Question

Luke invested $86,000 in an account paying an interest rate of 5, one eigth5 8 1 ​ % compounded continuously. alexander invested $86,000 in an account paying an interest rate of 5, start fraction, 5, divided by, 8, end fraction5 8 5 ​ % compounded annually. after 8 years, how much more money would alexander have in his account than luke, to the nearest dollar?

Answer 1

The question involves calculating the future value of two investments with different compound interest rates: one compounded continuously and one annually. The comparison is made after 8 years to determine which investment yields more.

Step-by-step explanation:

The student's question pertains to comparing the future values of two investments with different compound interest rates. Both Luke and Alexander invested $86,000, but Luke's account has an interest rate of 5 ⅛ % compounded continuously, while Alexander's account has the same rate compounded annually.

To calculate the future value of Luke's investment, we use the formula for continuous compounding:
A = Pert, where A is the amount of money accumulated after n years, including interest, P is the principal amount ($86,000), e is Euler's number (approximately 2.71828), r is the annual interest rate (0.05125 as a decimal), and t is the time in years (8).

For Alexander's account, compounded annually, we use the compound interest formula:
A = P(1 + r)t with the same variables defined as before.

Once we calculate the future values for both investments, we subtract Luke's total from Alexander's total to find out how much more money Alexander would have in his account after 8 years. The result is rounded to the nearest dollar for comparison.

Answer 2

After 8 years, there is no significant difference between the amounts in Alexander's and Luke's accounts. The difference is essentially $0.

Let's calculate the amounts for Luke
`(\(A_L\))` and Alexander
`(\(A_A\))` after 8 years.

For Luke (continuous compounding):

`\[A_L = 86000 \cdot e^((0.05625 \cdot 8)).\]`

For Alexander (annual compounding):

`\[A_A = 86000 \cdot \left(1 + (0.05625)/(1)\right)^(1 \cdot 8).\]`

Calculating these values:


`\[A_L \approx 86000 \cdot e^((0.45)) \approx 86000 \cdot 1.567736 \approx 134853.42.\]`


`\[A_A \approx 86000 \cdot \left(1 + 0.05625\right)^8 \approx 86000 \cdot 1.567736 \approx 134853.42.\]\\`
Now, find the difference
`\(A_A - A_L\):`

`\[134853.42 - 134853.42 \approx 0.\]\\`