Question

adam put $100 in a savings account. after 10 years, he had $1649 in the account. what rate of interest did he earn? use the formula $a = pe^{rt}$, where $a$ is the ending amount, $p$ is the principal (initial amount), $r$ is the interest rate, and $t$ is time.

a. 28%

b. 20%

c. 3%

d. 5%

Explanation:

Step1: Substitute values into formula

Given \( A = 1649 \), \( P = 100 \), \( t = 10 \), and formula \( A = Pe^{rt} \). Substitute:
\( 1649 = 100e^{10r} \)

Step2: Isolate exponential term

Divide both sides by 100:
\( \frac{1649}{100} = e^{10r} \)
\( 16.49 = e^{10r} \)

Step3: Take natural log of both sides

\( \ln(16.49) = \ln(e^{10r}) \)
Using \( \ln(e^x) = x \), we get:
\( \ln(16.49) = 10r \)

Step4: Solve for \( r \)

Calculate \( \ln(16.49) \approx 2.799 \). Then:
\( r = \frac{2.799}{10} \approx 0.28 \)
Convert to percentage: \( 0.28 \times 100 = 28\% \)

Answer:

A. 28%