Suppose an E. coli culture is growing exponentially at 37 ◦C. After 20 minutes at that temperature, there are 1.28×10^7 E. coli cells. After 60 minutes, there are 2.4×10^7 cells. How long does it take for the culture to have double the amount of cells that it had started with?
I do not know where to begin. I tried using the equation P(x)=Po e^kt but I don't know what to do because there would be two values for t.
You can do it that way, but I think simpler math will do..
amountfinal=amoutintial*(e)^.692 time/doubletime
amountfinal/amountintial=e^.692(t/doublettime)
so you are given two points...
2.4E7/1.28E7 = e^.692*40/timedouble
I am assuming the "after 60minutes"means 40 min after the first reading.
take the ln of each side
ln( 2.4/1.28)=.692*40/timedouble
solve for time to double.
Can I ask why you are using 0.692 in the exponent?
To solve this problem, you need to find the value of time when the culture has double the amount of cells it started with. Let's break down the problem step by step:
Step 1: Find the values for "Po" and "P(x)".
In this case, "Po" represents the initial number of cells, and "P(x)" represents the number of cells at a given time.
Given information:
After 20 minutes: P(20) = 1.28×10^7 cells
After 60 minutes: P(60) = 2.4×10^7 cells
Step 2: Use the exponential growth equation.
The general formula for exponential growth is: P(x) = Po * e^(kt), where:
- P(x) is the population at time x
- Po is the initial population
- k is the growth rate constant
- t is the time elapsed
We need to find the value of "t" when the population (P(x)) doubles from the initial population (Po).
Step 3: Set up the equation.
Since we want to find the value of "t" when the population doubles, we can write the equation as follows:
2 * Po = Po * e^(kt)
The initial population cancels out, and we are left with:
2 = e^(kt)
Step 4: Take the natural logarithm.
To solve for "t," we need to isolate it by taking the natural logarithm (ln) of both sides of the equation:
ln(2) = ln(e^(kt))
Using the logarithmic property ln(e^(kt)) = kt, we have:
ln(2) = kt
Step 5: Solve for "t."
Divide both sides of the equation by "k" to solve for "t":
t = ln(2) / k
Step 6: Calculate the value of "k."
To calculate "k," we can use the information provided for two different time points (20 minutes and 60 minutes).
For the first time point, when t = 20 minutes, we have:
1.28×10^7 = Po * e^(20k)
For the second time point, when t = 60 minutes, we have:
2.4×10^7 = Po * e^(60k)
Divide the second equation by the first equation to eliminate Po:
2.4×10^7 / 1.28×10^7 = e^(60k) / e^(20k)
Simplifying, we get:
1.875 = e^(60k - 20k)
1.875 = e^(40k)
Take the natural logarithm (ln) of both sides:
ln(1.875) = ln(e^(40k))
Using the logarithmic property ln(e^(40k)) = 40k, we have:
ln(1.875) = 40k
Step 7: Solve for "k."
Divide both sides of the equation by 40 to solve for "k":
k = ln(1.875) / 40
Step 8: Substitute the value of "k" into the equation to find "t."
With the value of "k" from Step 7, substitute it into the equation derived in Step 5 to find "t":
t = ln(2) / (ln(1.875) / 40)
Using a calculator, solve this equation to find the value of "t." The answer will give you the time it takes for the culture to double the amount of cells that it had started with.