Question

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

Answer 1

`color(red)("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:

`color(red)("You have not stated the number of years.")`
`color(red)("The use of the word 'annually' implies more than 1 year")`

Let the number of years be `n`

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

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Suppose the number of years is 1 then")`

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

`color(brown)("Subtract 144 from both sides")`

`" "144xx ?/100=270`

`color(brown)("Multiply both sides by "100/144)`

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

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Suppose "n>1)color(magenta)(" "->" Needs algebra")`

`color(brown)("Notice that " (1+x/100) ->((100+x)/100))`

`color(brown)("Taking logs of both sides")`

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

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

`" "log((100+x)/100)=(log(414)-log(144))/n`

`" "log(100+x)-log(100)=(log(414)-log(144))/n`

`" "log(100+x)=(log(414)-log(144))/n + log(100)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(brown)("But "log(100)=2`

`" "log(100+x)=(log(414)-log(144))/n + 2`

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

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