A bacteria culture has an initial population of 600. After 4 hours the population has grown to 1200. Assuming the culture grows at a rate proportional to the size of the population, find the function representing the population size after t hours and determine the size of the population after 8 hours.

Question

A bacteria culture has an initial population of 600. After 4 hours the population has grown to 1200. Assuming the culture grows at a rate proportional to the size of the population, find the function representing the population size after t hours and determine the size of the population after 8 hours.

Answer 1

To find the function representing the population size after t hours, we can use the information given that the culture grows at a rate proportional to its current population.

Let P(t) represent the population size after t hours. We know that the population starts at 600, so we have P(0) = 600.

We are also given that after 4 hours, the population has grown to 1200. So we have P(4) = 1200.

To find the function, we can use the general form of exponential growth, which is P(t) = P(0) * e^(kt), where e is the base of the natural logarithm and k is the constant of proportionality.

Using the given information, we can substitute the known values into the equation to find the exponential growth function for this bacteria culture:

P(0) = 600
P(4) = 1200

Substituting these values into the equation, we get:

600 * e^(4k) = 1200

Dividing both sides by 600, we have:

e^(4k) = 2

To solve for k, we take the natural logarithm of both sides:

ln(e^(4k)) = ln(2)

Using the logarithm property that ln(e^x) = x, we can simplify the equation:

4k = ln(2)

Dividing both sides by 4, we have:

k = ln(2)/4

Now that we know the value of k, we can substitute it back into the equation for P(t):

P(t) = 600 * e^(kt)

P(t) = 600 * e^((ln(2)/4)t)

Now we can determine the size of the population after 8 hours by substituting t = 8 into the equation:

P(8) = 600 * e^((ln(2)/4)*8)

P(8) = 600 * e^(2ln(2))

Using the property that e^(a*b) = (e^a)^b, we can simplify further:

P(8) = 600 * (e^(ln(2)))^2

Since e^(ln(x)) = x, we have:

P(8) = 600 * 2^2 = 600 * 4 = 2400.

Therefore, the size of the population after 8 hours is 2400.

Answer 2

Use

amount = a e^(kt) where a is the initial value, k is a constant, and t is in hours

1200 = 600 e^(k(4)
2 = e^(4k)
4k = ln 2
k = ln2 /4

amount = 600 e^((ln2/4)t)

when t=8
amount = 600 e^(2ln2)
= 2400

well, duh
Since the doubling period seems to be 4 hours
in 8 hours they would have doubled twice
600 --> 1200 --> 2400