To determine when the car will be worth $4,000 using the given depreciation function \( y = 21,000(1 - 0.15)^x \), we need to set \( y \) equal to $4,000 and solve for \( x \).
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Set the function equal to $4,000: \[ 21,000(1 - 0.15)^x = 4,000 \]
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Divide both sides by 21,000: \[ (1 - 0.15)^x = \frac{4,000}{21,000} \]
This simplifies to: \[ (0.85)^x = \frac{4}{21} \]
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Now, calculate \( \frac{4}{21} \): \[ \frac{4}{21} \approx 0.1905 \]
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To solve for \( x \), use logarithms: \[ x \log(0.85) = \log(0.1905) \] \[ x = \frac{\log(0.1905)}{\log(0.85)} \]
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Calculating this:
- \( \log(0.1905) \approx -0.7192 \)
- \( \log(0.85) \approx -0.0718 \)
Thus, \[ x \approx \frac{-0.7192}{-0.0718} \approx 10.0 \]
So the car will be worth approximately $4,000 after about 10 years.
Conclusion:
Based on the calculations, Statement #1 is the correct one: The car will have a value of $4,000 after roughly 10.2 years (very close to the calculated 10 years).
If you have the option to verify with GeoGebra, plotting the function and locating the intersection point with the line \( y = 4,000 \) confirms this time frame visually as well.