Exponential growth problem?

Question

Ten years ago your friend invested in a plot of land with the idea of eventually building a summer cottage on it. His plan was to place a "nest egg" of $12,000 in a long term Certificate of Deposit which paid 8% interest over a period of 20 years, compounded monthly. At the end of the 20 years, he would use the money to build the cottage.
The problem was the plot had 8 prairie dogs on it 10 years ago when he acquired the ground.
This year he made a count of the prairie dogs and found there were 102 on the same plot! A biologist friend assured him that the prairie dogs were growing exponentially according to the law of natural growth, and something needed to be done.
So....
1) What would be the FORMULA for the exponential growth of the prairie dog population.
2)approximately how many years from now will the population reach 365 prairie dogs.
3) approximately how large the population will be 10 years from now

Answer 1

Just when I thought you were talking about a compounding interest problem...

1) Let P(t) be the prairie dog population, where t is in years, with t=0 corresponding to the time that he acquired the property. Then P(t)=8*exp(kt), where k needs to be determined.

To find k, use P(10)=102. Then 102=8*exp(10t), or 12.75=exp(10k).

Take the natural log of both sides:
ln 12.75 = 10k, so k=(ln 12.75)/10=0.25455.

Therefore, P(t)=8*exp(0.25455t).

2) Solve the equation 365=8*exp(0.25455t) for t. Start with

365/8=exp(0.25455t)
Take ln:
ln 45.625=0.25455t
Divide:
t=(ln 45.625)/0.25455=15.0085 years. But that's since the time he acquired the property 10 years ago, so it's about 5 years from now.

3) Plug t=20 into the formula for P(t), since 10 years from now is 20 years from the time the property was acquired.

Answer 2

The answer is obvious.

If it weren't obvious, I'd help.

Good luck!

Answer 3

ITS SEEM LIKE I MISSING TWO INFORMATION, AT LEAST IF YOU WANT A SOLUTION OF THE FORM

F(X)=K1*EXP(K2*X+K3) + K4