SOLUTION: The fish population of a lake is modeled by the function P=(10)/(1+4e^(-0.8t)) where, P is the number of fish in thousands and t is the years since the lake was stocked.

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Question 194023: The fish population of a lake is modeled by the function
P=(10)/(1+4e^(-0.8t))
where, P is the number of fish in thousands and t is the years since the lake was stocked.
What is the fish population (correct up to two decimal places) after 3 years is? [Remember: If P=3.15, then there are 3,150 fish)
How many years will it take for the fish population to reach 7,000? (correct up to two decimal places)

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
P is the number of fish in thousands
(a)
t+=+3
P=10+%2F+%281+%2B+4e%5E%28-.8t%29%29
P=10+%2F+%281+%2B+4e%5E%28-2.4%29%29
P=10+%2F+%281+%2B+4%2A.0907%29
P=10+%2F+%281+%2B+.3629%29
P=10+%2F+%281.3629%29
P+=+7.34
the population is 7,340 after 3 years
(b)
P+=+7
P=10+%2F+%281%2B4e%5E%28-.8t%29%29
7=10+%2F+%281%2B4e%5E%28-.8t%29%29
Multiply both sides by %281%2B4e%5E%28-.8t%29%29
7%2A%281%2B4e%5E%28-.8t%29%29+=+10
1+%2B+4e%5E%28-.8t%29+=+10%2F7
4e%5E%28-.8t%29+=+3%2F7
e%5E%28-.8t%29+=+3%2F28
Take the ln of both sides
-.8t+=+ln%28.10714%29
-.8t+=+-2.2336
t+=+2.79 years