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")Building the starting point- increasing
Suppose the initial sum borrowed was P (principle sum)
Suppose the related interest was x%->x/100x%→x100
Then 1 years interest is x%xxPx%×P
Add this to the initial deposit and you have: P+xPP+xP
Factor out the P and you P(1+x%)P(1+x%)
Do this over nn years and you have P(1+x%)^nP(1+x%)n
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color(blue)("Modifying increasing so that it becomes depreciation")Modifying increasing so that it becomes depreciation
All we need to do is change +x%+x% to -x%−x% giving:
P(1-x%)^nP(1−x%)n
Using the above and substituting the values in the question we have:
color(green)("The depreciation equation")The depreciation equation
color(green)(P(1-x%)^n" "->" "$20000(1-x%)^7=$1590P(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) Going beyond the question and solving for x
divide both sides by $20000
(1-x%)^7=($1590)/($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%
