Given the model $f(x) = 224 \times (0.883)^x$, where $f(x)$ represents the number of fish in thousands, and $x$ represents the number of years after 1973, we need to estimate the year when the number of fish reached 97 thousand.

To find this, we need to solve the equation: $97 = 224 \times (0.883)^x$

Let's solve for $x$:

1. Divide both sides by 224: $\frac{97}{224} = (0.883)^x$

2. Simplify the fraction: $0.433 = (0.883)^x$

3. Take the natural logarithm of both sides to solve for $x$: $\ln(0.433) = \ln((0.883)^x)$

4. Use the property of logarithms $\ln(a^b) = b \ln(a)$: $\ln(0.433) = x \ln(0.883)$

5. Solve for $x$: $x = \frac{\ln(0.433)}{\ln(0.883)}$

Now, calculate the values: $\ln(0.433) \approx -0.837$ $\ln(0.883) \approx -0.124$

So, $x \approx \frac{-0.837}{-0.124} \approx 6.75$

This means the number of fish reached 97 thousand approximately 6.75 years after 1973. To find the specific year: $1973 + 6.75 \approx 1980$

Therefore, the number of these fish reached 97 thousand in the year 1980.

Would you like more details on any step or have any questions?

Here are some related questions for further practice:

1. How would the model change if the initial number of fish were different?
2. What would be the year if the number of fish was 50 thousand?
3. How do exponential decay models apply to other real-world scenarios?
4. What is the significance of the base in an exponential decay model?
5. How can logarithms help solve other types of exponential equations?
6. What does the constant 0.883 represent in this context?
7. How would the model be affected by environmental changes affecting fish population?
8. How can we verify the accuracy of this model with real-world data?

Tip: When solving exponential equations, always isolate the exponential term before applying logarithms for easier calculations.