Answer:
1. $640
2. About 10.3 years later
Explanation:
This is a compound decay problem. The formula is
![F=P(1-r)^t](https://img.qammunity.org/2020/formulas/mathematics/middle-school/gr5gjckb1ufyzaqg4ieyzqxeh0r8kz3er4.png)
Where
F is the future amount
P is the initial amount
r is the rate of decrease (in decimal), and
t is the time in years
Question 1:
We want to find F after 2 years of a phone initially costing 1000. So,
P = 1000
r = 20% or 0.2
t = 2
plugging into the formula, we solve for F:
![F=P(1-r)^t\\F=1000(1-0.2)^2\\F=1000(0.8)^2\\F=640](https://img.qammunity.org/2020/formulas/mathematics/middle-school/qt1dwspafg9n0ky0v6ihqkqfs9ugu75yir.png)
The phone is worth $640 after 2 years
Question 2:
We want to find when will the phone be worth 10% of original.
10% of 1000 is 0.1 * 1000 = 100
So, we want to figure this out for future value of 100, so F = 100
We know, P = 1000 r = 0.2 and t is unknown.
Let's plug in and solve for t (we need to use logarithms):
![F=P(1-r)^t\\100=1000(1-0.2)^t\\100=1000(0.8)^t\\(100)/(1000)=0.8^t\\0.1=0.8^t\\ln(0.1)=ln(0.8^t)\\ln(0.1)=t*ln(0.8)\\t=(ln(0.1))/(ln(0.8))\\t=10.32](https://img.qammunity.org/2020/formulas/mathematics/middle-school/8rx8t6h6i2efktlxjzpdxjjx0tparvdvyg.png)
So, after 10.32 years, the phone would be worth less than 10% of original value.