an equation for the depreciation of a car is given by y = a(1 - r)^t, where y = current value of the car, a…

an equation for the depreciation of a car is given by y = a(1 - r)^t, where y = current value of the car, a = original cost, r = rate of depreciation, and t = time, in years. the current value of a car is $12,282.50. the car originally cost $20,000 and depreciates at a rate of 15% per year. how old is the car?\n1/4 years\n1/3 years\n3 years\n4 years

an equation for the depreciation of a car is given by y = a(1 - r)^t, where y = current value of the car, a = original cost, r = rate of depreciation, and t = time, in years. the current value of a car is $12,282.50. the car originally cost $20,000 and depreciates at a rate of 15% per year. how old is the car?\n1/4 years\n1/3 years\n3 years\n4 years

Answer

Explanation:

Step1: Substitute given values into formula

Given $y = 12282.50$, $A=20000$, $r = 0.15$ into $y = A(1 - r)^t$, we get $12282.50=20000(1 - 0.15)^t$.

Step2: Simplify the equation

First, simplify the right - hand side. $1 - 0.15=0.85$, so the equation becomes $12282.50 = 20000\times0.85^t$. Then divide both sides by 20000: $\frac{12282.50}{20000}=0.85^t$, and $0.614125 = 0.85^t$.

Step3: Take the logarithm of both sides

Take the natural logarithm of both sides: $\ln(0.614125)=\ln(0.85^t)$. Using the property of logarithms $\ln(a^b)=b\ln(a)$, we have $\ln(0.614125)=t\ln(0.85)$.

Step4: Solve for t

$t=\frac{\ln(0.614125)}{\ln(0.85)}$. Calculate $\ln(0.614125)\approx - 0.487$ and $\ln(0.85)\approx- 0.1625$. Then $t=\frac{-0.487}{-0.1625}=3$.

Answer:

C. 3 years