Salim purchases a new car for $21,000, and it depreciates in value each year. The function y=21,000(1−0.15)x

Question

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 #

Answer 1

To determine when the car will be worth $4,000 using the given depreciation function, we can set the equation for the car's value equal to $4,000:

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

First, simplify the equation:

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

Calculate the right side:

\[ \frac{4,000}{21,000} \approx 0.1905 \]

Now the equation becomes:

\[ (0.85)^x \approx 0.1905 \]

To solve for \(x\), we can take the logarithm of both sides. Using natural logarithms (ln):

\[ \ln((0.85)^x) = \ln(0.1905) \]

This simplifies to:

\[ x \cdot \ln(0.85) = \ln(0.1905) \]

Now, solve for \(x\):

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

Calculating this will give us the number of years it takes for the car to reach a value of $4,000:

1. Calculate \(\ln(0.1905)\):

   • \(\ln(0.1905) \approx -1.6576\)

2. Calculate \(\ln(0.85)\):

   • \(\ln(0.85) \approx -0.1625\)

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

\[ x \approx \frac{-1.6576}{-0.1625} \approx 10.19 \text{ (approximately 10.2 years)} \]

Based on this calculation, Statement #1 is accurate:

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