Question

In a culture, the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{bacteria}\mathrm{count}\mathrm{at}\mathrm{any}\mathrm{time}t\mathrm{be}N. \\ \mathrm{Given}:\hspace{0.17em}\frac{dN}{dt}\alpha N \\ \Rightarrow \frac{dN}{dt}=\lambda N \\ \Rightarrow \frac{1}{N}dN=\lambda dt \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{N}dN=\int \lambda dt \\ \Rightarrow \log N=\lambda t+\log \mathrm{C}.....\left(1\right) \\ \mathrm{Initially}\mathrm{when}t=0,\mathrm{then}N=100000\left(\mathrm{Given}\right) \\ \therefore \log 100000=0+\log \mathrm{C} \\ \Rightarrow \log \mathrm{C}=\log 100000 \\ \mathrm{After}2\mathrm{hours}\mathrm{number}\mathrm{increased}\mathrm{by}10\% \\ \mathrm{Therefore},\mathrm{increased}\mathrm{number}=100000\left(1+10\%\right)=110000 \\ \mathrm{Given}:t=2,N=110000 \\ \mathrm{Putting}t=2,N=110000\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log 110000=2\lambda +\log 100000 \\ \Rightarrow \frac{1}{2}\log \frac{11}{10}=\lambda \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}\mathrm{of}\log C\mathrm{and}\lambda \mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ \log N=\frac{t}{2}\log \left(\frac{11}{10}\right)+\log 100000.....\left(2\right) \\ \mathrm{Now}, \\ \mathrm{Let}t=T\mathrm{when}N=200000 \\ \mathrm{Substituting}\mathrm{these}\mathrm{values}\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ \log 200000=\frac{T}{2}\log \left(\frac{11}{10}\right)+\log 100000 \\ \Rightarrow \log 2=\frac{T}{2}\log \frac{11}{10} \\ \Rightarrow T=2\frac{\log 2}{{\displaystyle \log \frac{11}{10}}} \\ \therefore \mathrm{The}\mathrm{count}\mathrm{will}\mathrm{reach}200000\mathrm{in}2\frac{\log 2}{{\displaystyle \log \frac{11}{10}}}\mathrm{hours}. \end{gathered}$$