if $500 was invested at 8%/a, compounded annually, in one account, and $600 was invested at 6%/a, compounded annually, in another account, then when would the amounts in both accounts be equal?
The amount in the first account after n years is:
B1=500*(1.08)^n,
and in the second account:
B2=600*(1.06)^n.
These are equal when:
B1=B2,
or:
500*(1.08)^n=600*(1.06)^n
Rearranging:
(1.08/1.06)^n=600/500,
taking logs:
n log(1.08/1.06)=log(600/500),
or:
n=log(600/500)/log(1.08/1.06)=9.75 years,
But as the interest is paid yearly they will in fact never be equal, but
are most nearly equal after 10 years, when the accounts contain
$1079.46 and $1074.50 respectively.
RonL