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

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 $20,000 and depreciates at a rate of 15% per year. how old is the car?\n$\frac{1}{4}$ years\n$\frac{1}{3}$ 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 $20,000 and depreciates at a rate of 15% per year. how old is the car?\n$\frac{1}{4}$ years\n$\frac{1}{3}$ years\n3 years\n4 years

Answer

Explanation:

Step1: Substitute given values into formula

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

Step2: Simplify the equation

First, simplify $(1 - 0.15)$ to $0.85$. The equation becomes $\frac{12282.50}{20000}=0.85^t$. Calculate $\frac{12282.50}{20000}=0.614125$, so $0.614125 = 0.85^t$.

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 get $\ln(0.614125)=t\ln(0.85)$.

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:

3 years