Question #32dad

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Answer 1 · 2018-01-15T17:10:42

$P(1+x%)^n$

$n~~ 7.761....$ years

Rounded to 8 years

$P(1+x%)^n$ for yearly calculation cycle $" "...Equation(1)$

Note that for monthly calculation cycle you would have:
$P(1+(x%)/12)^(12n)$

Using $Equation(1)$ we have:

$P(1+x%)^n color(white)("d") -> color(white)("d")$2500(1+3.2/100)^n=($2500+$716.46)$

$(103.2/100)^n= (cancel($)3216.46)/(cancel($)2500)$ Notice the units cancel out.

Take logs of both sides

$nxx ln(103.3/100)=ln( 3216.46/2500)$

$n=ln( 3216.46/2500)-: ln(103.3/100)$

$n~~ 7.761....$