We are given two investment options. Option 1 involves investing $5000 for 9 years at 10% compounded annually. Option 2 involves investing $5000 for 9 years compounded continuously at 9%. We need to calculate the final value of each investment and determine which investment will earn more money.

1. Problem Description

We are given two investment options. Option 1 involves investing $5000 for 9 years at 10$5000 for 9 years compounded continuously at 9%. We need to calculate the final value of each investment and determine which investment will earn more money.

2. Solution Steps

First, we calculate the final value of Option

1. The formula for compound interest is:

$A = P(1 + r/n)^{nt}$

where:

$A$ = the future value of the investment/loan, including interest

$P$ = the principal investment amount (the initial deposit or loan amount)

$r$ = the annual interest rate (as a decimal)

$n$ = the number of times that interest is compounded per year

$t$ = the number of years the money is invested or borrowed for

For Option 1:

$P = 5000$

$r = 0.10$

$n = 1$ (compounded annually)

$t = 9$

$A_1 = 5000(1 + 0.10/1)^{(1)(9)} = 5000(1.10)^9 = 5000(2.357947691) = 11789.738455 \approx 11789.74$

Next, we calculate the final value of Option

2. The formula for continuous compounding is:

$A = Pe^{rt}$

where:

$A$ = the future value of the investment/loan, including interest

$P$ = the principal investment amount (the initial deposit or loan amount)

$r$ = the annual interest rate (as a decimal)

$t$ = the number of years the money is invested or borrowed for

$e$ = Euler's number (approximately 2.71828)

For Option 2:

$P = 5000$

$r = 0.09$

$t = 9$

$A_2 = 5000e^{(0.09)(9)} = 5000e^{0.81} = 5000(2.247907926) = 11239.53963 \approx 11239.54$

Comparing the final values, $A_1 = 11789.74$ and $A_2 = 11239.54$. Since $11789.74 > 11239.54$, Option 1 will earn more money.

3. Final Answer

Option 1: $11789.74

Option 2: $11239.54

Which investment will earn more money? Option 1