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Question

John bought a car for $25,000 in 2010. The car depreciates at a constant rate of 21% per year. 

 

A.) write an equation to represent the balance the car x years after 2010.

B.)Does this represent growth or decay? Explain. 

C.)What will the value of the truck be ion 2017?

D.)in what year will the car be $12,500?

Answer 1

(A)

 

The typical set-up for this kind of equation is:

 

V(x) = a(r)^x

 

where a = initial value, r = rate of growth, x = time that has passed, and V(x) = value at time x. Usually, to write the equation, you'll be given a and r. For this problem, the initial value of the car is a = 25000. Since the car depreciates at a rate of 21% per year, each year it's worth 100 - 21 = 79% of what it was worth the year before. So, r = 0.79. Plug those values in to write your equation:

 

V(x) = 25000(0.79)^x

 

 

(B)

 

This is an example of exponential decay, since the car's value is decaying (going down) over time. You can tell just by looking at the equation: whenever r < 1, the value is decaying; whenever r > 1, the value is growing.

 

 

(C)

 

To find the value of the truck in 2017, figure out how many years have passed since 2010 (x). 2017 - 2010 = 7 years, so x = 7. Plug that in and solve for V(7):

 

V(7) = 25000(0.79)⁷

 

V(7) = 25000(0.192)

V(7) = 4800.98

 

The value of the car in 2017 is V(7) = $4,800.98.

 

 

(D)

 

To find what year the car will be valued at $12,500, plug V(x) = 12500 into the equation and solve for x:

 

12500 = 25000(0.79)^x

0.5 = 0.79^x

log_0.79(0.5) = x

2.94 = x

 

The car will be valued at $12,500 after x = 2.94 years, at the end of the year 2010 + 2 = 2012.