We are given that the value of the toy tractor grew exponentially between 1980 and 1988. To model this, we can use the exponential growth formula:

$V(t) = V_0 e^{kt}$

Where:

• $V(t)$ is the value of the item at time $t$,
• $V_0 = 226$ (the initial value in 1980),
• $k$ is the exponential growth rate,
• $t$ is the time in years from 1980,
• $e$ is Euler's number.

Step 1: Set up the equation for 1988.

In 1988, the value of the tractor is $V(8) = 492$, because 1988 is 8 years after 1980. Thus, we have:

$492 = 226 e^{8k}$

Step 2: Solve for $k$.

Divide both sides by 226 to isolate the exponential term:

$\frac{492}{226} = e^{8k}$

Now take the natural logarithm of both sides to get rid of the exponential:

$\ln\left(\frac{492}{226}\right) = 8k$

$k = \frac{1}{8} \ln\left(\frac{492}{226}\right)$

Now let's calculate $k$.The value of the exponential growth rate $k$ is approximately 0.097, rounded to the nearest thousandth.

Would you like more details or have any questions?

Here are five related questions to deepen your understanding:

1. How does exponential growth differ from linear growth?
2. What is the value of the toy tractor in 1995 using the same exponential model?
3. How would the equation change if the growth rate was negative (exponential decay)?
4. Can you apply this model to predict the value of a different collector’s item over time?
5. What factors in real life could affect the accuracy of this exponential growth model?

Tip: When working with exponential models, always ensure the growth rate is interpreted correctly—positive values indicate growth, while negative values indicate decay.