To solve the problem, we can follow these steps:
1. Write a function to model the value of the car over time.
The value of the car decreases by 15% each year, which means it retains 85% of its value after each year. We can express this using the following function, where \( V(t) \) is the value of the car at year \( t \):
\[ V(t) = V_0 \cdot (1 - r)^t \]
Here:
- \( V_0 = 24,250 \) (the initial value of the car)
- \( r = 0.15 \) (the rate at which the car loses value)
- \( t \) = number of years
So, the function becomes:
\[ V(t) = 24250 \cdot (0.85)^t \]
2. What will be the value of the car after 10 years?
To find the value of the car after 10 years, we plug \( t = 10 \) into the function:
\[ V(10) = 24250 \cdot (0.85)^{10} \]
Calculating \( (0.85)^{10} \):
\[ (0.85)^{10} \approx 0.196874 \]
Now calculate \( V(10) \):
\[ V(10) \approx 24250 \cdot 0.196874 \approx 4776.72 \]
So, the value of the car after 10 years will be approximately $4,776.72.
3. After how many years will the car be worth $20,000?
We need to find \( t \) such that:
\[ V(t) = 20000 \]
Using the value of the function:
\[ 20000 = 24250 \cdot (0.85)^t \]
To solve for \( t \), we first divide both sides by 24250:
\[ \frac{20000}{24250} = (0.85)^t \]
Calculating the left side:
\[ \frac{20000}{24250} \approx 0.8257 \]
Now we have:
\[ 0.8257 = (0.85)^t \]
Next, we take the logarithm of both sides to solve for \( t \):
\[ \log(0.8257) = t \cdot \log(0.85) \]
Now solve for \( t \):
\[ t = \frac{\log(0.8257)}{\log(0.85)} \]
Calculating both logarithms:
- \(\log(0.8257) \approx -0.0827\)
- \(\log(0.85) \approx -0.0706\)
Now substituting:
\[ t \approx \frac{-0.0827}{-0.0706} \approx 1.17 \]
Thus, it will take approximately 1.2 years for the car to be worth $20,000.
Summary of Answers:
- The function is \( V(t) = 24250 \cdot (0.85)^t \).
- The value of the car after 10 years is approximately $4,776.72.
- The car will be worth $20,000 after approximately 1.2 years.
13 answers