The value \( V \) of a certain automobile that is \( t \) years old can be modeled by \( V(t)=14,576(0.84) \). According to the model, when will the car be worth each of the following amounts? (a) \( \$ 7000 \) (b) \( \$ 6000 \) (c) \( \$ 4000 \)

The car will be worth \$7,000 after about 4.2 years. **Part (b): When will the car be worth \$6,000?** We'll follow a similar approach to find \( t \) when \( V(t) = \$6,000 \): \[ 14,576 \times (0.84)^t = 6,000 \] 1. **Divide both sides by 14,576:** \[ (0.84)^t = \frac{6,000}{14,576} \approx 0.412 \] 2. **Take the natural logarithm of both sides:** \[ t \times \ln(0.84) = \ln(0.412) \] \[ t = \frac{\ln(0.412)}{\ln(0.84)} \] 3. **Calculate the values:** \[ \ln(0.412) \approx -0.890 \] \[ \ln(0.84) \approx -0.174 \] \[ t \approx \frac{-0.890}{-0.174} \approx 5.12 \text{ years} \] **Conclusion:** **b) The car will be worth \$6,000 after approximately 5.1 years.** **Part (c): When will the car be worth \$4,000?** Now, we'll find \( t \) when \( V(t) = \$4,000 \): \[ 14,576 \times (0.84)^t = 4,000 \] 1. **Divide both sides by 14,576:** \[ (0.84)^t = \frac{4,000}{14,576} \approx 0.275 \] 2. **Take the natural logarithm of both sides:** \[ t \times \ln(0.84) = \ln(0.275) \] \[ t = \frac{\ln(0.275)}{\ln(0.84)} \] 3. **Calculate the values:** \[ \ln(0.275) \approx -1.287 \] \[ \ln(0.84) \approx -0.174 \] \[ t \approx \frac{-1.287}{-0.174} \approx 7.40 \text{ years} \] **Conclusion:** **c) The car will be worth \$4,000 after approximately 7.4 years.** **Summary:** - **\$7,000:** Approximately 4.2 years - **\$6,000:** Approximately 5.1 years - **\$4,000:** Approximately 7.4 years