Question 1199028

The  formula for compound interest is as follows:

{{{A= P(1 + r/n)^(n*t)}}}

where:
{{{P}}} : principal amount, or initial investment
{{{A}}} : amount after time {{{t}}}
{{{r}}} : interest rate
{{{n }}}: number of compounding periods, usually expressed in years
{{{t}}}:number of  years

given:

{{{P=1200}}}
{{{A=2200}}}
{{{r=0.045}}}
{{{n=12}}} (compounded monthly)

find:

{{{t}}}=


{{{A= P(1 + r/n)^(n*t)}}}

we want to know when will be

{{{P(1 + r/n)^(n*t)>A}}}

{{{1200(1 + 0.045/12)^(12*t)>2200}}}

{{{(1.00375)^(12*t)>2200/1200}}}

{{{(1.00375)^(12*t)>11/6}}}

use log

{{{log((1.00375)^(12*t))>log(11/6)}}}


{{{(12*t)log((1.00375))>log(11/6)}}}


{{{12*t>log(11/6)/log((1.00375))}}}


{{{t>log(11/6)/(12log((1.00375)))}}}


{{{t>13.4949}}}


so,

{{{t=14}}}


account will surpass ${{{2200}}} after {{{14}}} years