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?

Printable View

- Sep 24th 2006, 12:19 PMimthatgirlcompound interest
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?

- Sep 24th 2006, 02:06 PMCaptainBlack
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