13.450 If at the beginning of the time measurement there were $m_{0}$ g of substance $A$ and $2 m_{0}$ g of substance $B$, then after any number $t$ years, as a result of the radioactive decay of these substances, there will remain respectively $\boldsymbol{m}=\boldsymbol{m}_{0} \cdot 2^{-\lambda_{1} t}$ and $M=2 m_{0} \cdot 2^{-\lambda_{2} t}$, where $\lambda_{1}$ and $\lambda_{2}$ are constants depending on the nature of the substances. Calculate the half-life of each of these substances, i.e., find out after how many years only half of the initial amount of each substance will remain, if it is known that the half-life of substance $B$ is two times less than that of substance $A$, and that after 20 years the total mass of these substances decreases by 8 times.
Solution. Let $x$ and $2x$ be the half-lives of the second and first substances, respectively. According to the conditions,

$$
\left\{\begin{array}{l}
m_{0} \cdot 2^{-\lambda_{1} \cdot 2x} = \frac{m_{0}}{2}, \\
2m_{0} \cdot 2^{-\lambda_{2} \cdot x} = m_{0}, \\
m_{0} \cdot 2^{-\lambda_{1} \cdot 20} + 2m_{0} \cdot 2^{-\lambda_{2} \cdot 20} = \frac{3m_{0}}{8}
\end{array} \Leftrightarrow \left\{\begin{array}{l}
2^{-2\lambda_{1} x + 1} = 1 \\
2^{-\lambda_{2} \cdot x + 1} = 1 \\
8 \cdot 2^{-20\lambda_{1}} + 16 \cdot 2^{-20\lambda_{2}} = 3
\end{array}\right.\right.
$$

From the first two equations, it follows that $\lambda_{1} = \frac{1}{2x}, \lambda_{2} = \frac{1}{x}$.

Substituting these values into the third equation, we get: $16 \cdot 2^{-\frac{20}{x}} + 8 \cdot 2^{-\frac{10}{x}} - 3 = 0$.

Let $y = 2^{-\frac{10}{x}}$, then we have:

$16y^{2} + 8y - 3 = 0 \Rightarrow y = 2^{-2} \Rightarrow 2^{-\frac{10}{x}} = 2^{-2} \Rightarrow x = 5$.

Answer: 10 years, 5 years.