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.

Let \(P_0\) be the initial population and let the population after \(t\) years be \(P\). Then,
\(\frac{\mathrm{dP}}{\mathrm{dt}}=\mathrm{kP} \text {, where } \mathrm{k}>0 \)
\(\Rightarrow \frac{\mathrm{dP}}{\mathrm{P}}=\mathrm{kdt}\)
Integrating on both sides, we get
\(\log \mathrm{P}=\mathrm{kt}+\mathrm{c} \)
\(\text {When } \mathrm{t}=0, \mathrm{P}=\mathrm{P}_0 \)
\(\therefore \log \mathrm{P}_0=0+\mathrm{c} \)
\(\Rightarrow c=\log \mathrm{P}_0 \)
\(\log \mathrm{P}=\mathrm{kt}+\log \mathrm{P}_0 \)
\(\Rightarrow \log \frac{\mathrm{P}}{\mathrm{P}_0}=\mathrm{kt} \)
\(\therefore \log \mathrm{P}=\mathrm{kt}+\log \mathrm{P}_0 \)
\(\Rightarrow \log \frac{\mathrm{P}}{\mathrm{P}_0}=\mathrm{kt} \)
\(\text {When } \mathrm{t}=6 \text {hrs, } \mathrm{P}=2 \mathrm{P}_0 \)
\(\therefore \log \frac{2 \mathrm{P}_0}{\mathrm{P}_0}=6 \mathrm{k} \)
\(\Rightarrow \mathrm{k}=\frac{\log 2}{6} \)
\( \therefore \log \frac{P}{P_0}=\frac{\log 2}{6} t \)
\(\text {When } \mathrm{t}=18 \mathrm{hrs} \text {, we have } \)
\(\log \frac{P}{P_0}=\frac{\log 2}{6} \times 18 \)
\(=3 \log 2 \)
\(\therefore \log \frac{P}{P_0}=\log 8 \)
\(\Rightarrow \mathrm{P}=8 \mathrm{P}_0\)