Assume that the number of bacteria follows an exponential growth model: P(t)=P0ekt
. The count in the bacteria culture was 900 after 20 minutes and 1600 after 30 minutes.
(a) What was the initial size of the culture?
(b) Find the population after 80 minutes.
(c) How many minutes after the start of the experiment will the population reach 10000?
Po e^(20k) = 900
Po e^(30k) = 1600
e^(10k) = 16/9
k = 0.019
And you can take it from there, I assume.
To find the initial size of the culture (P0), we can use the given information.
(a)
We are given:
P(t) = 900 at t = 20 minutes
P(t) = 1600 at t = 30 minutes
Using the exponential growth model P(t) = P0 * e^(kt), we can set up two equations using the given information:
900 = P0 * e^(20k) Equation 1
1600 = P0 * e^(30k) Equation 2
To solve these equations, divide Equation 2 by Equation 1:
1600 / 900 = (P0 * e^(30k)) / (P0 * e^(20k))
Simplifying, we get:
1.7778 = e^(10k)
Now, take the natural logarithm (ln) of both sides to solve for k:
ln(1.7778) = ln(e^(10k))
ln(1.7778) = 10k
Divide both sides by 10:
ln(1.7778) / 10 = k
Now we have k, we can substitute it back into one of the original equations to solve for P0. Let's use Equation 1:
900 = P0 * e^(20k)
Substituting the value of k, we get:
900 = P0 * e^(20 * (ln(1.7778)/10))
Simplifying the exponent, we get:
900 = P0 * e^(2 * ln(1.7778))
900 = P0 * (e^(ln(1.7778))^2)
Using the property e^(ln(x)) = x:
900 = P0 * (1.7778^2)
900 = P0 * 3.1604
Dividing both sides by 3.1604:
P0 = 284.9305
Therefore, the initial size of the culture (P0) is approximately 284.9305.
(b) To find the population after 80 minutes, we can use the exponential growth model:
P(t) = P0 * e^(kt)
Substituting the values:
P(t) = 284.9305 * e^(80k)
We already found k in part a:
P(t) = 284.9305 * e^(80 * (ln(1.7778) / 10))
Simplifying the exponent, we get:
P(t) = 284.9305 * e^(8 * ln(1.7778))
P(t) = 284.9305 * (e^(ln(1.7778))^8)
Using the property e^(ln(x)) = x:
P(t) = 284.9305 * (1.7778^8)
P(t) = 284.9305 * 9.272
P(t) ≈ 2642.29
Therefore, the population after 80 minutes is approximately 2642.29 bacteria.
(c) To find out how many minutes after the start of the experiment the population reaches 10,000, we can set up the equation:
10000 = P0 * e^(kt)
Substituting the values:
10000 = 284.9305 * e^(k * t)
We know P0 and k from part a:
10000 = 284.9305 * e^(ln(1.7778)/10 * t)
Now, divide both sides by 284.9305:
10000 / 284.9305 = e^(ln(1.7778)/10 * t)
Simplifying, we get:
35.11 = e^(ln(1.7778)/10 * t)
Take the natural logarithm (ln) of both sides:
ln(35.11) = ln(e^(ln(1.7778)/10 * t))
ln(35.11) = (ln(1.7778)/10 * t)
Now solve for t by dividing both sides by (ln(1.7778)/10):
t = ln(35.11) / (ln(1.7778)/10)
t ≈ 96.45
Therefore, the population will reach 10,000 bacteria approximately 96.45 minutes after the start of the experiment.
To find the initial size of the bacteria culture (P0), we can use the given information that the count was 900 after 20 minutes. We'll substitute these values into the exponential growth model and solve for P0.
Given:
P(t) = P0 * e^(kt)
P(20) = 900
Substituting the values:
900 = P0 * e^(k * 20)
To isolate P0, divide both sides by e^(k * 20):
900 / e^(k * 20) = P0
Now we have the formula for P0 in terms of k.
To find the value of k, we'll use the information that the count was 1600 after 30 minutes. We'll substitute these values into the exponential growth model and solve for k.
Given:
P(t) = P0 * e^(kt)
P(30) = 1600
Substituting the values:
1600 = P0 * e^(k * 30)
To eliminate P0, divide both sides by the equation we found earlier for P0:
(1600 / 900) = (P0 * e^(k * 30)) / (P0 * e^(k * 20))
Simplifying:
(1600 / 900) = e^(k * (30-20))
Taking the natural logarithm (ln) of both sides:
ln(1600 / 900) = k * (30-20)
Now we can solve for k by dividing both sides by (30-20) and calculating the natural logarithm:
k = ln(1600/900) / 10
With the value of k found, we can now answer the questions.
(a) The initial size of the culture (P0) is given by:
P0 = 900 / e^(k * 20)
(b) To find the population after 80 minutes, substitute the values into the exponential growth model:
P(80) = P0 * e^(k * 80)
(c) To find how many minutes after the start of the experiment the population reaches 10000, substitute the values into the exponential growth model and solve for t:
10000 = P0 * e^(k * t)