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!
First you need to solve for k.
280 = 250 e^(10k)
1.12 = e^(10k)
ln 1.12 = 0.13929 = 10 k
k = 1.1333*10^-2
The doubling time T is given by
N/No = e^kT = 2
kT = ln 2 = 0.693
T = 0.693/k = 61.2 hours
To estimate the time required for the population to double in size, we can use the fact that when the population doubles, N = 2N₀, where N₀ is the initial population.
Since N = 280 when t = 10, we can substitute these values into the equation N = 250e^kt:
280 = 250e^(10k)
Next, we divide both sides of the equation by 250 to isolate the exponential term:
1.12 = e^(10k)
To solve for k, we take the natural logarithm (ln) of both sides of the equation:
ln(1.12) = ln(e^(10k))
Using the property of logarithms that ln(e^x) = x, we can simplify the equation to:
ln(1.12) = 10k
Now, divide both sides of the equation by 10 to solve for k:
k = ln(1.12) / 10 ≈ 0.11
At this point, we have found the value of k. To determine the time required for the population to double, we can use the fact that when the population doubles, N = 2N₀. Since N₀ is the initial population, when the population doubles, N will be 2N₀.
Let's denote the time required for doubling as t_d. Substituting N = 2N₀ and rearranging the original equation, we have:
2N₀ = 250e^(kt_d)
Now, divide both sides of the equation by N₀ to isolate the exponential term:
2 = (250 / N₀) * e^(kt_d)
Since we know that k ≈ 0.11, we can substitute this value into the equation:
2 = (250 / N₀) * e^(0.11t_d)
To find t_d, we can solve this equation for t_d.