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 .)
:
Substitute 3 for t in the given equation:
:
P = {{{10/(1+4e^(-0.8*3))}}}
P = {{{10/(1+4e^-2.4)}}}
Find e^-2.4 on a good calc:
P = {{{10/(1+4*.0907)}}}
P = {{{10/1.623}}}
P = 7.34 thousand 
:
The fish population reaches 5000 after years is __________ (correct up to two decimal places).
:
P = 5
5 = {{{10/(1+4e^(-0.8t))}}}
;
Multiply both sides by = {{{(1+4e^(-0.8t))}}} 
5{{{(1+4e^(-0.8t))}}} = 10
:
Divide both sides by 5:
{{{1+4e^(-0.8t)}}} = 2
:
Subtract 1 from both sides and you have:
{{{4e^(-0.8t)}}} = 1
Divide both sides by 4
{{{e^(-0.8t)}}} = .25

-.8t = ln(.25); remember the ln(e) = 1
:
-.8t = -1.3863
t = {{{(-1.3863)/(-.8)}}}
t = 1.733 years
:
:
Check solution by using 1.73 for t in the original equation and find P
P = {{{10/(1+4e^(-0.8*1.73))}}}
P = {{{10/(1+4e^-1.3863)}}}
P = {{{10/(1+4*.25)}}}
P = {{{10/(1+1)}}}
P = 5 which is 5000 fish