Take this all with a grain of salt, I don't work in dB on a daily basis. An RF guy could probably answer this blindfolded with their hands tied behind their back. But I'll take a crack with what I remember from university many years ago.
When you hear things described in decibels, that's almost always expressing a ratio. When it comes to a ratio of power, the formula you want to remember is: dB = 10 * log10(P_o / P_i). When it comes to a ratio of voltages the formula is dB = 20 * log10( P_out / P_in). This is due to the fact that Power is proportional to the square of Voltage, and exponents become multipliers in logarithmic space (because of the algebraic laws of exponents and logs).
Then it comes down to understanding what the ratio being described is. So when I see something that says "This amplifier has 30dB of feedback", what I take that to mean is the ratio of the power fed back to the total power output of the circuit, which is to say: 30 = 10 * log10( Power_feedback / Power_totalout). I could be wrong about that, but it's a decent guess. It matches a rule of thumb I remember, which is 3dB equates to 50%.
If 10 watts is fed back, and 10 watts is delivered to the load, the dB according to this logic would be 10 * log10 (10 / (10 + 10)) = 10 * log10( 0.5 ) = -0.3 * 10 = 3dB. Again according to this logic if its 30dB, that means the fraction of the output power which is fed back is:
-30 = 10 * log10(P_feedback / P_total)
So P_feedback / P_total = 10^(-30/10) = 0.001 = 0.1%
Hopefully someone else will either ratify this logic or refute it (in which case I will readily withdraw the answer).