Question #55088

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Answer 1 · 2018-02-20T12:24:16

Assumption: the calculation cycle is monthly (not stated!)

$14 %$ $14%$ per year compounded monthly

Let the principal sum be $P$ $P$
Let the numerator of the percent be $x$ $x$

2 years $->P(1+x/(12xx100))^(2xx12)=1440" "....Equation(1)$

3 years $->P(1+x/(12xx100))^(3xx12)=1656" "....Equation(2)$

The most straigh-forward way to 'get rid' of one of the unknowns ( $P$ ) is by division. It also simplifies the brackets to some extent.

$Eqn(2)-:Eqn(1)$

$[cancel(P)(1+x/(12xx100))^(3xx12)]/[cancel(P)(1+x/(12xx100))^(2xx12)]=1656/1440$

$(1+x/(12xx100))^(12) =1656/1440$

$((1200+x)/1200)^(12) =1656/1440$

Take logs of both sides. I choose $log_10$

Note that $log(a^b)=b log(a)$
and that $log(a/b)=log(a)-log(b)$

$log(((1200+x)/1200)^(12))=log(23/20)$

$12log(1200+x)-12log(1200)color(white)("d")=color(white)("d")log(23)-log(20)$

$log(1200+x)=(log(23)-log(20)+12log(1200))/12$

$1200+x=log^(-1)[(log(23)-log(20)+12log(1200))/12]$

$x=log^(-1)[(log(23)-log(20)+12log(1200))/12] -1200$

$x~~14.0579....$ by calculator

This will have rounding errors in it so lets call it $x=14$