Question
Salim purchased a new car for $21,000, and it depreciates in value each year. The function
y = 21, 000(1 − 0.15)
x
represents the depreciation. He also purchases a motorcycle, which
depreciates each year as well. Its depreciation is represented by the function
y = 14, 000(1 − 0.09)
x
. Graph these two functions using Desmos. Which statement best describes
when the car will be worth $4,000?
Statement #1: The car will have a value of $4,000 after roughly 10.2 years.
Statement #2: The car will have a value of $4,000 after roughly 13.3 years.
y = 21, 000(1 − 0.15)
x
represents the depreciation. He also purchases a motorcycle, which
depreciates each year as well. Its depreciation is represented by the function
y = 14, 000(1 − 0.09)
x
. Graph these two functions using Desmos. Which statement best describes
when the car will be worth $4,000?
Statement #1: The car will have a value of $4,000 after roughly 10.2 years.
Statement #2: The car will have a value of $4,000 after roughly 13.3 years.
Answers
GPT-4o mini
To determine when Salim's car will be worth $4,000 using the depreciation function \( y = 21,000(1 - 0.15)^x \), we can set \( y \) equal to $4,000 and solve for \( x \):
\[
4000 = 21000(1 - 0.15)^x
\]
Next, we simplify this equation:
\[
\frac{4000}{21000} = (1 - 0.15)^x
\]
Calculating the left side:
\[
\frac{4000}{21000} = \frac{4}{21} \approx 0.1905
\]
The equation now looks like this:
\[
0.1905 = (0.85)^x
\]
To solve for \( x \), we can take the logarithm of both sides:
\[
\log(0.1905) = x \log(0.85)
\]
Now, solving for \( x \):
\[
x = \frac{\log(0.1905)}{\log(0.85)}
\]
Using a calculator:
\[
\log(0.1905) \approx -0.720
\]
\[
\log(0.85) \approx -0.069
\]
Thus,
\[
x \approx \frac{-0.720}{-0.069} \approx 10.43
\]
So it will take approximately 10.43 years for the car to depreciate to $4,000.
Comparing this to the provided statements:
- Statement #1: The car will have a value of $4,000 after roughly 10.2 years.
- Statement #2: The car will have a value of $4,000 after roughly 13.3 years.
Statement #1 is closer to our calculated value of approximately 10.43 years. Therefore, **Statement #1** is the correct statement that best describes when the car will be worth $4,000.
\[
4000 = 21000(1 - 0.15)^x
\]
Next, we simplify this equation:
\[
\frac{4000}{21000} = (1 - 0.15)^x
\]
Calculating the left side:
\[
\frac{4000}{21000} = \frac{4}{21} \approx 0.1905
\]
The equation now looks like this:
\[
0.1905 = (0.85)^x
\]
To solve for \( x \), we can take the logarithm of both sides:
\[
\log(0.1905) = x \log(0.85)
\]
Now, solving for \( x \):
\[
x = \frac{\log(0.1905)}{\log(0.85)}
\]
Using a calculator:
\[
\log(0.1905) \approx -0.720
\]
\[
\log(0.85) \approx -0.069
\]
Thus,
\[
x \approx \frac{-0.720}{-0.069} \approx 10.43
\]
So it will take approximately 10.43 years for the car to depreciate to $4,000.
Comparing this to the provided statements:
- Statement #1: The car will have a value of $4,000 after roughly 10.2 years.
- Statement #2: The car will have a value of $4,000 after roughly 13.3 years.
Statement #1 is closer to our calculated value of approximately 10.43 years. Therefore, **Statement #1** is the correct statement that best describes when the car will be worth $4,000.