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23 November, 18:20

How long does it take to double a $1,000 investment that pays 6.5% annual interest, compounded monthly?

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Answers (2)
  1. 23 November, 21:24
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    Compound interest formula: A=P (1+interst) ^n

    P being the capital invested & n = the number of years, A the new capital

    A = 1000 (1+0.065) ^n. They want to know n when the new capital is doubled,

    2000=1000 (1.065) ^n & you will find n=11 years
  2. 23 November, 21:52
    0
    2p=p (1+0.065/12) ^12t

    2 = (1+0.065/12) ^12t

    Solve for t

    T=[log (2) / log (1+0.065/12) ]/12

    T=10.7 years
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