A population of fish in a newly dug pond grows according to the equation P = 4000/ (1 + 3e^-0.3t) where time t is given in years. In how many years will there be 2000 fish?

Answer 1

2000=4000/(1+ 3e^-.3t) solve for t

1+3e^-.3t = 4000/2000= 2

3e^-.3t = 2-1 = 1

e^-.3t = 1/3

-.3t = ln(1/3)

t = -(ln(1/3)/.3

t= about 0.3662 years to go to 2,000 fish population

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