Question 967716


To calculate this value exactly, a fairly simple formula can be utilized.

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

{{{P}}}=principal ({{{1000}}})
{{{r}}}=annual nominal interest rate ({{{.10}}})
{{{n}}}=number of compounding during year ({{{12}}}, because of {{{12}}} months)
{{{t}}}=number of years (unknown)
{{{A}}}=amount after {{{t}}} years (first {{{1000000}}})

This formula yields results when we have the time known, but in this case we need to isolate "{{{t}}}" for the time necessary for the principal to turn into ${{{1000000}}}. So we need to rearrange the formula into a fairly complex logarithmic formula:

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

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

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

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

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

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

Then substitute the numbers into the equation -

{{{t=log(1000000/1000)/((12*log(1+(0.10/12))))}}}

{{{t=log(1000)/(12*log(1.008333333333333))}}}

{{{t=69.36}}}

Finally we got the answer - it'll take about {{{69}}} years and a four months!!!