The general function P(t)= 640ekt is used to model a dying bird population, where Po = 640 is the initial population and t is time measured in days. Suppose the bird population was reduced to one quarter of its initial size after 9 days. How long will it take before there are only 40 birds left in the population? Simplify your answer as much as possible.
PLease help and show all work Thank you!
I am positive that you meant:
P(t) = 640 e^(kt)
given: when t = 9, P(t) = 160
640 e^(9k) = 160
e^(9k) = .25
9k = ln .25
k = ln.25/9 = -.154033
640 e^( -.154033 t) = 40
e^(-.154033 t) = .0625
-.154033 t = ln .0625
t = ln .0625/-.154033 = 18 days
It took 9 days to shrink to 1/4 its size.
40/640 = 1/16 = 1/4 * 1/4
So, it will take another 9 days to do that again.
To solve this problem, we need to find the value of k in the function P(t) = 640e^kt and then use it to determine the time it takes for the population to reduce to 40.
Given that the bird population was reduced to one quarter (1/4) of its initial size after 9 days, we can write the following equation:
P(9) = 640e^9k = (1/4) * 640
Simplifying the right side:
(1/4) * 640 = 160
Then, we have:
640e^9k = 160
Dividing both sides by 640:
e^9k = 160/640 = 1/4
Taking the natural logarithm of both sides:
ln(e^(9k)) = ln(1/4)
Using the logarithmic property that ln(a^b) = b * ln(a):
9k * ln(e) = ln(1/4)
Since ln(e) = 1:
9k = ln(1/4)
Now, we need to solve for k, so divide both sides by 9:
k = ln(1/4) / 9
Using a calculator to evaluate ln(1/4):
k ≈ -0.046
Now that we have the value of k, we can determine the time it takes for the population to reduce to 40.
We need to solve for t in the equation P(t) = 40:
640e^kt = 40
Substituting the value of k:
640e^(-0.046t) = 40
Dividing both sides by 640:
e^(-0.046t) = 40/640 = 1/16
Taking the natural logarithm of both sides:
ln(e^(-0.046t)) = ln(1/16)
Using the logarithmic property, we have:
-0.046t * ln(e) = ln(1/16)
Since ln(e) = 1:
-0.046t = ln(1/16)
Dividing both sides by -0.046:
t = ln(1/16) / -0.046
Using a calculator to evaluate ln(1/16):
t ≈ 29.716
Therefore, it will take approximately 29.716 days for the bird population to reduce to 40.
To find out how long it will take before there are only 40 birds left in the population, we need to first determine the value of the growth constant 'k' in the general function.
Given information:
Initial population (Po) = 640
Population after 9 days (P(9)) = 640/4 = 160
We can use these values to solve for 'k' using the equation:
P(t) = 640e^(kt) --(Equation 1)
Substituting 9 for t and 160 for P(9) in Equation 1, we get:
160 = 640e^(9k)
Dividing both sides by 640, we have:
160/640 = e^(9k)
Simplifying further, we get:
1/4 = e^(9k)
To solve for 'k', we can take the natural logarithm of both sides:
ln(1/4) = ln(e^(9k))
Using the property of logarithms (ln(e^x) = x), the equation becomes:
ln(1/4) = 9k
Now we can solve for 'k' by dividing both sides by 9:
k = ln(1/4)/9
Using a calculator to evaluate this expression, we find:
k ≈ -0.0646 (rounded to four decimal places)
Now that we have the value of 'k', we can use it to find how long it will take before there are only 40 birds left in the population.
Let's plug in the given population (P(t) = 40) into the general function, and solve for 't':
40 = 640e^(-0.0646t)
Dividing both sides by 640 and rearranging the equation, we get:
40/640 = e^(-0.0646t)
Simplifying further, we have:
1/16 = e^(-0.0646t)
Taking the natural logarithm of both sides, we get:
ln(1/16) = ln(e^(-0.0646t))
Using the property ln(e^x) = x, the equation becomes:
ln(1/16) = -0.0646t
Now, let's solve for 't' by dividing both sides by -0.0646:
t = ln(1/16)/-0.0646
Using a calculator to evaluate this expression, we find:
t ≈ 30.05 (rounded to two decimal places)
Therefore, it will take approximately 30.05 days for the bird population to reduce to 40.