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

Let \(x\) be the number of bacteria in certain culture at time \(t\). Then rate of increase is \(\frac{d x}{d t}\) which is proportional to \(x\).
\(\begin{array}{l}
\therefore \frac{\mathrm{dx}}{\mathrm{dt}} \propto \mathrm{x} \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{Kx} \Rightarrow \frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{Kdt} \Rightarrow \int \frac{\mathrm{dx}}{\mathrm{x}}=\mathrm{K} \int \mathrm{dt} \\
\therefore \log \mathrm{x}=\mathrm{Kt}+\mathrm{c} ...(1)\\
\text { Initially when } \mathrm{t}=0, \operatorname{let} \mathrm{x}=\mathrm{x}_{0} \\
\therefore \log \mathrm{x}=0+\mathrm{c} \Rightarrow \mathrm{c}=\log \mathrm{x}_{0} \\
\therefore \text { from }(1) \log \mathrm{x}=\mathrm{kt}+\log \mathrm{x}_{0} \\
\therefore \log \left(\frac{\mathrm{x}}{\mathrm{x}_{0}}\right)=\mathrm{Kt}...(2)
\end{array}\)
Given number doubles in 4 hrs i.e. when \(t=4, x=2 x_{0}\) \(\therefore \log \left(\frac{2 \mathrm{x}_{0}}{\mathrm{x}_{0}}\right)=4 \mathrm{~K} \Rightarrow \mathrm{K}=\frac{1}{4} \log 2\)
\(\therefore\) from (2) \(\log \left(\frac{x}{x_{0}}\right)=\frac{t}{4} \log 2\)
When \(\mathrm{t}=12\)
\(\log \left(\frac{x}{x_{0}}\right)=\frac{12}{4} \log 2=3 \log 2=\log 2^{3}=\log 8\)
\(\therefore \frac{x}{x_{0}}=8 \Rightarrow x=8 x_{0}\)
This problem can be alternatively solved as follows :
Let the intial number of bacteria be \(\mathrm{x}\).
\(\therefore\) Number of bacteria after \(4 \mathrm{hrs}=2 \mathrm{x}\)
Number of bacteria after \(8 \mathrm{hrs}=2(2 \mathrm{x})=4 \mathrm{x}\)
Number of bacteria after \(12 \mathrm{hrs}=2(4 \mathrm{x})=8 \mathrm{x}\)