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) $$ \$ 4000 $$

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To determine when the car will be worth a specific amount, we'll use the given depreciation model:

$$ V(t) = 14,576 \times (0.84)^t $$

Here, $$ V(t) $$ is the value of the car after $$ t $$ years, and $$ 0.84 $$ is the depreciation factor per year.

**Part (a): When will the car be worth \$7,000?**

We need to solve for $$ t $$ when $$ V(t) = \$7,000 $$:

$$
14,576 \times (0.84)^t = 7,000
$$

1. **Divide both sides by 14,576:**

$$
(0.84)^t = \frac{7,000}{14,576} \approx 0.480
$$

2. **Take the natural logarithm of both sides to solve for $$ t $$:**

$$
\ln\left(0.84^t\right) = \ln(0.480)
$$

$$
t \times \ln(0.84) = \ln(0.480)
$$

$$
t = \frac{\ln(0.480)}{\ln(0.84)}
$$

3. **Calculate the values:**

$$
\ln(0.480) \approx -0.733
$$
$$
\ln(0.84) \approx -0.174
$$

$$
t \approx \frac{-0.733}{-0.174} \approx 4.21 \text{ years}
$$

**Conclusion:**

**a) The car will be worth \$7,000 after approximately 4.2 years.**
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

The value of a certain automobile that is years old can be modeled by . According to the model,
when will the car be worth each of the following amounts?
(a)
(b)
©

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The value of a certain automobile that is years old can be modeled by . According to the model, when will the car be worth each of the following amounts? (a) (b) ©