Let P(t) represent the number of bacteria present at time t in 0.5 hours.

The statement that the number of bacteria doubles every 0.5 hours can be written as P(t+1) = 2*P(t)

P(0) = 1500

P(1) = 2*P(0)

P(2) = 2*P(1) = 2*2*P(0) = 2^2 * P(0)

P(3) = 2*P(2) = 2*2^2*P(0) = 2^3 * P(0)

P(4) = 2*P(3) = 2*2^3*P(0) = 2^4 * P(0)

We conjecture the following formula

P(t) = 2^t P(0)

Let’s verify that this function satisfies the condition that every 0.5 hours it’s size doubles:

P(t + 1) = 2^(t+1) P(0) = 2*2^t P(0) = 2* P(t)

Now we have to find the size of the bacterial population after 5 hours and for that purpose we have to covert 5 hours into half an hours:

5 hours = 2*5 half hours = 10 halfhours

P(10) = 2^10 P(0) = (2^10)* 1500

P(10) = 1,536,000 = 1.536*10^6

after 5 hours where will be 1.536*10^6 bacteria.