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Posted by on Friday, January 25, 2013 at 12:39pm.

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?

  • math - , Friday, January 25, 2013 at 1:34pm

    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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