The doubling period of a bacterial population is 20 minutes. At time $t=120$ minutes, the bacterial population was 80000 . What was the initial population at time $t=0$ ? Find the size of the bacterial population after 5 hours.

Step 1 ：The doubling period of the bacterial population is 20 minutes, which means the population doubles every 20 minutes. This is an example of exponential growth, and the formula for exponential growth is \(P(t) = P_0 * 2^{(t/T)}\), where \(P(t)\) is the population at time \(t\), \(P_0\) is the initial population, \(t\) is the time, and \(T\) is the doubling period.

Step 2 ：Given that \(P(120) = 80000\) and \(T = 20\), we can substitute these values into the formula to find \(P_0\), the initial population.

Step 3 ：The initial population \(P_0\) is calculated to be 1250. This is the population at time \(t=0\).

Step 4 ：Now, to find the size of the bacterial population after 5 hours, we can substitute \(t=5*60\) (since 5 hours is 300 minutes) into the exponential growth formula.

Step 5 ：The size of the bacterial population after 5 hours is 40960000.

Step 6 ：Final Answer: The initial population at time \(t=0\) was \(\boxed{1250}\). The size of the bacterial population after 5 hours is \(\boxed{40960000}\).