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.

(1 point)
Statement #

1 answer

To determine when the car will be worth $4,000 based on the depreciation function \( y = 21,000(1 - 0.15)^x \), we need to set \( y \) to $4,000 and solve for \( x \):

\[ 4000 = 21000(1 - 0.15)^x \]

Dividing both sides by 21,000:

\[ \frac{4000}{21000} = (0.85)^x \]

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

Now we apply the logarithm to both sides to solve for \( x \):

\[ \log\left(\frac{4}{21}\right) = x \cdot \log(0.85) \]

Solving for \( x \):

\[ x = \frac{\log\left(\frac{4}{21}\right)}{\log(0.85)} \]

Now, performing the calculations:

  1. Calculate \( \log\left(\frac{4}{21}\right) \):

    • \( \frac{4}{21} \approx 0.1905 \)
    • \( \log(0.1905) \approx -0.720 \)
  2. Calculate \( \log(0.85) \):

    • \( \log(0.85) \approx -0.069 \)

Now plug these into the equation for \( x \):

\[ x \approx \frac{-0.720}{-0.069} \approx 10.43 \]

Thus, the car will be worth $4,000 after approximately 10.43 years.

Comparing our result:

  • 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.

The correct choice is Statement #1, since it is closer to our calculated value of approximately 10.43 years.