1. **State the problem:** We want to find the year when the value of Sean's car, initially 20000, depreciates to approximately 5000 given a 19% annual depreciation rate. 2. **Formula used:** The value of the car after $t$ years with depreciation rate $r$ is given by: $$ V = P(1 - r)^t $$ where $P$ is the initial value, $r$ is the depreciation rate, and $t$ is the number of years. 3. **Apply values:** Here, $P = 20000$, $r = 0.19$, and we want $V \approx 5000$. 4. **Set up the equation:** $$ 5000 = 20000(1 - 0.19)^t $$ $$ 5000 = 20000(0.81)^t $$ 5. **Divide both sides by 20000:** $$ \frac{5000}{20000} = \cancel{\frac{20000}{20000}}(0.81)^t $$ $$ 0.25 = (0.81)^t $$ 6. **Take natural logarithm on both sides:** $$ \ln(0.25) = \ln((0.81)^t) $$ $$ \ln(0.25) = t \ln(0.81) $$ 7. **Solve for $t$:** $$ t = \frac{\ln(0.25)}{\ln(0.81)} $$ 8. **Calculate values:** $$ \ln(0.25) \approx -1.3863 $$ $$ \ln(0.81) \approx -0.2097 $$ 9. **Final calculation:** $$ t \approx \frac{-1.3863}{-0.2097} \approx 6.61 $$ 10. **Interpretation:** The value of the car will be closest to 5000 after about 7 years. **Answer: C 7**