The doubling period of a baterial population is 20 minutes. At time t=100 minutes, the baterial population was 80000. What was the initial population at time t=0?
Find the size of the baterial population after 4 hours.
To answer these questions, we need to use the formula for exponential growth. The formula is given by:
N(t) = N(0) * 2^(t/d)
where:
- N(t) is the population size at time t
- N(0) is the initial population size
- t is the time elapsed
- d is the doubling period
Let's solve the first question first:
1. Doubling period of the bacterial population is 20 minutes and at time t=100 minutes, the population was 80000.
Using the formula, we have:
80000 = N(0) * 2^(100/20)
Simplifying the equation:
80000 = N(0) * 2^5
Divide both sides of the equation by 2^5:
N(0) = 80000 / 32
N(0) = 2500
Therefore, the initial population at time t=0 was 2500.
Now let's move on to the second question:
2. Finding the size of the bacterial population after 4 hours.
We know that 4 hours is equivalent to 4 * 60 = 240 minutes.
Using the formula, we have:
N(t) = N(0) * 2^(t/d)
N(t) = 2500 * 2^(240/20)
N(t) = 2500 * 2^12
Calculating the value of 2^12:
N(t) = 2500 * 4096
N(t) = 10,240,000
Therefore, the size of the bacterial population after 4 hours is 10,240,000.
To find the initial population at time t=0, we can use the formula for exponential growth:
N = N0 * 2^(t/d)
where:
N = final population size (80000)
N0 = initial population size at time t=0 (to be found)
t = time passed (100 minutes)
d = doubling period (20 minutes)
Let's solve for N0:
80000 = N0 * 2^(100/20)
80000 = N0 * 2^5
Dividing both sides of the equation by 2^5, we get:
N0 = 80000 / 2^5
N0 = 80000 / 32
N0 = 2500
Therefore, the initial population at time t=0 was 2500.
Now, let's find the size of the bacterial population after 4 hours. Since 1 hour consists of 60 minutes, 4 hours would be 4 * 60 = 240 minutes.
Using the same formula:
N = N0 * 2^(t/d)
where:
N0 = initial population size at time t=0 (2500)
t = time passed (240 minutes)
d = doubling period (20 minutes)
Let's solve for N:
N = 2500 * 2^(240/20)
N = 2500 * 2^12
Calculating 2^12, we get:
N = 2500 * 4096
N ≈ 10,240,000
Therefore, the size of the bacterial population after 4 hours is approximately 10,240,000.
Note the correct spelling of your subject.
Let the initial population be No and the doubling time be T (= 20 minutes)
No*2^(t/T) = No*2^5 = 80,000
32 No = 80,000
No = 2500
After 4 hours (t = 240 minutes),
N = No*2^(240/20) = 2500*2^12
= 10.24*10^6