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