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) )

SOLVED IT!

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

F = Cubetr(10^(59/20) x 1^0.5 x Fc^3 - 1)

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:


F = F_c {( \frac 1 {(10^ {5.4}} -1)} ^{1/6} \\
F = F_c (-0.999996019) ^ {(1/6)}


At this point you see the problem clearly - the quantity that you're trying to take the sixth root of is negative!

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:


59 = 20 log ( \frac 1 {sqrt {1 + (\frac F {F_c})^{\frac 1 6}}})


Please confirm.