To model the number of fish in the lake using an exponential decay function, we can start with the general formula for exponential decay:
\[ N(t) = N_0 \times e^{-kt} \]
where:
- \( N(t) \) is the quantity at time \( t \),
- \( N_0 \) is the initial quantity,
- \( k \) is the decay constant,
- \( t \) is the time in years,
- \( e \) is the base of the natural logarithm.
In this case, we have:
- \( N_0 = 2000 \) (the initial number of fish),
- the decay rate is 3%, which can be represented as a decay constant of \( k = 0.03 \).
Thus, the exponential decay function for the number of fish in the lake can be modeled as:
\[ N(t) = 2000 \times e^{-0.03t} \]
Next, we need to find the number of fish in the lake in 2025. Since 2015 is our starting point, we compute \( t \) as follows:
\[ t = 2025 - 2015 = 10 \text{ years} \]
Now we substitute \( t = 10 \) into the decay function:
\[ N(10) = 2000 \times e^{-0.03 \times 10} \]
Calculating \( -0.03 \times 10 \):
\[ -0.03 \times 10 = -0.3 \]
So we have:
\[ N(10) = 2000 \times e^{-0.3} \]
Now we need to compute \( e^{-0.3} \). Using a calculator, we find:
\[ e^{-0.3} \approx 0.740818 \]
Now we can plug this value back into the function:
\[ N(10) \approx 2000 \times 0.740818 \approx 1481.636 \]
Rounding to the nearest whole fish, we find:
\[ N(10) \approx 1482 \]
Thus, the estimated number of fish in the lake in 2025 is approximately 1482 fish.