It's not difficult, just always having some doubts...

Power

$latex L_{dB} = 10 log_{10} \left( \dfrac{P_1}{P_0} \right)$

10 dB increase for a factor 10 increase in the ratio

3 dB = doubling

40 dB = 10000 times

Amplitude

$latex L_{dB} = 10 log_{10} \left( \dfrac{A_1^2}{A_0^2} \right) = 20 log_{10} \left( \dfrac{A_1}{A_0} \right)$

dBm

dBm is an absolute value obtained by a ratio with 1 mW:

$latex L_{dBm} = 10 log_{10} \left( \dfrac{P_1}{1 mW} \right)$

• 0 dBm = 1 mW

• 3 dBm ≈ 2 mW