# 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

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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