Anonymous
Anonymous asked in Science & MathematicsMathematics · 11 months ago

pre calc help?

You want to make an investment in a continuously compounding account over a period of 20 years. What interest rate is required for your investment to double in that time period? Round the logarithm value to the nearest hundredth and the answer to the nearest tenth.You want to make an investment in a continuously compounding account over a period of 20 years. What interest rate is required for your investment to double in that time period? Round the logarithm value to the nearest hundredth and the answer to the nearest tenth.

Update:

The letters r and θ represent polar coordinates. Write the following equation using rectangular coordinates (x, y).

r= 5/1+cosθ

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  • Mike G
    Lv 7
    11 months ago
    Favorite Answer

    2 = e^(20r)

    ln2 = 20r

    r = 0.034657

    Answer r = 3.5%

    [For a gross error check use the rule of 72

    72/20 = 3.6%]✓

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  • 11 months ago

    FV = future value of investment

    PV = present value of investment

    e = base of natural logarithms = 2.718281828

    Y = number of years of investment = 20 years

    r = annual interest rate = to be determined

    FV = PV*e^Yr

    FV/PV = e^Yr

    FV/PV = 2 (investment value doubles)

    2 = e^Yr

    ln(2) = ln(e^Yr)

    ln(2) = Yr[ln(e)]

    ln(2) = Yr(1)

    ln(2) = Yr

    ln(2)/Y = r

    r = ln(2)/Y

    r = ln(2)/20

    r = 0.034657359 = 0.03 logarithm value

    r = 3.5% interest rate

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  • 11 months ago

    The formula for continuous compounding is:

    A = Pe^(rt)

    To double the investment, A/P = 2, i.e. e^(rt) = 2, and since t = 20, we have:

    e^(20t) = 2

    Since t is a real number, 20t is a real number, so we have:

    20t = ln(2)

    t = ln(2)/20

    t =~ 0.034657359027997265470861606072909

    So the interest rate should be about 3.4657%

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