SOLUTION: A small lake is stocked with a certain species of fish. The fish population is modeled by the function P=10/1+4e^-0.8t where, P is the number of fish in thousands and t is me

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Question 115300: A small lake is stocked with a certain species of fish. The fish population is modeled by the function
P=10/1+4e^-0.8t
where, P is the number of fish in thousands and t is measured in years since the lake was stocked.
The fish population (correct up to two decimal places) after 3 years is______
(Remark: don't forget to use the correct units for P .)
The fish population reaches 5000 after years is __________ (correct up to two decimal places).

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 modeled by the function
P = 10%2F%281%2B4e%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.
:
The fish population (correct up to two decimal places) after 3 years is______
(Remark: don't forget to use the correct units for P .)
:
Substitute 3 for t in the given equation:
:
P = 10%2F%281%2B4e%5E%28-0.8%2A3%29%29
P = 10%2F%281%2B4e%5E-2.4%29
Find e^-2.4 on a good calc:
P = 10%2F%281%2B4%2A.0907%29
P = 10%2F1.623
P = 7.34 thousand
:
The fish population reaches 5000 after years is __________ (correct up to two decimal places).
:
P = 5
5 = 10%2F%281%2B4e%5E%28-0.8t%29%29
;
Multiply both sides by = %281%2B4e%5E%28-0.8t%29%29
5%281%2B4e%5E%28-0.8t%29%29 = 10
:
Divide both sides by 5:
1%2B4e%5E%28-0.8t%29 = 2
:
Subtract 1 from both sides and you have:
4e%5E%28-0.8t%29 = 1
Divide both sides by 4
e%5E%28-0.8t%29 = .25
-.8t = ln(.25); remember the ln(e) = 1
:
-.8t = -1.3863
t = %28-1.3863%29%2F%28-.8%29
t = 1.733 years
:
:
Check solution by using 1.73 for t in the original equation and find P
P = 10%2F%281%2B4e%5E%28-0.8%2A1.73%29%29
P = 10%2F%281%2B4e%5E-1.3863%29
P = 10%2F%281%2B4%2A.25%29
P = 10%2F%281%2B1%29
P = 5 which is 5000 fish