Question

suppose that $2600 is initially invested in an account at a fixed interest rate, compounded continuously. suppose also that, after six years, the amount of money in the account is $3392. find the interest rate per year. write your answer as a percentage. do not round any intermediate computations, and round your percentage to the nearest hundredth. % per year

Explanation:

Step1: Recall continuous compounding formula

The formula for continuous compounding is \( A = Pe^{rt} \), where \( A \) is the final amount, \( P \) is the principal (initial investment), \( r \) is the annual interest rate (in decimal), and \( t \) is the time in years. We know \( P = 2600 \), \( A = 3392 \), and \( t = 6 \). We need to solve for \( r \).

Step2: Substitute known values into the formula

Substitute \( P = 2600 \), \( A = 3392 \), and \( t = 6 \) into \( A = Pe^{rt} \):
\( 3392 = 2600e^{6r} \)

Step3: Solve for \( e^{6r} \)

Divide both sides by 2600:
\( \frac{3392}{2600} = e^{6r} \)
Simplify \( \frac{3392}{2600} \): \( \frac{3392\div 4}{2600\div 4}=\frac{848}{650}=\frac{424}{325}\approx 1.304615 \)
So, \( 1.304615 = e^{6r} \)

Step4: Take the natural logarithm of both sides

Take \( \ln \) of both sides:
\( \ln(1.304615) = \ln(e^{6r}) \)
Since \( \ln(e^{x}) = x \), this simplifies to:
\( \ln(1.304615) = 6r \)

Step5: Solve for \( r \)

Calculate \( \ln(1.304615) \approx 0.265 \) (using a calculator). Then:
\( r = \frac{\ln(1.304615)}{6} \approx \frac{0.265}{6} \approx 0.044167 \)

Step6: Convert \( r \) to a percentage

Multiply \( r \) by 100 to get the percentage:
\( r \approx 0.044167\times 100 = 4.4167\% \approx 4.42\% \) (rounded to the nearest hundredth)

Answer:

\( 4.42\% \)