Given that Jason takes out a loan of $8,000 and owes $9,294.67 after three years with continuous compounding, we can use the formula for continuous compounding, which is:
\[ P(t) = P_0 e^{rt} \]
where:
- \( P(t) \) is the amount owed at time \( t \),
- \( P_0 \) is the initial principal (the amount of the loan),
- \( r \) is the annual interest rate,
- \( t \) is the time in years,
- \( e \) is the base of natural logarithms.
We know:
- \( P_0 = 8000 \),
- \( P(3) = 9294.67 \),
- \( t = 3 \).
We can set up the equation as follows:
\[ 9294.67 = 8000 e^{3r} \]
Now, we can solve for \( r \):
- Divide both sides by 8000:
\[ e^{3r} = \frac{9294.67}{8000} \]
- Calculate the right side:
\[ e^{3r} = 1.16183375 \]
- Take the natural logarithm (ln) of both sides:
\[ 3r = \ln(1.16183375) \]
- Calculate \( \ln(1.16183375) \):
\[ 3r \approx 0.14819 \]
- Divide by 3 to solve for \( r \):
\[ r \approx \frac{0.14819}{3} \approx 0.04939667 \]
Now convert \( r \) to a percentage:
\[ r \approx 0.0494 \times 100 \approx 4.94% \]
Rounded to two decimal places, the annual interest rate is approximately 4.94%.
The final continuous compounding formula for the loan is:
\[ P(t) = 8000 e^{0.0494t} \]
Among the provided options, the closest and correct representation of this model is:
P=8,000e^(0.05t)
So, the correct statement is:
P=8,000e0.05t