1. **State the problem:** We have a car purchased for 20700 dollars, which depreciates at 8% per year. We want to find how many years it will take for the car's value to decrease to 7200 dollars. 2. **Formula used:** The value of the car after $t$ years with depreciation rate $r$ is given by the exponential decay formula: $$ V = P(1 - r)^t $$ where: - $V$ is the value after $t$ years, - $P$ is the initial value, - $r$ is the depreciation rate (as a decimal), - $t$ is the time in years. 3. **Plug in the known values:** $$ 7200 = 20700(1 - 0.08)^t $$ which simplifies to $$ 7200 = 20700(0.92)^t $$ 4. **Isolate the exponential term:** Divide both sides by 20700: $$ \frac{7200}{20700} = \cancel{\frac{20700}{20700}}(0.92)^t $$ $$ \frac{7200}{20700} = (0.92)^t $$ 5. **Simplify the fraction:** $$ \frac{7200}{20700} = \frac{7200 \div 300}{20700 \div 300} = \frac{24}{69} $$ 6. **Take the natural logarithm of both sides:** $$ \ln\left(\frac{24}{69}\right) = \ln\left((0.92)^t\right) $$ Using the logarithm power rule: $$ \ln\left(\frac{24}{69}\right) = t \ln(0.92) $$ 7. **Solve for $t$:** $$ t = \frac{\ln\left(\frac{24}{69}\right)}{\ln(0.92)} $$ 8. **Calculate the values:** $$ \ln\left(\frac{24}{69}\right) \approx \ln(0.3478) \approx -1.056 $$ $$ \ln(0.92) \approx -0.08338 $$ 9. **Divide:** $$ t \approx \frac{-1.056}{-0.08338} \approx 12.67 $$ 10. **Round to the nearest year:** $$ t \approx 13 \text{ years} $$ **Final answer:** It will take approximately 13 years for the car's value to depreciate to 7200 dollars.