Solve the following equation: $\log _{2010}(2009 x)=\log _{2009}(2010 x)$.
Solution. The domain of interpretation of the problem is: $x>0$. Switching to base 2009:

$$
\frac{\log _{2009}(2009 x)}{\log _{2009} 2010}=\log _{2009}(2010 x)
$$

We can multiply by the non-zero denominator:

$$
\log _{2009}(2009 x)=\log _{2009} 2010 \cdot \log _{2009}(2010 x)
$$

Apply the known relationship for the logarithm of a product:

$$
\begin{aligned}
\log _{2009} 2009+\log _{2009} x & =\log _{2009} 2010 \cdot\left(\log _{2009} 2010+\log _{2009} x\right), \\
1+\log _{2009} x & =\log _{2009}^{2} 2010+\log _{2009} 2010 \cdot \log _{2009} x, \\
\log _{2009} x & =-\frac{\log _{2009}^{2} 2010-1}{\log _{2009} 2010-1} \\
\log _{2009} x & =-\log _{2009} 2010-1, \\
\log _{2009} x & =\log _{2009} 2010^{-1}-\log _{2009} 2009, \\
\log _{2009} x & =\log _{2009} \frac{1}{2009 \cdot 2010} .
\end{aligned}
$$

Due to the strict monotonicity of the $\log _{2009} x$ function, this holds if and only if

$$
x=\frac{1}{2009 \cdot 2010}
$$

which satisfies the constraint. (It can be verified by checking that the calculations are correct.)