Question #84d74

If we apply the reverse process of increasing (as in compound percentage) then we should have what we need.

#color(blue)("Building the starting point- increasing")#

Suppose the initial sum borrowed was P (principle sum)
Suppose the related interest was #x%->x/100#

Then 1 years interest is #x%xxP#
Add this to the initial deposit and you have: #P+xP#

Factor out the P and you #P(1+x%)#

Do this over #n# years and you have #P(1+x%)^n#
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#color(blue)("Modifying increasing so that it becomes depreciation")#

All we need to do is change #+x%# to #-x%# giving:

#P(1-x%)^n#

Using the above and substituting the values in the question we have:

#color(green)("The depreciation equation")#
#color(green)(P(1-x%)^n" "->" "$20000(1-x%)^7=$1590#

#color(white)(.)#
#color(white)(.)#

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#color(blue)(" Going beyond the question and solving for "x)#

divide both sides by $20000

#(1-x%)^7=($1590)/($20000)#

Note that you can treat units of measurement the same way you do numbers. Very useful in applied maths or physics.

#(1-%)^7=(cancel($)159cancel(0))/(cancel($)2000cancel(0))#

Take logs of both sides

#7log(1-x%)=log(159)-log(2000)#

#log(1-x%)=(log(159)-log(2000))/7#

#log( 1-x%)=-0.15709.....#

#log^(-1)=0.69648....#

In the 'olden days' #log^(-1)# was called antilog.

So #1-x%" "=" "1-x/100" "=" "0.69648...#

#x%=1-0.69648... ~~ 30.35%#