Answer:
D
Step-by-step explanation:
We know that the equation for the depreciation of the car is:
![y=A(1-r)^t](https://tex.z-dn.net/?f=y%3DA%281-r%29%5Et)
Where y is the current cost, A is the original cost, r is the rate of depreciation, and t is the time, in years.
We are told that the value of the car now is half of what it originally cost. So:
![y=\frac{1}{2}A](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B2%7DA)
Substitute this for y:
![\frac{1}{2}A=A(1-r)^t](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DA%3DA%281-r%29%5Et)
We also know that the rate of depreciation is 10% or 0.1. Substitute 0.1 for r:
![\frac{1}{2}A=A(1-0.1)^t](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DA%3DA%281-0.1%29%5Et)
So, let's solve for t. Divide both sides by A:
![\frac{1}{2}=(1-0.1)^t](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%3D%281-0.1%29%5Et)
Subtract within the parentheses:
![\frac{1}{2}=(0.9)^t](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%3D%280.9%29%5Et)
Take the natural log of both sides:
![\ln(\frac{1}{2})=\ln((0.9)^t})](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B1%7D%7B2%7D%29%3D%5Cln%28%280.9%29%5Et%7D%29)
Using the properties of logarithms, we can move the t to the front:
![\ln(\frac{1}{2})=t\ln(0.9)](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B1%7D%7B2%7D%29%3Dt%5Cln%280.9%29)
Divide both sides by ln(0.9):
![t=\frac{\ln(\frac{1}{2})}{\ln(0.9)}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Cln%28%5Cfrac%7B1%7D%7B2%7D%29%7D%7B%5Cln%280.9%29%7D)
Use a calculator. So, the car is approximately:
![t\approx6.6\text{ years old}](https://tex.z-dn.net/?f=t%5Capprox6.6%5Ctext%7B%20years%20old%7D)
Our answer is D.
And we're done!