A town has a population of 4670. 8 months later is has a population of 4895. What would the population be after 2 years if it continued to grow at the same rate?

To determine the population after 2 years, assuming the population continues to grow at the same rate, we can follow these steps:

### Step 1: Calculate the growth rate over the 8 months.
- Initial population: \( P_0 = 4670 \)
- Population after 8 months: \( P_1 = 4895 \)
- Growth over 8 months: \( 4895 - 4670 = 225 \)

The growth rate \( r \) over 8 months is:
\[
r = \frac{\text{growth over 8 months}}{\text{initial population}} = \frac{225}{4670} \approx 0.0482 \text{ (or 4.82%)}
\]

### Step 2: Calculate the annual growth rate.
Since 8 months is \(\frac{2}{3}\) of a year, we can find the equivalent annual growth rate:
\[
\text{Annual growth rate} = \frac{r}{\frac{2}{3}} = \frac{0.0482}{\frac{2}{3}} \approx 0.0723 \text{ (or 7.23%)}
\]

### Step 3: Calculate the population after 2 years.
The population after 2 years (\( P_2 \)) can be estimated using the formula:
\[
P_2 = P_0 \times (1 + \text{annual growth rate})^2
\]
\[
P_2 = 4670 \times (1 + 0.0723)^2
\]
\[
P_2 = 4670 \times (1.0723)^2 \approx 4670 \times 1.1497 \approx 5370
\]

So, the estimated population after 2 years is approximately **5370**.

Would you like any further details on this calculation? Here are 5 related questions to consider:

1. How would the population change if the growth rate were compounded monthly?
2. What if the population had decreased instead—how would the calculation differ?
3. How does this growth rate compare to exponential growth over the same period?
4. What would the population be after 5 years at the same growth rate?
5. How does the calculation change if we consider leap years?

**Tip:** When calculating growth over time, it's essential to distinguish between different compounding periods (e.g., monthly, annually) to get accurate results.

Learn how to calculate population growth using the example of a town's population increasing over 8 months. Estimate the town's population after 2 years based on the growth rate. Suitable for high school students learning about exponential growth and compound interest.

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