Given data:
Initial population = 1,500
For every 33 minutes the population doubles.
The first step will be to write a function that models this statement
The model is given as

The next step will be to find the population of the bacteria at time t=66 minutes.
This will be obtained by substituting t=66 into the equation


=>

=>

Therefore the population of the bacteria after 66 minutes will be 6,000