The rate of growth of bacteria in a culture i | vedclass

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(A) Let $B(t)$ be the number of bacteria at time $t$. The rate of growth is given by $\frac{dB}{dt} = \lambda B$.Integrating this,we get $B(t) = B_0 e

Vedclass · Accepted Answer

$4$

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(A) Let $B(t)$ be the number of bacteria at time $t$. The rate of growth is given by $\frac{dB}{dt} = \lambda B$.Integrating this,we get $B(t) = B_0 e^{\lambda t}$,where $B_0 = 1000$.Given that at $t = 2$,$B(2) = 1000 + 20\% 	ext{ of } 1000 = 1200$.So,$1200 = 1000 e^{2\lambda} \Rightarrow e^{2\lambda} = \frac{6}{5} \Rightarrow 2\lambda = \log_{e}\left(\frac{6}{5}ight) \Rightarrow \lambda = \frac{1}{2} \log_{e}\left(\frac{6}{5}ight)$.We are given $B(T) = 2000$ where $T = \frac{k}{\log_{e}\left(\frac{6}{5}ight)}$.Using $B(T) = B_0 e^{\lambda T}$,we have $2000 = 1000 e^{\lambda T} \Rightarrow 2 = e^{\lambda T} \Rightarrow \log_{e} 2 = \lambda T$.Substituting $\lambda$ and $T$: $\log_{e} 2 = \left(\frac{1}{2} \log_{e}\left(\frac{6}{5}ight)ight) 	imes \left(\frac{k}{\log_{e}\left(\frac{6}{5}ight)}ight) = \frac{k}{2}$.Thus,$k = 2 \log_{e} 2$.Finally,$\left(\frac{k}{\log_{e} 2}ight)^{2} = \left(\frac{2 \log_{e} 2}{\log_{e} 2}ight)^{2} = 2^{2} = 4$.

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