The population \(\mathrm{P}(\mathrm{t})\) of a certain mouse species at time \(\mathrm{t}\) satisfies the differential equation \(\frac{d P(t)}{d t}=0 \cdot 5 P(t)-450 \cdot\) If \(\mathrm{P}(0)=850\), then the time at which the population becomes zero is

\(\frac{\mathrm{dp}(\mathrm{t})}{\mathrm{dt}}=0.5 \mathrm{p}(\mathrm{t})-450\)
\(\frac{\mathrm{dp}(\mathrm{t})}{\mathrm{dt}}=\frac{1}{2} \mathrm{p}(\mathrm{t})-450 \Rightarrow \frac{\mathrm{dp}(\mathrm{t})}{\mathrm{dt}}=\frac{\mathrm{p}(\mathrm{t})-900}{2}\)
\(\therefore 2 \int \frac{\mathrm{d}[\mathrm{p}(\mathrm{t})]}{\mathrm{p}(\mathrm{t})-900}=\int \mathrm{dt} \Rightarrow 2 \log |\mathrm{p}(\mathrm{t})-900|=\mathrm{t}+\mathrm{c}\)
When \(\mathrm{t}=0, \mathrm{p}(\mathrm{t})=850\)
\(\therefore 2 \log |850-900|=0+\mathrm{c} \Rightarrow \mathrm{c}=2 \log 50\)
\(\therefore 2 \log |\mathrm{p}(\mathrm{t})-900|=\mathrm{t}+2 \log 50\)
When \(\mathrm{p}(\mathrm{t})=0\), we write
\(2 \log |0-900|=\mathrm{t}+2 \log 50\)
\(\therefore \mathrm{t}=2 \log \left|\frac{900}{50}\right|\)
\(\mathrm{t} \quad=2 \log 18\)