Let's solve the problem step-by-step using the formula for continuous compound interest:
[tex]\[ A = P e^{rt} \][/tex]
where:
- [tex]\( A \)[/tex] is the ending amount,
- [tex]\( P \)[/tex] is the principal (initial amount),
- [tex]\( r \)[/tex] is the interest rate,
- [tex]\( t \)[/tex] is the time in years,
- [tex]\( e \)[/tex] is the base of the natural logarithm, approximately equal to 2.71828.
Given:
- [tex]\( A = 1649 \)[/tex],
- [tex]\( P = 100 \)[/tex],
- [tex]\( t = 10 \)[/tex].
We need to find the rate [tex]\( r \)[/tex]. Rearrange the formula to solve for [tex]\( r \)[/tex]:
[tex]\[ 1649 = 100 e^{10r} \][/tex]
Divide both sides by 100:
[tex]\[ 16.49 = e^{10r} \][/tex]
Take the natural logarithm (ln) of both sides to solve for [tex]\( r \)[/tex]:
[tex]\[ \ln(16.49) = \ln(e^{10r}) \][/tex]
Using the property of logarithms that [tex]\(\ln(e^x) = x\)[/tex]:
[tex]\[ \ln(16.49) = 10r \][/tex]
Now, solve for [tex]\( r \)[/tex]:
[tex]\[ r = \frac{\ln(16.49)}{10} \][/tex]
Calculate the natural logarithm of 16.49:
[tex]\[ \ln(16.49) \approx 2.79801 \][/tex]
Divide by 10:
[tex]\[ r \approx \frac{2.79801}{10} \][/tex]
[tex]\[ r \approx 0.279801 \][/tex]
Convert [tex]\( r \)[/tex] to a percentage by multiplying by 100:
[tex]\[ r \times 100 \approx 0.279801 \times 100 \][/tex]
[tex]\[ r \approx 27.9801\% \][/tex]
The closest option to 27.9801\% is 28\%.
Therefore, the rate of interest Adam earned is:
D. 28%
