The rate of growth of bacteria in a  culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time t=0. The number of bacteria is increased by 20% in 2 hours. If the population of bacteria is 2000 after \frac{k}{log_{e}\left ( \frac{6}{5} \right )} hours, then \left ( \frac{k}{log_{e}2} \right )^{2} is equal to :
 
Option: 1 2
Option: 2 4
Option: 3 8
Option: 4 16

Answers (1)

Initial bacteria count = 1000

20% bacteria increased in 2 hours = 1200

\\\frac{\mathrm{dB}}{\mathrm{dt}}=\lambda \mathrm{B}\\ \Rightarrow \int_{1000}^{1200} \frac{\mathrm{dB}}{\mathrm{B}}=\lambda \int_{0}^{2} \mathrm{dt} \\\Rightarrow \lambda=\frac{1}{2} \ell \mathrm{n}\left(\frac{6}{5}\right)

\\\int_{1000}^{2000} \frac{\mathrm{dB}}{\mathrm{B}}=\frac{1}{2} \ln \left(\frac{6}{5}\right) \int_{0}^{\mathrm{T}} \mathrm{dt}\\ \Rightarrow \mathrm{T}=\frac{2 \ell \mathrm{n} 2}{\ln \left(\frac{6}{5}\right)} \\ \Rightarrow \mathrm{k}=2 \ln 2

\\\left(\frac{\mathrm{k}}{\log _{\mathrm{e}} 2}\right)^{2}=\left(\frac{\mathrm{2\ln 2}}{\log _{\mathrm{e}} 2}\right)^{2}=4\\\ln 2a\text{ and }\log_e 2\;\text{is same thing}

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