Question

a car is purchased for $41,000. each year it loses 25% of its value. after how many years will the car be worth $10,200 or less? (use the calculator provided if necessary.) write the smallest possible whole number answer.

Explanation:

Step1: Define the depreciation formula

The value of the car after \( n \) years follows an exponential decay model. The formula for the value \( V \) of the car after \( n \) years is \( V = P(1 - r)^n \), where \( P \) is the initial price, \( r \) is the annual depreciation rate, and \( n \) is the number of years. Here, \( P = 41000 \), \( r = 0.25 \) (since 25% = 0.25), and we want to find \( n \) when \( V \leq 10200 \). So the inequality is \( 41000(1 - 0.25)^n \leq 10200 \).

Step2: Simplify the inequality

First, simplify \( (1 - 0.25) = 0.75 \). So the inequality becomes \( 41000(0.75)^n \leq 10200 \). Divide both sides by 41000: \( (0.75)^n \leq \frac{10200}{41000} \approx 0.24878 \).

Step3: Solve for \( n \) using logarithms

Take the natural logarithm of both sides: \( \ln(0.75^n) \leq \ln(0.24878) \). Using the logarithm power rule \( \ln(a^b)=b\ln(a) \), we get \( n\ln(0.75) \leq \ln(0.24878) \). Since \( \ln(0.75) \) is negative (because \( 0.75 < 1 \)), when we divide both sides by \( \ln(0.75) \), the inequality sign flips. So \( n \geq \frac{\ln(0.24878)}{\ln(0.75)} \).

Calculate \( \frac{\ln(0.24878)}{\ln(0.75)} \approx \frac{-1.395}{-0.2877} \approx 4.85 \).

Step4: Determine the smallest whole number \( n \)

Since \( n \) must be a whole number and \( n \geq 4.85 \), the smallest whole number \( n \) is 5. We can check: For \( n = 4 \), \( V = 41000(0.75)^4 = 41000\times0.31640625 = 12972.65625 \), which is more than 10200. For \( n = 5 \), \( V = 41000(0.75)^5 = 41000\times0.2373046875 = 9729.4921875 \), which is less than 10200.

Answer:

5