In a culture,the bacteria count is 1,00,000 i | vedclass

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(A) Let $N$ be the number of bacteria at time $t$. Given that the rate of growth is proportional to the number present,we have $\frac{dN}{dt} = kN$. I

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$\frac{2 \log 2}{\log(1.1)}$

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(A) Let $N$ be the number of bacteria at time $t$. Given that the rate of growth is proportional to the number present,we have $\frac{dN}{dt} = kN$. Integrating this,we get $\ln N = kt + C$,or $N(t) = N_0 e^{kt}$. Initially,at $t = 0$,$N_0 = 1,00,000$. After $2$ hours,the count increases by $10 \%$,so $N(2) = 1,00,000 + 0.10 	imes 1,00,000 = 1,10,000$. Substituting these values: $1,10,000 = 1,00,000 e^{2k} \implies e^{2k} = 1.1 \implies 2k = \ln(1.1) \implies k = \frac{\ln(1.1)}{2}$. We want to find $t$ such that $N(t) = 2,00,000$. $2,00,000 = 1,00,000 e^{kt} \implies 2 = e^{kt} \implies \ln 2 = kt$. Substituting $k$: $\ln 2 = \left(\frac{\ln(1.1)}{2}ight) t$. Solving for $t$: $t = \frac{2 \ln 2}{\ln(1.1)} = \frac{2 \log 2}{\log(1.1)}$.

In a culture,the bacteria count is $1,00,000$ initially. The number increases by $10 \%$ in the first $2$ hours. In how many hours will the count reach $2,00,000$,if the rate of growth of bacteria is proportional to the number present?

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