SOLUTION: A small lake is stocked with a certain species of fish. The fish population is modelled by the function P(t) = 10/1+5e^(-0.8t) where P is the number of fish in hundreds and t is me

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Question 1154016: A small lake is stocked with a certain species of fish. The fish population is modelled by the function P(t) = 10/1+5e^(-0.8t) where P is the number of fish in hundreds and t is measured in months since the lake was stocked. After how many months will the fish population reach 500 fish?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A small lake is stocked with a certain species of fish.
The fish population is modelled by the function P%28t%29+=+10%2F%28%281%2B5e%5E%28-0.8t%29%29%29 where
P is the number of fish in hundreds and
t is measured in months since the lake was stocked.
After how many months will the fish population reach 500 fish?
:
P is the number of fish in hundreds, therefore 500 fish is 5
5+=+10%2F%28%281%2B5e%5E%28-0.8t%29%29%29
multiply both sides by %281%2B5e%5E%28-0.8t%29%29
5%281%2B5e%5E%28-0.8t%29%29+=+10
divide both sides by 5
1%2B5e%5E%28-0.8t%29+=+2
subtract 1 from both sides
5e%5E%28-0.8t%29+=+1
divide both sides by 5
e%5E%28-0.8t%29%29+=+.2
using nat logs
ln%28e%5E%28-.8t%29%29+=+ln%28.2%29
-.8t*ln(e) = ln(.2)
ln of e is 1, find the ln of .2
-.8t = -1.6094
t = %28-1.6094%29%2F-.8
t ~ +2 months