A 
2
​
 =(A 
1
​
 +20,000)(1+ 
2×100
8
​
 ) 
2×4
 
𝐴
2
=
(
𝐴
1
+
20
,
000
)
×
(
1.04
)
8
A 
2
​
 =(A 
1
​
 +20,000)×(1.04) 
8

The equation you've provided is:

\[
A_2 = (A_1 + 20,000) \times (1.04)^8
\]

This can be interpreted as an exponential growth or compound interest formula, where:

- \(A_2\) represents the final amount.
- \(A_1\) represents the initial amount.
- \(20,000\) is an added value or increment.
- \(1.04\) is the growth factor.
- \(8\) is the number of periods (perhaps years, months, etc.).

To solve for \(A_2\), you need to know the value of \(A_1\). Once \(A_1\) is known, you can substitute and compute the final value for \(A_2\).

### Example Calculation (assuming \(A_1 = 50,000\)):

\[
A_2 = (50,000 + 20,000) \times (1.04)^8
\]
First, sum the values inside the parentheses:

\[
A_2 = 70,000 \times (1.04)^8
\]
Now calculate \( (1.04)^8 \):

\[
(1.04)^8 \approx 1.368569
\]

Finally:

\[
A_2 \approx 70,000 \times 1.368569 = 95,800.43
\]

So, \(A_2 \approx 95,800.43\).

Would you like to provide the value of \(A_1\), or do you have any other questions? 

### Related Questions:
1. What is the interpretation of \(A_1\) in real-world problems like investment?
2. How does changing the growth factor (1.04) impact the final result?
3. What if the number of periods (\(8\)) were different, say 5 or 10?
4. How can this formula be used for financial planning, like retirement?
5. Can we derive the formula for continuous compounding?

### Tip:
When working with exponential growth, small changes in the growth rate or time period can lead to significant changes in the result.

Explore how to solve an exponential growth problem using the formula A2 = (A1 + 20,000) × (1.04)^8. This solution demonstrates how small changes in growth factors can significantly affect outcomes. Suitable for high school students learning compound interest and exponential growth.

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