Frogs have been breeding like flies at the Enormous State University (ESU) campus! Each year, the pledge class of the Epsilon Delta fraternity is instructed to tag all the frogs residing on the ESU campus. Two years ago (t = 0) they managed to tag all 58,000 of them (with little Epsilon Delta Fraternity tags). This year's pledge class discovered that all the tags had all fallen off, and they wound up tagging a total of 92,800 frogs.
Find an exponential model for the frog population.
I come up with f(t)=58000(1.264911064^t) - can someone please tell me if this is right, or if I'm at least on the right track?
Looks good to me.
But lots of time science folks like to use natural logs and base-e exponentials. So, since
log 1.264911064 = .23500
1.264911064 = e^.23500
So, your function becomes
f(t) = 58000 e^.235t
To cut through the word salad of this question, we know:
when t = 0 , f = 58000
when t = 2, f = 92800 , and our model is an exponential function
the default base of an exponential equation is e,
f = c e^kt , were c is the starting number
so we can start with:
f = 58000 e^kt
when t = 2, k = 92800
92800 = 58000 e^2k
1.6 = e^2k
take ln of both sides
2k lne = ln1.6, recall that lne = 1
2k = ln 1.6/2 = appr .235
so frogs = 58000 e^(.235t) , t in years
there are many ways to express this function,
yours is also correct
I could have picked 2 as base,
then
92800 = 58000 (2)^(kt)
1.6 = 2^(2k)
2k ln2 = ln1.6
k = ln 1.6/(2ln2) = appr .339
frogs = 58000 (2)^(.339t)
check:
f = 58000(2)^(2(.339)) = 92795 , close considering I only used 3 decimals for my k.
Wow, simultaneous posting times.
I wish there was a way to know if a fellow tutor is working on the same problem at the same time as another tutor. This would eliminate the unnecessary time that both spent on the solution.
The only positive is that in most cases we end up with the same solution, so it is a good check for the student.
As well as for the tutors. :-)
Thank you both!
To find an exponential model for the frog population, we can use the given information about the number of frogs tagged in different years.
Let's assume that the frog population follows an exponential growth pattern, where the number of frogs increases by a constant percentage each year.
First, let's consider the initial population at t = 0. We know that 58,000 frogs were tagged in that year.
Next, we have the information about the number of frogs tagged this year, which is 92,800.
To find the growth rate, we can divide the number of frogs tagged this year by the number of frogs tagged two years ago:
Growth rate = (92,800 - 58,000) / 58,000
Simplifying this, we have:
Growth rate = 34,800 / 58,000
Growth rate ≈ 0.6
Now we can use this growth rate in the exponential model.
The general form of an exponential model is given by:
f(t) = P0 * (1 + r)^t
Where:
f(t) represents the frog population at time t
P0 is the initial population (at t = 0)
r is the growth rate
t is the time elapsed
Plugging in the values we have:
f(t) = 58,000 * (1 + 0.6)^t
Simplifying this, we get:
f(t) ≈ 58,000 * (1.6)^t
So, the exponential model for the frog population is approximately f(t) = 58,000 * (1.6)^t.
Note: It's important to double-check the accuracy of the given information and assumptions to ensure the model is appropriate for the specific scenario.