Solve the following equation:

$$
\log (7 x-9)^{2}+\log (3 x-4)^{2}=2
$$
Our equation can also be written as:

$$
\log (7 x-9)+\log (3 x-4)=1
$$

which means

$$
(7 x-9)(3 x-4)=10
$$

or

$$
21 x^{2}-55 x+26=0
$$

from which

$$
x_{1}=2, \quad x_{2}=\frac{13}{21}
$$

(János Polyák, Zenta.)