Question

1. $3,000 is invested into an account compounding interest continuously. after 4 years, there is $3,315.51 in the account. find the interest rate.

Explanation:

The formula for continuous compounding is \( A = Pe^{rt} \), where \( A \) is the final amount, \( P \) is the principal amount, \( r \) is the annual interest rate (in decimal), and \( t \) is the time in years.

Step 1: Identify the given values

We know that \( P = 3000 \), \( A = 3315.51 \), and \( t = 4 \). We need to solve for \( r \).

Step 2: Substitute the values into the formula

Substitute \( P \), \( A \), and \( t \) into the formula \( A = Pe^{rt} \):
\[
3315.51 = 3000e^{4r}
\]

Step 3: Divide both sides by 3000

\[
\frac{3315.51}{3000} = e^{4r}
\]
Calculate \( \frac{3315.51}{3000} \):
\[
1.10517 = e^{4r}
\]

Step 4: Take the natural logarithm of both sides

\[
\ln(1.10517) = \ln(e^{4r})
\]
Since \( \ln(e^{x}) = x \), this simplifies to:
\[
\ln(1.10517) = 4r
\]

Step 5: Solve for \( r \)

First, calculate \( \ln(1.10517) \approx 0.100 \) (using a calculator). Then:
\[
0.100 = 4r
\]
Divide both sides by 4:
\[
r = \frac{0.100}{4} = 0.025
\]

Step 6: Convert to a percentage

To convert \( r \) from a decimal to a percentage, multiply by 100:
\[
r = 0.025 \times 100 = 2.5\%
\]

Answer:

The interest rate is \( 2.5\% \) per year.