In a certain culture of bacteria, their rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.

Answer: The bacteria become 8 times the original number in 12 hours.

Let \(x\) be the number of bacteria at time \(t\) hours. Then \(\dfrac{dx}{dt} \propto x\), i.e. \(\dfrac{dx}{dt} = kx\).

Separating the variables: \(\dfrac{dx}{x} = k\,dt\). Integrating, \(\log x = kt + c\).

At \(t = 0\), let \(x = x_0\): \(c = \log x_0\). So \(\log\dfrac{x}{x_0} = kt\), i.e. \(x = x_0e^{kt}\).

Number doubles in 4 hours: \(2x_0 = x_0e^{4k}\), so \(e^{4k} = 2\).

At \(t = 12\): \(x = x_0e^{12k} = x_0\left(e^{4k}\right)^3 = x_0 \cdot 2^3 = 8x_0\).

Hence the number of bacteria increases 8 times in 12 hours.