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Spike1984 · Dec 6, 2007, 08:40 AM

Revising for exams and stumped by rearranging this question to find the value of F.

Any ideas anyone?

$59dB =20log(\frac{1}{(1+(\frac{F}{Fc})^6)^{\frac{1}{2}} })$

Be Grateful of any help!

Spike1984 · Dec 6, 2007, 08:41 AM

OK the Math think didn't work so here it is

59dB = 20Log(1/ (1+(F/Fc)^6)^0.5))

asterisk_man · Dec 6, 2007, 09:18 AM

can't you start with the fact that $a^{\log_a b}=b$ ? After that it's all algebra I think.

Spike1984 · Dec 6, 2007, 09:41 AM

Thanks Asterisk_man. Just before I checked on here I managed to have a brainwave and solved it. Just by reducing the powers down. I got it down to:

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

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

asterisk_man · Dec 6, 2007, 11:23 AM

is your equation
$59dB=20\log {\frac 1 {{\left(1+{\left(\frac F {Fc}\right)}^6\right)}^{\frac 1 2}}}$

And what base is your log? I'm assuming base 10. So log(10)=1.

Capuchin · Dec 6, 2007, 04:07 PM

I edited the original question so the formula displays correctly.