Question

A brand new music system is worth $18,000 in 2001. If the music system's value decreases by 6% every year, what will its value be in 2003?

Answer 1

`15904.8`

Explanation:

You can use the formula:

`y=i*(1+-x)^t`

`"i is the initial value."`

`"x is the change. Use negative for decay, and positive for growth." (+-)`

`"t is time."`

Plug in:

`18000*(1-0.06)^2`,
(2 because 2003-2001 is 2).

Solve to get `15904.8`.

Have a nice day!

Answer 2

The value of the music system in 2003 is `"$15,904.80".`

Explanation:

The function we will use is

`A = P(1+r)^t`

where

`A=` the final value of the music system
`P=` the initial value of the music system
`r=` the rate of growth per year
`t=` the number of years

(Since we're dealing with decay instead of growth, `r` will be negative.)

To solve for `A`, we plug in the other values we know:

`A=18000(1-0.06)^2`
`color(white)(A)=18000(0.94)^2`
`color(white)(A)=18000(0.8836)`
`color(white)(A)=15904.8`

So the depreciated value of the music system is `"$15,904.80".`

Bonus:

Why does this formula work? Let's think of how much the music system would be worth after 1 year `(`call this `A_1).` This is 6% less than its initial value of $18,000 (a.k.a. 94% of $18,000):

`A_1=0.94 xx 18000`

The value of the music system after a 2nd year `(A_2)` is then 94% of its value after one year -- that is, 94% of `A_1`:

`A_2=0.94 xx A_1`
`color(white)(A_2)=0.94 xx 0.94 xx 18000`
`color(white)(A_2)=(0.94)^ 2 xx 18000`

It's not hard to see that, if we extend this to a 3rd year, the value for `A_3` will be `(0.94)^3xx18000,` and if we generalize it to `t` years, the depreciated value is `(0.94)^txx18000.` Since 0.94 was `1-0.06` (a.k.a. 1 minus the rate of decay) and 18000 was our initial value `P,` this gets us back to the form of the equation we started with:

`A=(1-r)^txxP`