For an annual deposit of A=$1000 (at the end of the year) at an annual interest rate of i=7% compounded yearly, the future value [tex]F=\frac{A((1+i)^n-1)}{i}[/tex] where n=number of years => [tex]20000=\frac{1000((1+.07)^n-1)}{.07}[/tex] on simplification [tex]1.4=(1.07)^n-1[/tex] [tex](1.07)^n=2.4[/tex] take logs and solve for n [tex]n=log(2.4)/log(1.07)[/tex] [tex]n=12.939[/tex] years, to the nearest 0.001 year
