A colony of bacteria is growing exponentially, increasing in size from 200 to 500 cells in 100 minutes will it take the colony to double in size?
p = ae^kt
500 = 200e^100k
2.5 = e^100k
ln 2.5 = 100k
k = ln2.5/100
p = 200e^(ln2.5/100 t)
= 200*2.5^t/100
to double,
2 = 2.5^t/100
ln 2 = (t/100)ln2.5
t = 100 * ln2/ln2.5 = 75.65 minutes
To determine how long it will take for the colony to double in size, we need to find the growth rate. In an exponential growth model, the equation can be represented as:
Nt = N0 * (1 + r)^t
Where:
Nt = final population size
N0 = initial population size
r = growth rate
t = time
In this case, the initial population size (N0) is 200 cells, and the final population size (Nt) is 500 cells.
Nt = N0 * (1 + r)^t
500 = 200 * (1 + r)^t
Dividing both sides by 200:
2.5 = (1 + r)^t
To determine the time it takes for the colony to double in size, we need to find the value of "t" that satisfies the equation. Taking the logarithm of both sides can help us solve for "t".
log(2.5) = log((1 + r)^t)
Using the property that log(a^b) = b * log(a):
log(2.5) = t * log(1 + r)
Now, we can solve for "t" by substituting the values into the equation. The growth rate (r) is unknown, so we can rearrange the equation to isolate r:
log(2.5) / t = log(1 + r)
Now we can find the value of log(2.5) / t and then solve for r by using the antilog:
r = antilog(log(2.5) / t) - 1
Let's calculate the growth rate (r):
r = antilog(log(2.5) / 100) - 1
r ≈ 0.0187
Since the growth rate (r) is now known, let's find the time it takes to double the colony size by substituting the value of r into the equation:
2 = (1 + 0.0187)^t
Taking the logarithm of both sides:
log(2) = log((1 + 0.0187)^t)
Using the property mentioned earlier:
log(2) = t * log(1 + 0.0187)
Now, solve for "t" by dividing both sides of the equation:
t = log(2) / log(1 + 0.0187)
Using a calculator:
t ≈ 37.1 minutes
So, it will take approximately 37.1 minutes for the colony to double in size.
To determine how long it will take for the colony to double in size, we need to find the doubling time. The doubling time is the amount of time it takes for a population or quantity to double in size.
In an exponential growth model, the general formula is:
N = N0 * (2^(t / T))
Where:
N = final population/size
N0 = initial population/size
t = time
T = doubling time
In this case, we know the initial population (N0) is 200 cells, the final population (N) is 500 cells, and the time (t) is 100 minutes. We need to find the doubling time (T).
We can rearrange the formula to solve for T:
T = t / log2(N/N0)
Plugging in the values:
T = 100 / log2(500/200)
We can use natural logarithm (ln) to calculate this, as log2(x) = ln(x) / ln(2):
T = 100 / (ln(500/200) / ln(2))
After evaluating this expression:
T ≈ 100 / (0.678 / 0.693)
T ≈ 100 / 0.978
T ≈ 102.25 (rounded to nearest hundredth)
Therefore, the doubling time for this colony of bacteria is approximately 102.25 minutes.
To answer the original question, how long will it take for the colony to double in size? It will take approximately 102.25 minutes.