To determine how much more you would earn in the first investment than the second, we need to calculate the future values of both investments using the formula for compound interest:

$A = P \left(1 + \frac{r}{n}\right)^{nt}$

Where:

• $A$ is the amount of money accumulated after $n$ years, including interest.
• $P$ is the principal amount (initial investment).
• $r$ is the annual interest rate (decimal).
• $n$ is the number of times that interest is compounded per year.
• $t$ is the number of years.

Given that the interest is compounded annually (so $n = 1$), the formula simplifies to:

$A = P(1 + r)^t$

First Investment (10% compounded annually):

• $P = 23,000$
• $r = 0.10$
• $t = 40$

$A_1 = 23,000(1 + 0.10)^{40}$ $A_1 = 23,000(1.10)^{40} \approx 23,000 \times 45.259 \approx 1,040,957$

Second Investment (5% compounded annually):

• $P = 23,000$
• $r = 0.05$
• $t = 40$

$A_2 = 23,000(1 + 0.05)^{40}$ $A_2 = 23,000(1.05)^{40} \approx 23,000 \times 7.040 \approx 161,918$

Difference in Earnings:

$\text{Difference} = A_1 - A_2 = 1,040,957 - 161,918 \approx 879,039$

So, you would earn approximately $879,039 more from the first investment than from the second investment.

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Would you like more details or explanations?

Related Questions:

1. How does compounding frequency affect the total earnings of an investment?
2. What would be the result if the compounding period was quarterly instead of annually?
3. What would the difference be if the time period was 30 years instead of 40 years?
4. How do you convert an interest rate into a decimal for compound interest calculations?
5. How does the interest rate affect the growth of the investment over time?

Tip:

The effect of compounding becomes more significant over longer time periods and with higher interest rates, resulting in exponentially larger earnings.