[tex] \$1000 [/tex] are deposited in an account with an [tex] 8.5 \% [/tex] interest rate, compounded continuously. What is the balance after 5 years?

Now enter [tex] t [/tex], or the number of years in the bank.

[tex] t = [?] [/tex]

[tex] \square [/tex]

Enter



Answer :

To determine the balance of the account after 5 years with an interest rate of 8.5% compounded continuously, we use the formula for continuous compounding. The formula is:

[tex]\[ A = P \cdot e^{rt} \][/tex]

where:
- [tex]\( A \)[/tex] is the amount of money accumulated after [tex]\( t \)[/tex] years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial deposit).
- [tex]\( r \)[/tex] is the annual interest rate (expressed as a decimal).
- [tex]\( t \)[/tex] is the time the money is invested for, in years.
- [tex]\( e \)[/tex] is the base of the natural logarithm, approximately equal to 2.71828.

Given the values:
- [tex]\( P = \$ 1000 \)[/tex]
- [tex]\( r = 0.085 \)[/tex] (which is 8.5% expressed as a decimal)
- [tex]\( t = 5 \)[/tex] years

We substitute these values into the formula:

[tex]\[ A = 1000 \cdot e^{0.085 \cdot 5} \][/tex]

First, calculate the exponent:

[tex]\[ 0.085 \times 5 = 0.425 \][/tex]

So, the equation becomes:

[tex]\[ A = 1000 \cdot e^{0.425} \][/tex]

Using the natural constant [tex]\( e \)[/tex] (approximately 2.71828), we can calculate [tex]\( e^{0.425} \)[/tex]:

[tex]\[ e^{0.425} \approx 1.52959 \][/tex]

Now, multiply this value by the initial deposit [tex]\( P \)[/tex]:

[tex]\[ A = 1000 \cdot 1.52959 = 1529.59 \][/tex]

Thus, the balance after 5 years is approximately:

[tex]\[ \boxed{1529.59} \][/tex]