Salim purchases 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 GeoGebra. 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.

(1 point)
Statement #

1 answer

To determine when the car will be worth $4,000 using the depreciation function \( y = 21,000(1 - 0.15)^x \), you can set the equation equal to $4,000 and solve for \( x \):

\[ 21,000(1 - 0.15)^x = 4,000. \]

First, simplify the equation:

\[ (1 - 0.15)^x = \frac{4,000}{21,000}, \]

which simplifies to:

\[ (0.85)^x = \frac{4}{21} \approx 0.1905. \]

Next, take the logarithm of both sides:

\[ \log((0.85)^x) = \log(0.1905). \]

Using the power rule of logarithms:

\[ x \cdot \log(0.85) = \log(0.1905). \]

Now, solve for \( x \):

\[ x = \frac{\log(0.1905)}{\log(0.85)}. \]

Calculate the values:

  1. Calculate \( \log(0.1905) \) and \( \log(0.85) \) using a calculator:

    • \( \log(0.1905) \approx -0.719 \)
    • \( \log(0.85) \approx -0.0682 \)
  2. Now plug these values into the formula: \[ x \approx \frac{-0.719}{-0.0682} \approx 10.54. \]

From the result, we see that after approximately 10.5 years, the car will be worth $4,000.

Conclusion

The best statement to describe when the car will be worth $4,000 is:

Statement #1: The car will have a value of $4,000 after roughly 10.2 years.

(This statement is closer to the calculated value of 10.5 years compared to Statement #2.)