Answer:
This situation represents [a.) exponential decay
The rate of growth or decay, r, is equal to c.) 0.2
So the value of the computer each year is g.) 80% of the value in the previous year.
It will take [i.) 3 years for the value of the computer to reach $512.
Step-by-step explanation:
Since the value depreciates each year, it represents exponential decay.
The rate of growth or decay, r, is equal to
Depreciates 20%, so r = 0.2.
So the value of the computer each year is ... of the value in the previous year,
Depreciates 20%, so it is 100% - 20% = 80% of the value in the previous year.
It will take x years for the value of the computer to reach $512.
The value of the computer after x years is given by:
![V(x) = V(0)(1-r)^x](https://tex.z-dn.net/?f=V%28x%29%20%3D%20V%280%29%281-r%29%5Ex)
In which V(0) is the initial value and r is the decay rate. So
![V(x) = 1000(1-0.2)^x](https://tex.z-dn.net/?f=V%28x%29%20%3D%201000%281-0.2%29%5Ex)
![V(x) = 1000(0.8)^x](https://tex.z-dn.net/?f=V%28x%29%20%3D%201000%280.8%29%5Ex)
We want to find x for which V(x) = 512. So
![V(x) = 1000(0.8)^x](https://tex.z-dn.net/?f=V%28x%29%20%3D%201000%280.8%29%5Ex)
![512 = 1000(0.8)^x](https://tex.z-dn.net/?f=512%20%3D%201000%280.8%29%5Ex)
![(0.8)^x = \frac{512}{1000}](https://tex.z-dn.net/?f=%280.8%29%5Ex%20%3D%20%5Cfrac%7B512%7D%7B1000%7D)
![(0.8)^x = (0.512)](https://tex.z-dn.net/?f=%280.8%29%5Ex%20%3D%20%280.512%29)
Applying log to both sides:
![\log{(0.8)^x} = \log{(0.512)}](https://tex.z-dn.net/?f=%5Clog%7B%280.8%29%5Ex%7D%20%3D%20%5Clog%7B%280.512%29%7D)
![x\log{0.8} = \log{(0.512)}](https://tex.z-dn.net/?f=x%5Clog%7B0.8%7D%20%3D%20%5Clog%7B%280.512%29%7D)
![x = \frac{\log{0.512}}{\log{0.8}}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B%5Clog%7B0.512%7D%7D%7B%5Clog%7B0.8%7D%7D)
![x = 3](https://tex.z-dn.net/?f=x%20%3D%203)
So it will take 3 years.