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?

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!

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#