The doubling period of a bacterial population is 10 minutes. At time t=90
minutes, the bacterial population was 80000.
What was the initial population at time t=0?
Incorrect
Find the size of the bacterial population after 3 hours
t0 = 80000 * (1/2)^9
t180 = 80000 + 2^9
count = initial * 2^(t/10) , where t is in minutes
80000 = initial * (2^9)
initial = 8000/2^9 = 156.25
(mathematically possible, but we can't have partial bacteria, but anyway.....)
count after 3 hours, t = 180
count = 156.25 (2^18) = 40,960,000
correction...hit the wrong key
t180 = 80000 × 2^9
To find the initial population at time t=0, we need to use the doubling time formula:
N = N₀ * (2^(t / T))
Where:
N₀ is the initial population
N is the final population
t is the time in minutes
T is the doubling period in minutes
Given that the doubling period is 10 minutes and the population at t=90 minutes is 80000, we can substitute these values into the formula to find N₀:
80000 = N₀ * (2^(90 / 10))
Simplifying the equation:
80000 = N₀ * (2^9)
To find N₀, we can rearrange the equation:
N₀ = 80000 / (2^9)
Calculating this, we find:
N₀ = 80000 / 512
N₀ = 156.25
So, the initial population at time t=0 is approximately 156.
Now, let's find the size of the bacterial population after 3 hours (180 minutes).
Using the formula again:
N = N₀ * (2^(t / T))
Substituting the values:
N = 156.25 * (2^(180 / 10))
Calculating this, we find:
N ≈ 156.25 * 512
N ≈ 80000
Therefore, the size of the bacterial population after 3 hours is approximately 80000.