1) The number of bacteria N in a culture is given by N=250e^kt where t is the time in hours. If N = 280 when t=10 estimate the time required for the population to double in size.
280 = 250e^10k
divide by 250
1.12 = e^10k
ln 1.12 / 10 = .011 K = 0.11
Now I do not know what to do!
To estimate the time required for the population to double in size, we need to find the value of t when the number of bacteria N is twice the initial population size.
Since we know that N=250e^kt, we can set this equation equal to 2N and solve for t.
2N = 250e^kt
Substituting the given values N=280 and t=10:
2(280) = 250e^(10 * k)
Simplifying:
560 = 250e^(10 * k)
Dividing both sides by 250:
2.24 = e^(10 * k)
To solve for k, we need to take the natural logarithm (ln) of both sides:
ln 2.24 = ln e^(10 * k)
Using the property of logarithms, ln e^(10 * k) is equal to (10 * k):
ln 2.24 = 10 * k
Now divide both sides by 10 to isolate k:
k = ln 2.24 / 10
Using a calculator, we can find the value of k to be approximately 0.079.
Now, to estimate the time required for the population to double, we need to find the value of t when N is twice the initial population size of 250:
2N = 2(250) = 500
Substituting the known values of N=500 and k=0.079 into the original equation:
500 = 250e^(0.079 * t)
Dividing both sides by 250:
2 = e^(0.079 * t)
To solve for t, we need to take the natural logarithm (ln) of both sides:
ln 2 = 0.079 * t
Now divide both sides by 0.079 to isolate t:
t = ln 2 / 0.079
Using a calculator, we can find the value of t to be approximately 8.8 hours.
Therefore, it would take approximately 8.8 hours for the population to double in size from 250 to 500 bacteria.