# 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?

Oct 7, 2015

I found $F \left(t\right) = 1250 {e}^{- 0.04 t}$

#### 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:
$F \left(t\right) = 1250 {e}^{- 0.04 t}$
In 2008 we will have $t = 6$ and:
$F \left(6\right) = 1250 {e}^{- 0.04 \cdot 6} = 983$ fish

Oct 7, 2015

$p \left(t\right) = 1250 \cdot {0.96}^{t}$ where $t$ is time in years from $2002$

$p \left(6\right) \approx 1250 \cdot 0.7828 \approx 978$

#### Explanation:

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

$p \left(t\right) = 1250 \cdot {0.96}^{t}$

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

${0.96}^{6} \approx 0.7828$

so $p \left(6\right) \approx 1250 \cdot 0.7828 \approx 978$

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

$\ln p \left(t\right) = \ln \left(1250 \cdot {0.96}^{t}\right) = \ln \left(1250\right) + t \ln \left(0.96\right)$

Then taking exponents:

$p \left(t\right) = {e}^{\ln p \left(t\right)} = {e}^{\ln \left(1250\right) + t \ln \left(0.96\right)} = 1250 {e}^{\ln \left(0.96\right) t}$

$\ln \left(0.96\right) \approx - 0.040822$

So:

$p \left(t\right) \approx 1250 {e}^{- 0.040822 t}$