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

Let the bacteria count at any time t be N. 

\[\text{ Given:- }\hspace{0.167em} \frac{dN}{dt}\alpha N\]

\[ \Rightarrow \frac{dN}{dt} = \lambda N\]

\[ \Rightarrow \frac{1}{N}dN = \lambda dt\]

Integrating both sides, we get

\[\int\frac{1}{N}dN = \int\lambda dt\]

\[ \Rightarrow \log N = \lambda t + \log C . . . . . . . . . (1)\]

\[\text{ Initially when }t = 0,\text{ then }N = 100000 .............\left(\text{Given }\right)\]

\[ \therefore \log 100000 = 0 + \log C\]

\[ \Rightarrow \log C = \log 100000\]

After 2 hours number increased by 10 % 

Therefore, increased number = 100000 1 + 10 % = 110000

\[\text{ Given: }t = 2, N = 110000\]

\[\text{ Putting }t = 2, N = 110000\text{ in }(1),\text{ we get }\]

\[\log 110000 = 2\lambda + \log 100000\]

\[ \Rightarrow \frac{1}{2}\log \frac{11}{10} = \lambda\]

\[\text{Substituting the values of }\log C\text{ and }\lambda\text{ in }(1),\text{ we get }\]

\[\log N = \frac{t}{2}\log \left( \frac{11}{10} \right) + \log 100000 . . . . . . . (2)\]

Now, 

Let t = T when N = 200000

Substituting these values in (2), we get 

\[\log 200000 = \frac{T}{2}\log \left( \frac{11}{10} \right) + \log 100000\]

\[ \Rightarrow \log 2 = \frac{T}{2}\log \frac{11}{10}\]

\[ \Rightarrow T = 2\frac{\log 2}{\log\frac{11}{10}}\]

\[ \therefore\text{ The count will reach 200000 in }2\frac{\log 2}{\log\frac{11}{10}}\text{ hours .}\]