Question

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.

Answer 1

Let the original count of bacteria be N and the count of bacteria at any time t be P.
Given: $\frac{dP}{dt}\alpha P$
$$\begin{gathered} \Rightarrow \frac{dP}{dt}=aP \\ \Rightarrow \frac{dP}{P}=adt \\ \Rightarrow \log \left|P\right|=at+C.....\left(1\right) \\ \mathrm{Now}, \\ P=N\mathrm{at}t=0 \\ \mathrm{Putting}P=N\mathrm{and}t=0\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log \left|N\right|=C\mathit{} \\ \mathrm{Putting}C=\log \left|N\right|\mathit{}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log \left|P\right|=at+\log \left|N\right| \\ \Rightarrow \log \left|\frac{P}{N}\right|=at.....\left(2\right) \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}, \\ \log \left|\frac{3N}{N}\right|=5a \\ \Rightarrow a=\frac{1}{5}\log \left|3\right|=\frac{1}{5}\times 1.0986=0.21972 \\ \mathrm{Putting}a=0.21972\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ \log \left|\frac{P}{N}\right|=0.21972t.....\left(3\right) \\ \Rightarrow e^{0.21972t}=\frac{P}{N}.....\left(4\right) \\ \mathrm{Putting}t=10\mathrm{in}\left(4\right)\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{bacteria}\mathrm{after}10\mathrm{hours},\mathrm{we}\mathrm{get} \\ {e}^{0.21972\times 10}=\frac{P}{N} \\ \Rightarrow e^{2.1972}=\frac{P}{N} \\ \Rightarrow \frac{P}{N}=9 \\ \Rightarrow P=9N \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{time}\mathrm{taken}\mathrm{when}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{bacteria}\mathrm{becomes}10\mathrm{times}\mathrm{of}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{initial}\mathrm{population},\mathrm{we}\mathrm{have} \\ P=10N \\ \therefore \log \left|\frac{10N}{N}\right|=\frac{1}{5}t\log 3 \\ \Rightarrow t=\frac{5\log 10}{\log 3} \end{gathered}$$