3}-\sqrt{2})^{2}=2 \cdot \log _{4}(5-\sqrt{24})
\]

and \(4-2=2\), this pair indeed satisfies both equations. Hence, the pair \((2 ; 4)\) is the solution.