The number N of bacteria in a culture is given by the model N=250e^kt where t is the time (in hours). If N=280 when t=10 estimate the time for the population to double in size.
To estimate the time for the population to double in size, we need to find the value of t when N=2N = 2(280) = 560.
Substitute these values into the equation N=250e^kt:
560 = 250e^k(10)
Divide both sides by 250:
2.24 = e^10k
To solve for k, take the natural logarithm of both sides:
ln(2.24) = ln(e^10k)
Use the property of logarithms to simplify:
ln(2.24) = 10k
Solve for k:
k = ln(2.24)/10
Now, we can substitute the value of k we found back into the original equation to find the time when N=560:
N = 250e^(ln(2.24)/10)*(t_2)
Setting N = 560:
560 = 250e^(ln(2.24)/10)*(t_2)
Divide both sides by 250 and multiply by 10:
2.24 = e^(ln(2.24)/10)*(t_2)
Take the natural logarithm of both sides:
ln(2.24) = ln(e^(ln(2.24)/10)*(t_2))
Use the property of logarithms to simplify:
ln(2.24) = (ln(2.24)/10)*(t_2)
Solve for t_2:
t_2 = (ln(2.24))/(ln(2.24)/10)
Simplifying:
t_2 = 10
Therefore, it will take approximately 10 hours for the population to double in size.
To estimate the time for the population to double in size, we need to find the value of t when N is twice its initial value.
Given the model N = 250e^(kt) and the information that N = 280 when t = 10, we can substitute these values into the equation to solve for k:
280 = 250e^(k*10)
Divide both sides of the equation by 250:
280/250 = e^(k*10)
1.12 = e^(k*10)
Now, take the natural logarithm (ln) of both sides of the equation to isolate k:
ln(1.12) = ln(e^(k*10))
Using the natural logarithm property ln(e^x) = x, we have:
ln(1.12) = k * 10
Solve for k:
k = ln(1.12) / 10
Now, to estimate the time for the population to double, we need to find the value of t when N is twice its initial value (2 * 250 = 500):
500 = 250e^(kt)
Divide both sides of the equation by 250:
2 = e^(kt)
Take the natural logarithm (ln) of both sides of the equation to isolate kt:
ln(2) = ln(e^(kt))
Using the natural logarithm property ln(e^x) = x, we have:
ln(2) = kt
Now, substitute the value of k we found earlier:
ln(2) = (ln(1.12) / 10) * t
Solve for t:
t = 10 * ln(2) / ln(1.12)
Calculating this expression will give us an estimate of the time for the population to double in size.
Plug in N and t and solve for k!
ln280=(ln250)*k*(10)
Then, plug in the k value and N=2 and solve for t!