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Algebra II

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I'm pretty confused about these problems. We're learning growth and decay, but there are quite a few formulas.

1. Suppose $500 is invested at 6% annual interest compounded twice a year. When will the investment be worth $1000?

2. Suppose $500 is invested at 6% annual interest compounded continuously. When will the investment be worth $1000?

... I'm really confused

  • Algebra II - ,

    If the annual interest is 6 percent every half year the amount in the account is multiplied by 1.03 every half year.
    so after n half years the account will be worth:
    $500 * (1.03)* (1.03) * (1.03) etc n times
    or
    $500 * (1.03)^n
    so do the problem in half years
    $500 * (1.03)^n = 1000
    1.03^n = 2
    take the log of both sides remembering that log x^y = y log x
    n * .012837 = .301030
    n = 23.45 half years = 11.7 years
    -----------------------------------
    Now for the continuous problem you either have to know the formula from your text book or know calculus.
    Since I do not have a textbook, I will figure it out and find the formula
    dy/dt = .06 y
    dy/y = .06 dt
    ln y = .06 t + c
    y = e^(.06 t + c) = C e^.06 t
    when t = 0, y = $500 so
    THIS NEXT LINE IS PROBABLY IN YOUR TEXT
    y = 500 e^.06 t
    so we want
    1000 = 500 e^.06 t
    ln 2 = .06 t
    .693147 = .06 t
    so t = 11.55 years
    Just a little better than compounding twice a year

  • Algebra II - ,

    By the way, a rule of thumb is that your money doubles in ten years at 7% interest.

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