# Question 84d74

Feb 23, 2017

$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.

$\textcolor{b l u e}{\text{Building the starting point- increasing}}$

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: $P + x P$

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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$\textcolor{b l u e}{\text{Modifying increasing so that it becomes depreciation}}$

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:

$\textcolor{g r e e n}{\text{The depreciation equation}}$
color(green)(P(1-x%)^n" "->" "$20000(1-x%)^7=$1590

$\textcolor{w h i t e}{.}$
$\textcolor{w h i t e}{.}$

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$\textcolor{b l u e}{\text{ Going beyond the question and solving for } x}$

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..... ${\log}^{- 1} = 0.69648 \ldots .$In the 'olden days' ${\log}^{- 1}\$ was called antilog.

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

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