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Topic review (newest first)

Jacobpm
2006-01-03 10:00:10

I don't need any more help with this one, thanks anyway

Jacobpm
2006-01-03 09:00:40

Suppose that in any given year, the population of a certain endangered species is reduced by 25%. If the population is now 7500, in how many years will the population be 4000?

Well, I know that the formula for this is P(t) = Ce^(kt).

I just can't figure out the value of k.

i did 7500 x .25 = 1875... 7500 - 1875 = 5625...
so i set up 7500x = 5625 and got .75.
so, essentially.. the population is growing at .75 per year.
so i used .75 as k and:

4000 = 7500e^(.75t)
4000/7500 = e^(.75t)
8/15 = e^(.75t)
ln(8/15) = .75t lne
ln(8/15) = .75t
t = [ln(8/15)] / .75
t = -.8381448792 years

This can't be correct because time isn't negative, and if you do it logically, it would be around 2. something years:

7500 x .75 = 5625 ----> 1 year
5625 x .75 = 4218.75 ----> 2 yrs
4218.75 x .75 = 3164.0625 ----> 3 yrs

So it's somewhere between 2 and 3 years... closer to 2 years.

and if i set up 4218.75x = 4000
x = .9481481481

so i know after 2 years.. i can multiply by .9481481481 and i get 4000..
but i don't know how to convert that into years.. since that is a percentage.

BLEH.. brain is about to explode.. don't know what to do.. lol

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