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

Any ideas anyone?




Be Grateful of any help!

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

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

can't you start with the fact that ? After that it's all algebra I think.

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)

is your equation


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

I edited the original question so the formula displays correctly.