The count in a bacteria culture was 600 after 10 minutes and 1100 after 30 minutes. What was the initial size of the culture?Find the doubling period. Find the population after 80 minutes. When will the population reach 13000?
We can use any positive base we want, but since they are talking about doubling period, let's use
number = a (2)^(t/k) , where k will be the doubling period and t is in minutes, and a is the initial number
given: when t = 10 , number = 600
600 = a(2)^(10/k) ----- #1
when t= 30 , number = 1100
110 = a(2)^(30/k) ---- #2
divide #1 by #2
600/1100 = (a(2)^(10/k)) / (a(2)^(30/k))
6/11 = 2^(10/k - 30/k)
6/11 = 2^(-20/k)
take log of both sides
log 6 - log11 = -20/k log2 , using the rules of logs
-20/k = (log6 - log11)/log2 = -.874469...
k = 20/.874469 ..
= 22.871
So the doubling period is 22.871 minutes
we have to find the initial a
when t=10 , number = 600 , k= 22.871
600 = a(2)^(10/22.871)
a = 443
so amount = 443 (2)^(t/22.871)
when t = 80
number = 443 (2)^(80/22.871) = 50005 or appr 5000
when is number = 13000 ?
13000 = 443 2^(t/22.871)
29.34537.. = 2^(t/22.871)
log 29.34537 = (t/22.871 log2
t/22.871 = log 29.34../log2 = 4.87506..
t = 22.871(4.875..) = 111.4975 min
= appr 111.5 minutes
let's check our 5000 answer
t=0 , number = 443
t = appr 23 minutes , number = appr 886
t = appr 46 minutes, number = appr 1772
t = 69 minutes , number = appr 3544
t = 92 minutes , number = appr 7088
we had t - 80 for a number of 5000
makes sense
To find the initial size of the culture, we can use the formula for exponential growth:
N = N0 * (2^(t/d))
where:
N is the final count
N0 is the initial count
t is the time duration
d is the doubling period
We know that N0 is what we need to find, N is 600 after 10 minutes, and t is 10 minutes. Let's substitute these values:
600 = N0 * (2^(10/d))
To find the doubling period, we can use the difference between the two time periods:
doubling period = 30 minutes - 10 minutes = 20 minutes
Now, let's find the population after 80 minutes. We can use the same formula:
N = N0 * (2^(t/d))
Substituting the values, we have:
N = N0 * (2^(80/20)) = N0 * (2^4) = N0 * 16
To calculate, we need the initial count N0. Given that N is 1100 after 30 minutes, we can rearrange the formula to find N0:
1100 = N0 * (2^(30/d))
We know that d is 20 minutes, so let's substitute this value:
1100 = N0 * (2^(30/20)) = N0 * (2^(3/2)) = N0 * sqrt(2^3) = N0 * 2^(3/2)
Now let's solve for N0:
N0 = 1100 / (2^(3/2))
Lastly, to find when the population reaches 13000, we can rearrange the formula as:
13000 = N0 * (2^(t/d))
We know N0, and we need to solve for t. Substituting the values:
13000 = (1100 / (2^(3/2))) * (2^(t/20))
Simplifying:
13000 = 1100 * (2^((t/20) - (3/2))) = 1100 * (2^(t/20 - 3/2))
Now, let's solve for t.
To find the initial size of the bacteria culture, we can apply the concept of exponential growth. The formula for exponential growth is:
N(t) = N0 * 2^(t/d)
Where:
N(t) = final population size at time t
N0 = initial population size
t = time passed
d = doubling period
Given that the count was 600 after 10 minutes and 1100 after 30 minutes, we can use these values to set up two equations:
600 = N0 * 2^(10/d) ...(Equation 1)
1100 = N0 * 2^(30/d) ...(Equation 2)
Dividing Equation 2 by Equation 1, we get:
(1100/600) = 2^(30/d) / 2^(10/d)
Simplifying further:
(1100/600) = 2^((30-10)/d)
(1100/600) = 2^(20/d)
Now, we can solve for d by taking the logarithm of both sides:
log(1100/600) = log(2^(20/d))
Using logarithm properties, we can rewrite this as:
log(1100/600) = (20/d) * log(2)
Solving for d, we get:
d = (20 * log(2)) / log(1100/600)
Using a scientific calculator, we find that d ≈ 25.5 minutes. Therefore, the doubling period is approximately 25.5 minutes.
To find the population after 80 minutes, we can use Equation 1:
N(80) = N0 * 2^(80/d)
Substituting the values we know:
N(80) = N0 * 2^(80/25.5)
Using a calculator, we find that N(80) ≈ 4316.
To find when the population will reach 13000, we can set up the equation:
13000 = N0 * 2^(t/d)
Solving for t, we have:
t = d * log(13000/N0) / log(2)
Substituting the known values:
t = 25.5 * log(13000/N0) / log(2)
Using a calculator, we can solve for t.