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    Spike1984's Avatar
    Spike1984 Posts: 5, Reputation: 1
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    #1

    Dec 6, 2007, 08:20 AM
    Rearranging a Formula.
    Got an equation that I am struggling to rearrange to find the value of F. I have tried numerous time to no avail!

    59dB = 20log(1 / ( (1+(F/Fc)^6)^0.5) )
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    Spike1984 Posts: 5, Reputation: 1
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    #2

    Dec 6, 2007, 09:43 AM
    SOLVED IT!

    59dB = 1 / (1 + (F/Fc)^6)^0.5

    F = Cubetr(10^(59/20) x 1^0.5 x Fc^3 - 1)
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    ebaines Posts: 12,131, Reputation: 1307
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    #3

    Dec 6, 2007, 09:59 AM
    I can simplify this for you ,but there is a fundamental problem here - the denominator of the log exprssion must be less than 1 (since the log of 1/denominator has to be greater than 1). But that means that (F/Fc) ^ (1/6) must be negative. And that can't be. So I suggest you go back and check that you wrote the equation correctly.

    However, if you want to go through the simpliifcation process anyway, here are the details:

    Start with:
    59dB = 20log(1 / ( (1+(F/Fc)^6)^0.5) )

    Divide both sides by 20:
    59/20 = log (1 / ( (1+(F/Fc)^6)^0.5) )

    Using the definition of logarithms:
    10^(59/20) = 1 / ( (1+(F/Fc)^6)^0.5)

    Square both sides:
    (10 ^ (59/20))^2 = 10 ^ (108/20) = 1 / (1+(F/Fc)^6)

    Take the inverse:
    1+(F/Fc)^6 = 10 ^ -(108/20) = 1/(10^(108/20))

    Massage:
    (F/Fc)^6 = 1/(10^(108/20)) -1
    F/Fc = (1/(10^(108/20)) -1)^1/6
    F = Fc * (1/(10^(108/20)) -1)^1/6

    Or, perhaps clearer:



    At this point you see the problem clearly - the quantity that you're trying to take the sixth root of is negative!
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Dec 6, 2007, 10:11 AM
    Quote Originally Posted by Spike1984
    SOLVED IT!

    59dB = 1 / (1 + (F/Fc)^6)^0.5

    F = Cubetr(10^(59/20) x 1^0.5 x Fc^3 - 1)
    You seem to have changed the equation you are trying to solve - is it supposed to be this:



    Please confirm.

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