# What is the percent of increase from $147 annually to$414?

May 23, 2016

$\textcolor{red}{\text{The time span not defined as 1 year}}$
For 1 year only color(blue)(-> "percentage is " 187.5%)
Or simple interest over whole period
'.....................................................
For more than 1 year where $n$ is number of years
and annually compounded interest.

" "color(blue)(x%=log^(-1)((log(414)-log(144))/n + 2) - 100)

#### Explanation:

$\textcolor{red}{\text{You have not stated the number of years.}}$
$\textcolor{red}{\text{The use of the word 'annually' implies more than 1 year}}$

Let the number of years be $n$

" "$144(1+x/100)^n=$414

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$\textcolor{b l u e}{\text{Suppose the number of years is 1 then}}$

" "144+(144xx ?/100)=414

$\textcolor{b r o w n}{\text{Subtract 144 from both sides}}$

" "144xx ?/100=270

$\textcolor{b r o w n}{\text{Multiply both sides by } \frac{100}{144}}$

color(blue)(" "?%=270xx100/144 = 187.5%)

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$\textcolor{b l u e}{\text{Suppose "n>1)color(magenta)(" "->" Needs algebra}}$

$\textcolor{b r o w n}{\text{Notice that } \left(1 + \frac{x}{100}\right) \to \left(\frac{100 + x}{100}\right)}$

$\textcolor{b r o w n}{\text{Taking logs of both sides}}$

" "log($144)+nlog((100+x)/100)=log($414)

$\text{ } n \log \left(\frac{100 + x}{100}\right) = \log \left(414\right) - \log \left(144\right)$

$\text{ } \log \left(\frac{100 + x}{100}\right) = \frac{\log \left(414\right) - \log \left(144\right)}{n}$

$\text{ } \log \left(100 + x\right) - \log \left(100\right) = \frac{\log \left(414\right) - \log \left(144\right)}{n}$

$\text{ } \log \left(100 + x\right) = \frac{\log \left(414\right) - \log \left(144\right)}{n} + \log \left(100\right)$

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color(brown)("But "log(100)=2

$\text{ } \log \left(100 + x\right) = \frac{\log \left(414\right) - \log \left(144\right)}{n} + 2$

$\text{ } 100 + x = {\log}^{- 1} \left(\frac{\log \left(414\right) - \log \left(144\right)}{n} + 2\right)$

" "color(blue)(x%=log^(-1)((log(414)-log(144))/n + 2) - 100)