Given that $f(t)$ is exponential, we can assume a general exponential form:

$f(t) = P_0 \cdot e^{kt}$

where:

• $P_0$ is the initial population at $t = 0$,
• $k$ is the growth rate.

Step 1: Set up the equations

From the given information:

1. $f(5) = 13$, which implies: $13 = P_0 \cdot e^{5k}$
2. $f(9) = 18$, which implies: $18 = P_0 \cdot e^{9k}$

Step 2: Divide the equations to solve for $k$

To eliminate $P_0$, we can divide the second equation by the first:

$\frac{18}{13} = \frac{P_0 \cdot e^{9k}}{P_0 \cdot e^{5k}}$

This simplifies to:

$\frac{18}{13} = e^{4k}$

Step 3: Solve for $k$

Take the natural logarithm on both sides:

$\ln \left(\frac{18}{13}\right) = 4k$

Then:

$k = \frac{\ln \left(\frac{18}{13}\right)}{4}$

Calculating this value:

$k \approx \frac{\ln(1.3846)}{4} \approx \frac{0.325}{4} \approx 0.08125$

Step 4: Solve for $P_0$

Now that we have $k$, we can substitute it back into one of the original equations to find $P_0$. Using $f(5) = 13$:

$13 = P_0 \cdot e^{5 \cdot 0.08125}$

Calculating the exponential term:

$13 = P_0 \cdot e^{0.40625} \approx P_0 \cdot 1.501$

Then:

$P_0 \approx \frac{13}{1.501} \approx 8.66$

Final Answer

The formula for $f(t)$, the population $t$ years after 1990, is approximately:

$f(t) = 8.66 \cdot e^{0.08125 \cdot t}$

This gives us the model for the population growth of the town.

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Would you like a deeper breakdown or to verify each calculation?

Related Questions

1. How would the model change if $f$ were assumed to be a linear function instead of exponential?
2. What would the population be in the year 2000 according to this model?
3. How does changing the value of $k$ affect the growth rate of $f(t)$?
4. Can this model be used for long-term predictions, say in 100 years? Why or why not?
5. How would we determine the half-life of the population if it were declining exponentially?

Tip

When dealing with exponential functions, dividing two instances at different times can often simplify solving for the growth rate $k$ without needing to know the initial population.