Rate of increase of bacteria in a culture is proportional to the number of bacteria present at that instant…
Rate of increase of bacteria in a culture is proportional to the number of bacteria present at that instant and it is found that the number doubles in 6 hours. The number of bacteria becomes times at the end of 18 hours.
9
6
8
3
Solution
Let $P_0$ be the initial population and let the population after $t$ years be $P$. Then,
$\begin{aligned}
& \frac{\mathrm{dP}}{\mathrm{dt}}=\mathrm{kP} \text {, where } \mathrm{k}>0 \\
& \Rightarrow \frac{\mathrm{dP}}{\mathrm{P}}=\mathrm{kdt}
\end{aligned}$
Integrating on both sides, we get
$\begin{aligned}
& \log \mathrm{P}=\mathrm{kt}+\mathrm{c} \\
& \text { When } \mathrm{t}=0, \mathrm{P}=\mathrm{P}_0 \\
& \therefore \quad \log \mathrm{P}_0=0+\mathrm{c} \\
& \Rightarrow c=\log \mathrm{P}_0 \\
& \log \mathrm{P}=\mathrm{kt}+\log \mathrm{P}_0 \\
& \Rightarrow \log \frac{\mathrm{P}}{\mathrm{P}_0}=\mathrm{kt} \\
& \begin{aligned}
\therefore \quad & \log \mathrm{P}=\mathrm{kt}+\log \mathrm{P}_0 \\
& \Rightarrow \log \frac{\mathrm{P}}{\mathrm{P}_0}=\mathrm{kt} \\
& \text { When } \mathrm{t}=6 \text { hrs, } \mathrm{P}=2 \mathrm{P}_0
\end{aligned} \\
& \therefore \quad \log \frac{2 \mathrm{P}_0}{\mathrm{P}_0}=6 \mathrm{k} \\
& \Rightarrow \mathrm{k}=\frac{\log 2}{6} \\
& \therefore \quad \log \frac{P}{P_0}=\frac{\log 2}{6} t \\
& \text { When } \mathrm{t}=18 \mathrm{hrs} \text {, we have } \\
& \log \frac{P}{P_0}=\frac{\log 2}{6} \times 18 \\
& =3 \log 2 \\
& \therefore \quad \log \frac{P}{P_0}=\log 8 \\
& \Rightarrow \mathrm{P}=8 \mathrm{P}_0 \\
&
\end{aligned}$