4. (4 points) The population of the city increased by $50\%$ annually for $x$ years, then decreased by $\frac{388}{9}\%$ annually for $y$ years, and then decreased by $\frac{50}{3}\%$ annually for $z$ years. As a result, the population doubled. It is known that $x, y, z$ are integers. Find them.
Solution: Let $a=50, b=-\frac{388}{9}, c=-\frac{50}{3}$. By the condition $\left(1+\frac{a}{100}\right)^{x} \cdot\left(1+\frac{b}{100}\right)^{y} \cdot\left(1+\frac{c}{100}\right)^{z}=2$.

Adding the fractions in the parentheses and factoring the numerator and denominator into prime factors:

$$
1+\frac{a}{100}=2^{-1} \cdot 3, \quad 1+\frac{b}{100}=2^{7} \cdot 3^{-2} \cdot 5^{-2}, \quad 1+\frac{c}{100}=2^{-1} \cdot 3^{-1} \cdot 5
$$

Equating the exponents in the prime factorization of the left and right sides of the equation (1), we obtain a system of linear equations for $x, y, z$:

$$
\left\{\begin{array}{l}
-x+7 y-z=1 \\
x-2 y-z=0 \\
-2 y+z=0
\end{array}\right.
$$

Solving the system, we get

Answer: $x=4, y=1, z=2$.