Asked by Jae M.
okay sorry but this is the last problem that I don't understand how to start...
A population of prairie dogs grows exponentially. The colony begins with 32 prairie dogs; 3 years later there are 200 prairie dogs. Give a formula for population, P, as a function of time.
help would be appreciated, thank you :)
A population of prairie dogs grows exponentially. The colony begins with 32 prairie dogs; 3 years later there are 200 prairie dogs. Give a formula for population, P, as a function of time.
help would be appreciated, thank you :)
Answers
Answered by
Reiny
p = 32 e^(kt) , where k is a constant
when t = 3, p = 200
32 e^(3k) = 200
e^(3k) = 6.25
3k = ln 6.25
k = appr .61086
p = 32 e^(.61086 t)
when t = 3, p = 200
32 e^(3k) = 200
e^(3k) = 6.25
3k = ln 6.25
k = appr .61086
p = 32 e^(.61086 t)
Answered by
Jae M.
can you please tell me how did you know which formula to use?
Answered by
Steve
exponential growth is always of the form
p = a*e^(kt)
that's why it's called exponential...
The starting amount (at t=0) is a, since e^0 = 1.
p = a*e^(kt)
that's why it's called exponential...
The starting amount (at t=0) is a, since e^0 = 1.
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