The fish population of lake henry is decreasing at a rate of 4% per year. in 2002 there were about 1,250 fish. write an exponential decay function to model this situation then find the population in 2008?

2 Answers
Oct 7, 2015

Answer:

I found #F(t)=1250e^(-0.04t)#

Explanation:

I tried considering 2002 as the time #t=0# with #t# representing years; so I can write a function #F# of number of fish in time as:
#F(t)=1250e^(-0.04t)#
In 2008 we will have #t=6# and:
#F(6)=1250e^(-0.04*6)=983# fish

Oct 7, 2015

Answer:

#p(t) = 1250 * 0.96^t# where #t# is time in years from #2002#

#p(6) ~~ 1250 * 0.7828 ~~ 978#

Explanation:

A loss of #4%# is equivalent to multiplying by #0.96#, so we can write:

#p(t) = 1250 * 0.96^t#

where #t# is time in years since #2002#

#0.96^6 ~~ 0.7828#

so #p(6) ~~ 1250 * 0.7828 ~~ 978#

If we want to express this in terms of #e^x#, then we can take logs first to find:

#ln p(t) = ln (1250*0.96^t) = ln(1250) + t ln (0.96)#

Then taking exponents:

#p(t) = e^(ln p(t)) = e^(ln(1250) + t ln (0.96)) = 1250 e^(ln(0.96)t)#

#ln(0.96) ~~ -0.040822#

So:

#p(t) ~~ 1250 e^(-0.040822t)#