A bacteria sample of certain number of bacteria is observed to grow exponentially in a given amount of time. Using exponential growth model, the rate…

$$\begin{aligned} & \frac{d P}{d \mathrm{t}}=k P \Rightarrow \int \frac{d P}{P}=\int k d \mathrm{t} \\ & \Rightarrow \log P=k \mathrm{t}+C \text { or } P=e^{k \mathrm{t}+C} \end{aligned} $$ $$\begin{aligned} & \log P=k \mathrm{t}+C \\ & \text { when } \mathrm{t}=0, P=1000 \Rightarrow C=\log 1000 \\ & \text { when } \mathrm{t}=1, P=2000 \Rightarrow \log 2000=k+\log 1000 \\ & \Rightarrow k=\log 2 \end{aligned} $$