The value V of a certain automobile that is t years old can be modeled by 

Upper V left parenthesis t right parenthesis equals 14 comma 580 left parenthesis 0.79 right parenthesis Superscript tV(t)=14,580(0.79)t. 

According to the​ model, when will the car be worth each of the following​ amounts?

​(a) 

​$70007000

​(b) 

​$50005000

​(c) 

​$4000

The value $ V(t) = 14,580(0.79)^t $ represents the value of an automobile in terms of the years $ t $. To find out when the car will be worth a specific amount, we need to solve for $ t $ in each case by setting $ V(t) $ equal to the given amount and solving the equation.

### Formula:
\[ V(t) = 14,580(0.79)^t \]

For each part, we follow the same steps:

1. Set the value $ V(t) $ equal to the given amount.
2. Divide both sides by 14,580.
3. Take the natural logarithm (ln) of both sides to solve for $ t $.
4. Use logarithmic properties to isolate $ t $.

---

### (a) When will the car be worth $7000?

Set $ V(t) = 7000 $:

\[
7000 = 14,580(0.79)^t
\]

Now solve for $ t $:

\[
\frac{7000}{14,580} = (0.79)^t
\]

\[
0.48 = (0.79)^t
\]

Take the natural logarithm of both sides:

\[
\ln(0.48) = t \cdot \ln(0.79)
\]

\[
t = \frac{\ln(0.48)}{\ln(0.79)}
\]

Now calculate $ t $.

---

### (b) When will the car be worth $5000?

Set $ V(t) = 5000 $:

\[
5000 = 14,580(0.79)^t
\]

Solve for $ t $:

\[
\frac{5000}{14,580} = (0.79)^t
\]

\[
0.343 = (0.79)^t
\]

Take the natural logarithm of both sides:

\[
\ln(0.343) = t \cdot \ln(0.79)
\]

\[
t = \frac{\ln(0.343)}{\ln(0.79)}
\]

Now calculate $ t $.

---

### (c) When will the car be worth $4000?

Set $ V(t) = 4000 $:

\[
4000 = 14,580(0.79)^t
\]

Solve for $ t $:

\[
\frac{4000}{14,580} = (0.79)^t
\]

\[
0.2744 = (0.79)^t
\]

Take the natural logarithm of both sides:

\[
\ln(0.2744) = t \cdot \ln(0.79)
\]

\[
t = \frac{\ln(0.2744)}{\ln(0.79)}
\]

Now calculate $ t $.

---

I will now calculate the values of $ t $ for each part.Here are the results for each case:

- (a) The car will be worth $7,000 after approximately **3.11 years**.
- (b) The car will be worth $5,000 after approximately **4.54 years**.
- (c) The car will be worth $4,000 after approximately **5.49 years**.

Would you like more details on any of these calculations?

### Related Questions:
1. How does the decay factor $0.79$ affect the value over time?
2. What would be the value of the car after 10 years?
3. How long will it take for the car to lose half of its value?
4. Can we model a different decay rate for faster depreciation?
5. How is logarithmic decay used in other real-life applications?

### Tip:
When solving exponential decay problems, always use logarithms to handle the exponent efficiently!

The value V of a certain automobile that is t years old can be modeled by V(t) = 14,580(0.79)^t. According to the model, when will the car be worth each of the following amounts? (a) $7000 (b) $5000 (c) $4000.

Solve the problem of determining when an automobile's value modeled by V(t) = 14,580(0.79)^t will reach $7000, $5000, and $4000. Learn how to apply logarithmic decay and exponential functions to real-world scenarios, suitable for high school students in Grades 9-12.

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