Two bacteria colonies are cultivated. The first colony has a doubling time of 2 hours and the second a doubling time of 3 hours. initially, the first colony contains 1,000 bacteria and the second colony 3,000 bacteria. At what time t will sizes of the colonies be equal?
Is doubling time = (ln2/k)?
Yes, that is the formula for doubling time in terms of the exponential growth facor k. But you don't need to use k to get the answer.
Populations are equal when
1000*2^(t/2) = 3000*2^(t/3)
3 = 2^[(t/2)-(t/3)] = 2^(t/6)
ln 3 = (t/6)*ln2
t/6 =ln3/ln2 = 1.58496
t = 9.51 hours
Yes, the formula for doubling time is commonly expressed as doubling time (T) = ln(2) / k, where k is the growth rate constant.
However, in this case, we already have the doubling times provided for the two bacteria colonies. So, we can use these doubling times directly to find the sizes of the colonies at a specific time when they become equal.
To solve this problem, we'll need to calculate the number of doubling periods for each colony and then determine the time when the colonies reach the same size.
Let's start with the first colony, which has a doubling time of 2 hours and an initial population of 1,000 bacteria.
Doubling time for the first colony = 2 hours
Initial population of the first colony = 1,000 bacteria
To find the number of doubling periods for the first colony, we can use the formula: Number of doubling periods = log (final population/ initial population) / log(2)
Since we want to calculate the time it takes for the first colony to catch up with the second colony (which has an initial population of 3,000 bacteria), we can set the final population as 3,000.
Number of doubling periods for the first colony = log (3,000 / 1,000) / log(2) = log(3) / log(2) = 1.585
Now, let's move on to the second colony, which has a doubling time of 3 hours and an initial population of 3,000 bacteria.
Doubling time for the second colony = 3 hours
Initial population of the second colony = 3,000 bacteria
Similarly, we can find the number of doubling periods for the second colony:
Number of doubling periods for the second colony = log (final population/ initial population) / log(2)
= log (3,000 / 3,000) / log(2)
= log(1) / log(2)
= 0 / log (2)
= 0
Now, we need to find the time it takes for the colonies to reach the same size.
Time taken for the first colony to catch up = Number of doubling periods for the first colony * Doubling time for the first colony
= 1.585 * 2
≈ 3.17 hours
Therefore, it will take approximately 3.17 hours for the sizes of the colonies to be equal.
Yes, the doubling time can be calculated using the formula t = (ln2) / k, where t is the doubling time and k is the growth rate constant.
To determine when the sizes of the colonies will be equal, we can set up equations for both colonies and solve for the time at which their sizes are equal.
Let's start with the first colony with a doubling time of 2 hours and an initial size of 1,000 bacteria.
The formula for the size of a colony at a given time is:
N(t) = N(0) * 2^(t / t_doubling)
where N(t) is the size of the colony at time t, N(0) is the initial size of the colony, and t_doubling is the doubling time.
For the first colony:
N1(t) = 1,000 * 2^(t / 2)
Now let's move on to the second colony with a doubling time of 3 hours and an initial size of 3,000 bacteria.
For the second colony:
N2(t) = 3,000 * 2^(t / 3)
We want to find the time t at which the sizes of the colonies are equal:
N1(t) = N2(t)
Substituting the equations for the colonies:
1,000 * 2^(t / 2) = 3,000 * 2^(t / 3)
To solve this equation, we can simplify:
2^(t / 2) = 3 * 2^(t / 3)
Using the property of exponents, we can rewrite the equation as:
2^(t / 2) = 2^(log2(3) * (t / 3))
Since the base is the same, the exponents can be equated:
t / 2 = log2(3) * (t / 3)
Now, we can solve for t:
Multiply both sides by 6 to remove fractions:
3t = 4log2(3)t
Divide both sides by t:
3 = 4log2(3)
Now, divide both sides by 4log2(3) to isolate t:
t = 3 / (4log2(3))
So, the time at which the sizes of the colonies will be equal is approximately t = 3 / (4log2(3)).