Math Qustion?

Question

The doubling period of a baterial population is 10 minutes. At time t = 90 minutes, the baterial population was 90000. What was the initial population at time t =0?
And Find the size of the baterial population after 5 hours?

Answer 1

The population can be modelled using the following equation.
P(t) = P(initial) * 2^(t/10)

This means at time t = 0, you have the initial population. At time t = 10, you have double (2^10/10 = 2^1 = 2 times) the population. At time t = 20, you have quadruple (2^20/2 = 2^2 = 4 times) the population, etc.

You know P(90) = 90000:
So plug this in:
P(90) = 90000 = P(initial) * 2^(90/10)

90000 = P(initial) * 2^9

Solve for P(initial):
P(initial) = 90000 / 512
P(initial) = 175.78125

Here are the details:
P(0) = 175.78125
P(10) = 351.5625
P(20) = 703.125
P(30) = 1,406.25
P(40) = 2,812.5
P(50) = 5,625
P(60) = 11,250
P(70) = 22,500
P(80) = 45,000
P(90) = 90,000

You can also do this working backwards, just dividing by 2. So at time t = 80, you have half the population (45,000) at t = 70 you have half of that (22,500), etc. Until you get to t = 0.

So the answer is:
Initial population = P(0) = 175.78125 ≈ 176 bacteria

To figure the population at 5 hours (300 minutes), just plug in t = 300:
P(300) = 175.78125 * 2^(300/10)
P(300) ≈ 188,743,680,000 or about 189 billion bacteria