Question 373821
Consider 
t(1) = 1,000 and t(n) = 1.06t(n-1).  The 2nd equation gives {{{t(n)/t(n-1) = 1.06}}}.  Then {{{(t(n)/t(n-1))*(t(n-1)/t(n-2))}}}*...*{{{(t(3)/t(2))*(t(2)/t(1)) = (1.06)^(n-1) = t(n)/t(1)}}}.  Hence
{{{t(n)/1000 = 1.06^(n-1)}}}, or {{{t(n) = 1000*1.06^(n-1)}}}.
We want {{{1000*(1.06)^(n-1)>3000}}}, or {{{(1.06)^(n-1)>3}}}.
Take ln of both sides:  {{{(n-1)ln1.06>ln3}}}, or {{{n-1 > ln3/ln1.06 = 18.8542}}}.  Thus n > 19.8542, so that {{{n>=20}}}.