Math Problem Statement

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

Solution

The equation you've provided is:

A2=(A1+20,000)×(1.04)8A_2 = (A_1 + 20,000) \times (1.04)^8

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

  • A2A_2 represents the final amount.
  • A1A_1 represents the initial amount.
  • 20,00020,000 is an added value or increment.
  • 1.041.04 is the growth factor.
  • 88 is the number of periods (perhaps years, months, etc.).

To solve for A2A_2, you need to know the value of A1A_1. Once A1A_1 is known, you can substitute and compute the final value for A2A_2.

Example Calculation (assuming A1=50,000A_1 = 50,000):

A2=(50,000+20,000)×(1.04)8A_2 = (50,000 + 20,000) \times (1.04)^8 First, sum the values inside the parentheses:

A2=70,000×(1.04)8A_2 = 70,000 \times (1.04)^8 Now calculate (1.04)8(1.04)^8:

(1.04)81.368569(1.04)^8 \approx 1.368569

Finally:

A270,000×1.368569=95,800.43A_2 \approx 70,000 \times 1.368569 = 95,800.43

So, A295,800.43A_2 \approx 95,800.43.

Would you like to provide the value of A1A_1, or do you have any other questions?

Related Questions:

  1. What is the interpretation of A1A_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 (88) 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.

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Compound Interest

Formulas

A2 = (A1 + 20,000) × (1.04)^8

Theorems

Compound Interest Formula

Suitable Grade Level

Grades 9-12