When power is increased to twice the initial power, what is the power increment in dB?

Question

When power is increased to twice the initial power, what is the power increment in dB?

Answer 1

It is 3db.

Proof:  10log(P/0.5) = 10(log(P) - log(0.5))

log(0.5) = -0.30

so 10log(P/0.5) = 10log(P) + 3.0dB

Answer 2

power to dB is

v (dB) = 10 log10 (P/Pref) Pref being the reference power

So take the reference power to be something & we want to know how many dB represents twice that so P/Pref = 2

v (dB) = 10 log10(P/Pref) = 10 * log10(2) = 10* (0.3010..)

approximately 3dB

Answer 3

I think (if I remember correctly) when you double the power (RMS) the decibel level increases about 3 dB.