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The problem involves finding the rate of growth, \( r \), for a fish population that follows an exponential growth model. The formula for this model is:

\[
P_n = P_0(1 + r)^n
\]

Where:
- \( P_n \) is the population after \( n \) years,
- \( P_0 \) is the initial population,
- \( r \) is the growth rate (as a decimal),
- \( n \) is the number of years.

### Given:
- Initial population \( P_0 = 3900 \),
- Population after 9 years \( P_n = 25000 \),
- \( n = 9 \).

We need to solve for \( r \).

### Step-by-step solution:

1. Start with the formula:
   \[
   25000 = 3900(1 + r)^9
   \]

2. Divide both sides by 3900 to isolate \( (1 + r)^9 \):
   \[
   \frac{25000}{3900} = (1 + r)^9
   \]

   \[
   6.41025641 = (1 + r)^9
   \]

3. Take the 9th root of both sides:
   \[
   1 + r = \left( 6.41025641 \right)^{\frac{1}{9}}
   \]

4. Subtract 1 to solve for \( r \):
   \[
   r = \left( 6.41025641 \right)^{\frac{1}{9}} - 1
   \]

Now let's compute this numerically.The rate of growth \( r \) is approximately **22.9%**, rounded to the nearest tenth.

Do you have any questions or need further details?

Here are five related questions to explore:
1. How would the result change if the population after 9 years was different?
2. Can this model be used for population decline? How would the formula change?
3. How does exponential growth compare to linear growth in this context?
4. What happens to the growth rate if the population grows faster than expected?
5. How would the formula adjust if the rate of growth were compounded quarterly instead of annually?

**Tip:** Always double-check your unit conversions when expressing growth rates as percentages!

Suppose a fish population is currently 3900 and 9 years later the population is 25000. Use the explicit exponential model P_n = P_0(1 + r)^n to find the rate of growth. Express the rate of growth as a percentage to the nearest tenth.

Find the growth rate of a fish population using the exponential model P_n = P_0(1 + r)^n. Starting from an initial population of 3900, learn how to calculate the rate of growth to reach 25000 after 9 years. Suitable for high school students (Grades 10-12).

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