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(D) Let $x$ be the number of bacteria present at time $t$.The rate of increase is proportional to the number present,so $\frac{dx}{dt} = kx$.Separatin

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$\frac{10^4}{8}$

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(D) Let $x$ be the number of bacteria present at time $t$.The rate of increase is proportional to the number present,so $\frac{dx}{dt} = kx$.Separating variables and integrating,we get $\ln x = kt + c$,or $x = e^{kt+c} = Ae^{kt}$,where $A = e^c$ is the initial number of bacteria at $t=0$.At $t=3$,$x = 10^4$,so $10^4 = Ae^{3k}$ ... $(i)$At $t=5$,$x = 4 \cdot 10^4$,so $4 \cdot 10^4 = Ae^{5k}$ ... (ii)Dividing (ii) by $(i)$: $\frac{4 \cdot 10^4}{10^4} = \frac{Ae^{5k}}{Ae^{3k}} \Rightarrow 4 = e^{2k} \Rightarrow e^k = 2$.Substitute $e^k = 2$ into $(i)$: $10^4 = A(e^k)^3 = A(2)^3 = 8A$.Therefore,$A = \frac{10^4}{8}$.

In a certain culture of bacteria,the rate of increase is proportional to the number present. If there are $10^4$ at the end of $3$ hours and $4 \cdot 10^4$ at the end of $5$ hours,then there were $\qquad$ in the beginning.

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In a certain culture of bacteria,the rate of increase is proportional to the number of bacteria present at that instant. It is found that there are $10,000$ bacteria at the end of $3$ hours and $40,000$ bacteria at the end of $5$ hours. Find the number of bacteria present in the beginning.

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