The rate of increase of bacteria in a culture | vedclass

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(C) Let $P_0$ be the initial population and $P$ be the population at time $t$. Given $\frac{dP}{dt} = kP$,where $k > 0$. Separating variables and inte

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$8$

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(C) Let $P_0$ be the initial population and $P$ be the population at time $t$. Given $\frac{dP}{dt} = kP$,where $k > 0$. Separating variables and integrating,we get $\ln P = kt + C$. At $t = 0$,$P = P_0$,so $C = \ln P_0$. Thus,$\ln \left( \frac{P}{P_0} ight) = kt$. Given that at $t = 6$,$P = 2P_0$,we have $\ln(2) = 6k$,so $k = \frac{\ln 2}{6}$. At $t = 18$,$\ln \left( \frac{P}{P_0} ight) = \left( \frac{\ln 2}{6} ight) 	imes 18 = 3 \ln 2 = \ln(2^3) = \ln 8$. Therefore,$\frac{P}{P_0} = 8$,which means the number of bacteria becomes $8$ times the initial number.

The rate of increase of bacteria in a culture is proportional to the number of bacteria present at that instant. It is found that the number doubles in $6 \text{ hours}$. The number of bacteria becomes $n$ times the initial number at the end of $18 \text{ hours}$. Find $n$.

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