Exponential growth and decay (word problem)
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Exponential growth and decay (word problem)

[From: ] [author: ] [Date: 12-04-29] [Hit: ]
-In exponential growth,Q(t) = Q(0)e^(kt) where k is a constant.k = ln(2)/3 = 0.Therefore Q(5) = 90e^(0.23104906*5) = 285.So the answer is 285+ black-nosed rabbits in 1995.......
could anyone explain this to me?

Conservationists tagged 90 black-nosed rabbits in a national forest in 1990. In 1993, they tagged 180 black-nosed rabbits in the same range. If the rabbit population follows the exponential law, how many rabbits will be in the range 5 years from 1990?

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In exponential growth, the quantity at time t from the initial measurement is given by:

Q(t) = Q(0)e^(kt) where k is a constant. In this case we can find k by taking:

Q(3) = Q(0)e^(k3) or

180 = 90*e^(3k)

2 = e^(3k) take the natural log (ln) of both sides

ln(2) = 3k or
k = ln(2)/3 = 0.23104906

Therefore Q(5) = 90e^(0.23104906*5) = 285.732189

So the answer is 285+ black-nosed rabbits in 1995.
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