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.

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

As well as for the tutors. :-)

Answer 5

Thank you both!