A new car is purchased for 20700 dollars. The value of the car depreciates at 7.25% per year. To the nearest year, how long will it be until the value of the car is 10800 dollars

Answer 1

P = P₀(1 - 0.0725)^t = P₀·0.9275^t = 20700·0.9275^t;

10800 = 20700·0.9275^t; 10800/20700 = 0.9275^t;

0.9275^t = 12/23; t·ln0.9275 = ln(12/23); t = ln(12/23)/ln(0.9275) = 8.64 years

Answer 2

P = P₀(1 - 0.0725)^t = P₀·0.9275^t = 20700·0.9275^t;

10800 = 20700·0.9275^t; 10800/20700 = 0.9275^t;

0.9275^t = 12/23; t·ln0.9275 = ln(12/23); t = ln(12/23)/ln(0.9275) = 8.64 years

Answer 3

Hi Ella! I hope you're doing well today.

This question involves the compound interest equation, but we will modify it for depreciation:

A = P(1-R)ⁿ

A is the value after depreciation; P is the principal, or original worth; R is the rate of depreciation; and n is the number of years. In the compound interest equation, the parentheses contain (1+R), but since we are losing value as time goes on, we subtract from 1.

Using this equation, we just plug in our numbers:

10800 = 20700(1-(0.0725))ⁿ

Make sure to change the rate, R, to decimal form when you put it in the equation! Since we want to know how long it will take for the car to depreciate to $10,800, we just solve for n:

10800/20700 = (0.9275)ⁿ

0.52174 = (0.9275)ⁿ

log(0.52174) = nlog(0.9275)

log(0.52174)/log(0.9275) = n

n = 8.644 years = 9 years

Recall the log rules that allow us to move exponents around:

x^y = y * log(x)

This rule makes the problem much easier, and allows us to solve for n! I hope this explanation helps and feel free to ask any questions!

Answer 4

So Ella L.,

If the car depreciates 7.25% yr^-1, that means it retains 92.75% of its value per year. Why do we immediately shift from value-lost to value-retained basis? It is because the depreciation only operates on the value-retained basis of the car. There is no mathematical particular interest to the value it has just lost. And if you brooded on it too much, it might depress you. Instead, think of it as "Wow, look how much value my car still has!".

So what happens now? Now, you set up an exponential function -- that's a convenient, short way of expressing those yearly adjustments to car value. In this case, the retained-value-fraction (or decimal) is the thing that must be repeatedly multiplied by, as the years roll by, to decrease the "book value". So we have:

V(t) = 20700 * (0.9275)ⁿ , where V is value at time t in years, and n is the number of elapsed years. So we want a V(t) to be 10800. Plug that in and solve:

10800 = V(t=n=?) = 20700 *(0.9275)ⁿ

We have to pry the value for n out of this equation. So we strip it away from associated "things" by the inverse mathematical operations, in the proper order! (the opposite of how PEMDAS assembled the value in the first place). First the *20700, pry away by dividing the equation through by 20700:

(10800/20700) ~= 0.521739 = (0.9275)ⁿ

Next, remove the exponentiation operation on n by the reverse operation: log or ln (your choice), I'll use log here:

log(0.521739) = log((0.9275)ⁿ) = n*log(0.9275)

-0.28255 = n * (-0.032686)

n~= 8.644 .

So, to the nearest year, that rounds up to 9 years.

To make a car-based analogy to solving an equation: the frame of the car is your unknown, and the total car is your final equation form. To see what the frame looks like, you start stripping away portions of the outer cladding = fenders, bumpers, wheels, etc. Also, at the appropriate time, big chunks of things like the engine. When you are done, voila, the frame!

Hope this helped you, -- Cheers, --Mr. d.