Question 130282
let x="years to equality"


2000(1.04)^x=1000(1.06)^x __ dividing by 1000 __ 2(1.04)^x=(1.06)^x


taking log __ log(2)+x(log(1.04))=x(log(1.06))


subtracting x(log(1.04)) __ log(2)=x(log(1.06))-x(log(1.04))


factoring __ log(2)=x(log(1.06)-log(1.04))


dividing by (log(1.06)-log(1.04)) __ (log(2))/(log(1.06)-log(1.04))=x