## SUBJECT 1.

Solve in $(0 ;+\infty)$ the equation $a^{x} \cdot \log _{a} x=b^{x} \cdot \log _{b} x$, where $a>b>1$.
Subject 1.

$a^{x} \log _{a} x=b^{x} \frac{\log _{a} x}{\log _{a} b} \Leftrightarrow x=1$.....................................2p

Or

$$
\begin{aligned}
& a^{x}=\frac{b^{x}}{\log _{a} b} \Leftrightarrow\left(\frac{a}{b}\right)^{x}=\log _{b} a \\
& x=\log _{\frac{a}{b}}\left(\log _{b} a\right) \in(0 ;+\infty)
\end{aligned}
$$