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

5 years = 5*365 days = 5*365*24 hours = 5*365*24*2 half an hours = 87600 half an hours

P(87600) = 2^87600 P(0) = 2^87600 * 1500

let

x = 2^87600 * 1500 apply logarithm with base 10 on both sides of the equation

lg x = 87600 lg2 + lg 1500

lg x = 87600 * 0.30102999566398114 + 3.176091259055681

lg x = 26373.404

x = 10^26373.4037

x = (10^0.4037) * (10^26373)

x = 2.53 * (10^26373)

after 5 years where will be 2.53 * (10^26373) bacteria.