\section*{Problem 2 - 091042}

For the real numbers \(a, b\) with \(a>0, b>0\) and \(a \neq 1, b \neq 1\), determine all numbers \(x\) that satisfy the equation \(\left(\log _{a} x\right)\left(\log _{b} x\right)=\log _{a} b\).
}

It holds that \(\ln _{b} a=\frac{\ln a}{\ln b}\) and thus

\[
\begin{aligned}
\left(\log _{a} x\right)\left(\log _{b} x\right)=\log _{a} b & \Longleftrightarrow \frac{\ln x}{\ln a} \cdot \frac{\ln x}{\ln b}=\frac{\ln b}{\ln a} \Longleftrightarrow\left(\frac{\ln x}{\ln b}\right)^{2}=\left(\log _{b} x\right)^{2}=1 \\
& \Longleftrightarrow x=b \vee x=\frac{1}{b}
\end{aligned}
\]

Second solution:

Assume a real number \(x\) satisfies the equation. Since \(\log _{b} x=\frac{\log _{a} x}{\log _{a} b}\), it follows that

\[
\log _{a} x \frac{\log _{a} x}{\log _{a} b}=\log _{a} b \quad \text { so } \quad\left(\log _{a} x\right)^{2}=\left(\log _{a} b\right)^{2}
\]

and therefore either

\[
\log _{a} x=\log _{a} b \quad \text { or } \quad \log _{a} x=-\log _{a} b=\log _{a} \frac{1}{b}
\]

From (1) it follows \(x=b\), from (2) \(x=\frac{1}{b}\). Therefore, only the numbers \(b\) and \(\frac{1}{b}\) can satisfy the given equation.

Indeed, these two numbers are solutions to the equation; for we have

\[
\log _{a} b \log _{b} b=\log _{a} b \cdot 1=\log _{a} b \quad, \quad \log _{a} \frac{1}{b} \log _{b} \frac{1}{b}=\left(-\log _{a} b\right) \cdot(-1)=\log _{a} b
\]

Adapted from \([2]\)