At what frequency will the frequency be attenuated by -20dB?

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For a high pass filter with a 1kΩ resistor and a 330nf capacitor at what frequency will the frequency be attenuated by -20dB?

I thought about using the equation gain=20log(Xc/Z) but I am not sure if this is correct, I saw it on a website and even if it is I can't seem to calculate the frequency.
 
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I get about 48.47 Hz.
 
Can you show me how you got that answer please.
 
-20 dB = 10*Log(Mag(R/(R+jXc)))
Xc = -1/(2*pi*F*3.3*10^-7)
Mag is short for magnitude = sqrt(R^2 + Xc^2)

I did it by trial and error substituting values for F.
 
First order RC filters have a power attenuation rate of 20dB per decade. So find the corner frequency from the part values and add a decade (in the appropriate direction).
 
gneill said:
First order RC filters have a power attenuation rate of 20dB per decade. So find the corner frequency from the part values and add a decade (in the appropriate direction).

The answer is 53.6Hz but by doing it this way I get an answer of 48.23Hz so I don't know where I have gone wrong
 
In your original post you phrased the question as, "at what frequency will the frequency be attenuated by -20dB?". I think everyone has interpreted that to mean at what frequency will the power be attenuated by 20 dB. Is it indeed the power ratio that we're looking for or is it the voltage ratio? Or something else?
 
gneill said:
In your original post you phrased the question as, "at what frequency will the frequency be attenuated by -20dB?". I think everyone has interpreted that to mean at what frequency will the power be attenuated by 20 dB. Is it indeed the power ratio that we're looking for or is it the voltage ratio? Or something else?

I don't know, what I wrote was the exact wording of the question.