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.
First determine k from the information provided.
280 = 250*e^10k
ln(28/25) = 10 k
k = 0.01133 hr^-1
Now solve this equation for t:
500 = 250e^0.01133t
ln2 = .01133 t
t = 61.2 hours
Ah, bacteria, the masters of exponential growth! Let's find out when they'll be doubling their population in size.
So, we have the formula N = 250e^kt, where N is the number of bacteria and t is the time in hours. We know that N = 280 when t = 10.
Now, let's find k by plugging in these values into the equation:
280 = 250e^(k*10)
To make things a bit easier, let's divide both sides of the equation by 250:
1.12 = e^(10k)
To solve for k, we'll take the natural logarithm (ln) on both sides:
ln(1.12) = ln(e^(10k))
Using the property of logarithms that ln(e^x) is equal to x, we can simplify it to:
ln(1.12) = 10k
Now, let's divide both sides of the equation by 10:
k = ln(1.12) / 10
Alright, we've got k! Now, let's figure out how long it'll take for the population to double.
When the population doubles, N will be twice its initial value, so we can write:
2 * 250 = 250e^(k * t_double)
Simplifying this equation:
2 = e^(k * t_double)
Now, let's take the natural logarithm of both sides again to isolate t_double:
ln(2) = k * t_double
Finally, let's substitute the value of k that we found earlier:
ln(2) = (ln(1.12) / 10) * t_double
Solving for t_double:
t_double = 10 * ln(2) / ln(1.12)
Calculating this expression, we find:
t_double ≈ 20.67 hours
So, according to my calculations, it will take approximately 20.67 hours for the bacteria population to double in size. Keep an eye on those little rascals!
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.
Let's consider the initial value of N to be N0 when t = 0. Then, N0 = 250e^0 = 250.
We know that N doubles in size, so the new value of N is 2N0 = 2 * 250 = 500.
Now we can substitute this into the given equation:
500 = 250e^kt
Simplifying the equation, we get:
2 = e^kt
To solve for k, we can take the natural logarithm (ln) of both sides of the equation:
ln(2) = ln(e^kt)
Using the property ln(e^x) = x, we get:
ln(2) = kt
Solving for k, we divide both sides of the equation by t:
k = ln(2) / t
Now we can use the given information N=280 when t=10 to find the value of k:
280 = 250e^(k*10)
Dividing both sides of the equation by 250:
e^(10k) = 280/250
Taking the natural logarithm of both sides:
10k = ln(280/250)
Now we can solve for k:
k = ln(280/250) / 10
Substituting this value of k back into the equation k = ln(2) / t, we can solve for t:
ln(2) / t = ln(280/250) / 10
Multiplying both sides of the equation by t:
ln(2) = (ln(280/250) / 10) * t
Dividing both sides of the equation by ln(280/250):
t = ln(2) / (ln(280/250) / 10)
Evaluating this expression, we can 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 time when the number of bacteria reaches twice the initial population.
Let's start by finding the value of k.
Given the equation N = 250e^kt, we know that N = 280 when t = 10. Substituting these values into the equation, we get:
280 = 250e^(k * 10)
Divide both sides of the equation by 250:
280 / 250 = e^(k * 10)
Simplify:
1.12 = e^(10k)
To isolate k, take the natural logarithm (ln) of both sides:
ln(1.12) = ln(e^(10k))
Using the property ln(e^x) = x, we can simplify further:
ln(1.12) = 10k
Now, divide both sides of the equation by 10:
k = ln(1.12) / 10
Now that we have the value of k, we can find the time it takes for the population to double. Doubling the initial population means N = 500 (2 * 250).
Substituting this value into the equation N = 250e^kt, we get:
500 = 250e^(k * t)
Divide both sides of the equation by 250:
2 = e^(k * t)
Take the natural logarithm of both sides to isolate t:
ln(2) = ln(e^(k * t))
Using the property ln(e^x) = x, we can simplify further:
ln(2) = k * t
Now, rearrange the equation to solve for t:
t = ln(2) / k
Substituting the value of k from above, we have:
t = ln(2) / (ln(1.12) / 10)
Calculating this using a calculator:
t ≈ 6.055 hours
Therefore, it would take approximately 6.055 hours for the population to double in size.