Question

Question #84d74

Answer 1

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

Explanation:

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%`