Example 10. Solve the equation

$$
5^{2 x-1}=7^{3-x}
$$
Solution. Both sides of the equation are positive; therefore, it can be logarithmized with base 5. We obtain the equation

$$
2 x-1=(3-x) \log _{5} 7
$$

equivalent to the original one.

Thus,

$$
x\left(2+\log _{5} 7\right)=1+3 \log _{5} 7
$$

i.e., $x=\frac{1+3 \log _{5} 7}{2+\log _{5} 7}$.