Question

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?

Answer 1

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

Answer 2

`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)`