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

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$1250$

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(A) Let $x$ be the number of bacteria present at time $t$.The rate of increase is proportional to the number of bacteria,so $\frac{dx}{dt} = kx$.Separating variables,we get $\frac{dx}{x} = k dt$.Integrating both sides,we get $\log x = kt + c$ ... $(i)$.Given that at $t = 3$,$x = 10,000$,we have $\log(10,000) = 3k + c$ ... $(ii)$.Given that at $t = 5$,$x = 40,000$,we have $\log(40,000) = 5k + c$ ... $(iii)$.Subtracting $(ii)$ from $(iii)$,we get $\log(40,000) - \log(10,000) = 2k$,which simplifies to $\log(4) = 2k$,so $k = \log(2)$.Substituting $k = \log(2)$ into $(ii)$,we get $\log(10,000) = 3\log(2) + c$.Thus,$c = \log(10,000) - \log(8) = \log(\frac{10,000}{8}) = \log(1250)$.Initially,at $t = 0$,$\log x = k(0) + c$,so $\log x = \log(1250)$.Therefore,the initial number of bacteria is $x = 1250$.

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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