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28 March, 17:30

how many years will it take an investment of 1000 to double if the interest rate is 6% compounded annually

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Answers (2)
  1. 28 March, 18:55
    0
    11.9 years

    Step-by-step explanation:

    The compound amount formula applicable here is A = P (1+r) ^t.

    Substituting the given data, $2000 = $1000 (1.06) ^t, or 2 = 1.06^t.

    Taking the log of both sides, log 2 = t log 1.06. Then t = 0.30103/0.02531, or t = 11.896.

    The investment will doube in 11.9 years.
  2. 28 March, 20:34
    0
    Years = {log (total) - log (Principal) } : log (1 + rate)

    Years = log (2,000) - log (1,000) / log (1.06)

    Years = 3.3010299957 - 3 / 0.025305865265

    Years =.3010299957 / 0.025305865265

    Years = 11.8956610473 or about 11.9 Years
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