anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
What is the limit as t goes to infinity of e^-0.159t?
anonymous
  • anonymous
Isn't it Zero?
anonymous
  • anonymous
So what does P(t) go to as t goes to infinity?

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anonymous
  • anonymous
I'm not quite sure?
anonymous
  • anonymous
Since e^-0.159t goes to 0, 5500e^(-0.159t) goes to zero as well, right. It's just 5500(0).
anonymous
  • anonymous
What happens to the 6000?
anonymous
  • anonymous
Since that second part goes to zero, you have 6000-0 which is just 6000. That's the limiting value
anonymous
  • anonymous
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.
anonymous
  • anonymous
Yeah.
anonymous
  • anonymous
Thank you! Do you know how to do the second part?
anonymous
  • anonymous
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.
anonymous
  • anonymous
So just simply solve for t?
anonymous
  • anonymous
Yeah. They are asking for the time the population will be 3000.
anonymous
  • anonymous
I know how to do it know! Thanks a lot!
anonymous
  • anonymous
Np

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