Exponential population decay word problem

[Jacobpm64]

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

 

[end3r7]

Should K be -.25? The time comes out to 2.514 then

 

[HallsofIvy]

There is no reason at all to think that k= 0.75! Don't just plug numbers in at random- think!
Each year, the population decreases by 0.25 and so must be 0.75 times the previous years population. That is, if this year's (t= 1) population is P, next year's is 0.75P: 0.75P= Pe^k or e^k= 0.75. Solve that for k.

Personally, I wouldn't use "e^kt" at all. The population is multiplied by 0.75 each year so its population after t years would be P₀(0.75)^t. You want to solve
7500(0.75)^t= 4000.

 

[Jacobpm64]

got it.. 2.1850811 years