The voltage gain of an amplifier, expressed in dB, is 20log(Vo/Vi). That's clear and unambiguous. You cannot relate that to power gain of the same amplifier unless you know the input impedance (say, resistance Ri) and the load impedance (say, resistance RL).This is because output power is Vo^2/RL and input power is Vi^2/Ri and the power gain is thus:

10log[(Vo^2/Vi^2).(Ri/RL)] = 10log(Vo^2/Vi^2)+10log(Ri/RL) = 20log(Vo/Vi)+10log(Ri/RL)

If input and output impedances are equal then log(Ri/RL) = log(1) = 0 and the power gain and the voltage gain, in dB, are equal. This is the case in, eg, transmission line repeater amplifiers where the inputs and outputs are connected to 50 Ohm transmission lines and input and output impedances must both be matched to the line to prevent relections and maximise signal transfer. In audio power amplifiers power gain is somewhat meaningless, as the input impedance is normally very high and input signal is very small, and the output impedance is only a few of ohms and output signal is large. So power output is normally quoted.

It's worth noting that dBm and dBW, essentially, represent specific powers, since they strictly refer to ratios with respect to 1mW and 1W, respectively. Thus for example, 3dBm = 2mW and 20dBW = 100W