We are given an exponential decay model for the population of an endangered fish species:

$A(t) = A_0 e^{kt}$

Where:

• $A(t)$ is the population at time $t$,
• $A_0$ is the initial population (9 years ago in this case),
• $k$ is the decay rate,
• $t$ is the time in years.

Step 1: Determine the decay rate $k$

We know that:

• 9 years ago ($t = -9$) the population was $A(-9) = 1500$,
• The population today ($t = 0$) is $A(0) = 1000$.

At $t = 0$: $A(0) = A_0 e^{k(0)} = A_0 = 1000$ Thus, $A_0 = 1000$, meaning the population 9 years ago was 1500, but relative to today's 1000.

Now, use the given population from 9 years ago ($A(-9) = 1500$) to solve for $k$:

$1500 = 1000 e^{k(-9)}$ Divide both sides by 1000:

$1.5 = e^{-9k}$ Take the natural logarithm (ln) of both sides:

$\ln(1.5) = -9k$ $k = \frac{-\ln(1.5)}{9}$

Now calculate $k$:

$k \approx \frac{-0.4055}{9} \approx -0.04506$

Step 2: Predict when the population will drop below 100

We now want to find the time $t$ when the population $A(t)$ drops below 100. Using the equation:

$A(t) = 1000 e^{kt}$

Set $A(t) = 100$:

$100 = 1000 e^{kt}$ Divide both sides by 1000:

$0.1 = e^{kt}$ Take the natural logarithm (ln) of both sides:

$\ln(0.1) = kt$ $t = \frac{\ln(0.1)}{k}$

Substitute the value of $k$:

$t = \frac{\ln(0.1)}{-0.04506}$

Calculate $t$:

$t \approx \frac{-2.3026}{-0.04506} \approx 51.1 \text{ years from today.}$

Conclusion:

According to the model, the fish population will drop below 100 in approximately 51 years from today.

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Do you need further details on any part of this solution?

Related Questions:

1. How would the model change if the population decreased more slowly?
2. What happens to the population if $k$ is a positive constant?
3. How would we model the fish population if conservation efforts slowed the decay?
4. Can the model account for environmental changes over time?
5. How accurate is this model if population dynamics change after hitting a certain threshold?

Tip:

Always make sure the initial conditions in your exponential models are accurate to avoid errors in long-term predictions.