Problem 108 - Elementary and Intermediate Algebra

Question

The effective rate of interest \(r\) earned by an investment is given by the formula \(r=\sqrt[n]{\frac{A}{P}}-1\) where \(P\) is the initial investment that grows to value \(A\) after \(n\) years. If a diamond buyer got \(\$ 4,000\) for a 1.73 -carat diamond that he had purchased 4 years earlier, and earned an annual rate of return of \(6.5 \%\) on the investment, what did he originally pay for the diamond?

Step-by-Step Solution

Answer

The diamond buyer originally paid approximately \( \$ 3116.28 \) for the diamond.

Step 1: Identify Given Values

We are given the following values from the problem: \( A = 4000 \) (the amount after 4 years), \( n = 4 \) (the number of years), and \( r = 6.5\% = 0.065 \) (the annual rate of return).

Step 2: Set Up the Formula for Original Price

To find the original price \( P \), we start with the given formula for the effective interest rate: \[ r = \sqrt[n]{\frac{A}{P}} - 1 \]Rearrange to solve for \( P \): \[ P = \frac{A}{(1 + r)^n} \]

Step 3: Plug In the Known Values

Substitute \( A = 4000 \), \( r = 0.065 \), and \( n = 4 \) into the rearranged formula:\[ P = \frac{4000}{(1 + 0.065)^4} \]

Step 4: Calculate the Power

Calculate the value of \((1 + 0.065)^4\):\[ (1 + 0.065)^4 = 1.065^4 \approx 1.2837 \]

Step 5: Calculate the Original Price

Divide \( A = 4000 \) by \( 1.2837 \) to find \( P \):\[ P = \frac{4000}{1.2837} \approx 3116.28 \]

Step 6: State the Original Price

The original price the diamond buyer paid for the diamond is approximately \( \$ 3116.28 \).

Key Concepts

Effective Interest Rate

The effective interest rate (EIR) is a crucial concept in finance for understanding how much money you actually earn or pay on an investment or loan over a specified period. It considers the effects of compound interest, which means interest on interest, providing a more accurate measure of financial growth or cost than the nominal interest rate.

The formula for calculating the effective interest rate using initial investment values is:

• \( r = \sqrt[n]{\frac{A}{P}} - 1 \)

where:

• \( A \) is the final amount obtained after a period of time
• \( P \) is the initial amount of the investment
• \( n \) is the number of years the investment is held
• \( r \) is the effective rate of interest

By using the EIR, investors can determine the true rate of return on an investment, considering the effect of compounding. It's useful when comparing investments or loans with different rates and compounding periods.

Investment Growth

Investment growth refers to the increase in the value of an asset or investment over time. The key principle behind investment growth is that the value of an investment increases due to earning interest or dividends, as well as the potential for capital appreciation.

For the diamond buyer example, investment growth can be seen as the increase in the diamond's value from the original purchase price to the amount obtained after 4 years.

• The formula used: \( P = \frac{A}{(1 + r)^n} \)

With this equation:

• \( A = 4000 \), representing the amount received from selling
• The effective annual rate \( r = 6.5\% \)
• \( n = 4 \), the time period in years

This calculation allows one to find the initial amount \( P \) needed to achieve a final amount \( A \) after growth periods.

Exponentiation in Algebra

Exponentiation is a fundamental operation in algebra, often taught early in mathematics. It involves raising a base number to a power, which is defined as repeated multiplication of the base.

• In the diamond investment problem, we use exponentiation to calculate compound growth over time: \((1 + r)^n\)

Here:

• \( (1 + r) \) is the base, reflecting the growth factor for each period
• \( n \) is the exponent, representing the number of periods over which growth occurs

When you raise a number to a power, you're essentially applying the growth rate continuously over several periods. For instance, calculating \( (1.065)^4 \) involves multiplying 1.065 by itself four times, demonstrating how the effect of compounding builds up over multiple years. Understanding how exponentiation works allows one to comprehend how investments grow over time with compounded interest.