Asked by Jackie

a) 1150 dollars invested at 9% annual interest rate (compounded yearly) or
b) 1475 invested at 6% annual interest (compounded yearly) after
When would the two investments have equal value?
Answered by Reiny
let the time be t

1150(1.09)^t = 1475(1.06)^t

46(1.09)^t = 59(1.06)^t
take log of both sides , and using simple log rules

log46 + tlog1.09 = log59 + tlog1.06
tlog1.09 - tlog1.06 = log 59 - lot 46
t(log 1.09 - log 1.06) = log 59 - log 46
t = (log59 -log46)/(log1.09 - log1.06) = 8.918

it would take appr 9 years

check:
1150(1.09^9 = 2497.68
1475(1.06)9 = 2491.98

using t = 8.918
1150(1.09)^8.918 = 2480.09
1475(1.06)^8.918 = 2480.13 , not bad
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