SOLUTION: Can someone please assist me in setting this equation up and explain why I am doing it in such a way... please? A lake is stocked with a certain species of fish, the fish populati

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Question 953045: Can someone please assist me in setting this equation up and explain why I am doing it in such a way... please?
A lake is stocked with a certain species of fish, the fish population is modeled by the function p=700/e^-0.8t
where P is the number of fish in thousands and t is measured in years since the lake was stocked
Find the fish population after 2 years?
After how many years will the fish population be 250000 fish?
THanks!

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A lake is stocked with a certain species of fish,
the fish population is modeled by the function p=700%2F%28e%5E%28-0.8t%29%29
where P is the number of fish in thousands
and t is measured in years since the lake was stocked
:
Find the fish population after 2 years?
t = 2,so we have
p=700%2F%28e%5E%28%28-0.8%2A2%29%29%29
p=700%2F%28e%5E%28-1.6%29%29
find e^-1.6 on your calc
p=700%2F.2019
p = 3,467 fish after 2 yrs
:
After how many years will the fish population be 250000 fish?
Write this as
250000 = 700%2F%28e%5E%28-.8t%29%29
multiply both sides by e^(-.8t)
250000%2Ae%5E%28-.8t%29+=+700
divide both sides by 250000
e%5E%28-.8t%29+=+700%2F250000
e%5E%28-.8t%29+=+.0028
find the nat log of both sides
ln%28e%5E%28-.8t%29%29+=+ln%28.0028%29
the log equiv of exponents, find the ln of .0028
-.8t*ln(e) = -5.878
the ln of e is 1 so we can just write
-.8t = -5.878
t = %28-5.878%29%2F%28-.8%29
t = 7.35 years to increase to 250000 fish
:
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did I explain well enough so you know what'sgoing on here? ankor@att.net