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
