Solve the equation $\frac{2003 x}{2004}=2003^{\log _{x} 2004}$.
Solution. Both sides are positive, so we can write the logarithm of both sides of the equation with base $x$ ($x>0$, $x \neq 1$):

$$
\log _{x} 2003+\log _{x} x-\log _{x} 2004=\log _{x} 2004 \cdot \log _{x} 2003
$$

We know that $\log _{x} x=1$, and let $\log _{x} 2003$ be $a$, and $\log _{x} 2004$ be $b$. Then we get the following equation:

$$
a+1-b=a b
$$

Transform it into a product:

$$
(a+1)(1-b)=0
$$

A product is zero if and only if one of its factors is zero.

If $a+1=\log _{x} 2003+1=0$, then $\log _{x} 2003=-1$ and $x=\frac{1}{2003}$.

If $1-b=1-\log _{x} 2004=0$, then $x=2004$.

We can verify by substitution that both values are solutions to the equation.