A population grows according to the equation P(t) =6000-5500e^(-0.159t) for t>0 or t=0, t measured in years. This population will approach a limiting value as time goes on. During which year will the population reach HALF of this limiting value?

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A population grows according to the equation P(t) =6000-5500e^(-0.159t) for t>0 or t=0, t measured in years. This population will approach a limiting value as time goes on. During which year will the population reach HALF of this limiting value?

Mathematics
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What is the limit as t goes to infinity of e^-0.159t?
Isn't it Zero?
So what does P(t) go to as t goes to infinity?

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I'm not quite sure?
Since e^-0.159t goes to 0, 5500e^(-0.159t) goes to zero as well, right. It's just 5500(0).
What happens to the 6000?
Since that second part goes to zero, you have 6000-0 which is just 6000. That's the limiting value
Okay so I basically just take the limit of each parts 6000 and -5500e^(-0.159t) and I end up with 6000 for the limiting value.
Yeah.
Thank you! Do you know how to do the second part?
You know the limiting value. So half of that is 3000. So P(t)=3000, you need t. So 3000=6000-5500e^(-0.159t). Isolate e^(-0.159t) then take the natural log of both sides.
So just simply solve for t?
Yeah. They are asking for the time the population will be 3000.
I know how to do it know! Thanks a lot!
Np

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