Confusion between Voltage gain & Voltage gain in decibels (dB)

The definition of voltage gain is $V_{\text{out}}/V_{\text{in}}$.

However, I read some articles about the gain in decibels, and I have a confusion now.

Here is an article about it: https://en.wikipedia.org/wiki/Gain#Voltage_gain

Here, I understand the definition of Power gain in decibels, which is $$\text{Gain} = 10 \log \left( {P_{\text{out}} \over P_{\text{in}}} \right)\text{ dB}$$

However, I can't understand why Voltage gain in decibels is $$20 \log \left( {V_{\text{out}} \over V_{\text{in}}} \right)\text{ dB}$$

If $ 20 \log \left( {V_{\text{out}} \over V_{\text{in}}} \right)\text{ dB} $ is derived from $$10 \log {\left( {V_{\text{out}}^2 \over R_{\text{out}}}\right) \over \left({V_{\text{in}}^2 \over R_{\text{in}}}\right) }\text{ dB}$$ then this is the power gain, not the voltage gain, isn't it? However, the Wikipedia says it is a formula for the Voltage gain in decibels. I thought the voltage gain in decibels would be $ 10 \log \left( {V_{\text{out}} \over V_{\text{in}}} \right)\text{ dB} $. Actually, the example section in that linked page uses voltage gain $ V_{\text{out}} \over V_{\text{in}} $.

Why did $V^2/R$ suddenly come out from the voltage gain in decibels?

Answer

As you say, the decibel is a unit of power ratio.

$ G\ [\mathrm{dB}] = 10 \log_{10}\left(\dfrac{P_1}{P_2}\right)$.

When the input and output impedances are equal and then we can express the gain in terms of voltage as

$ G\ [\mathrm{dB}] = 20 \log_{10}\left(\dfrac{V_1}{V_2}\right) $

I wouldn't call this the "voltage gain in decibels." I'd rather say it's the decibel gain, calculated from the voltage gain.

Sometimes, you will see a voltage gain expressed in decibel according to this formula even when the input and output impedances are different. There is no technical justification for this --- it's simply a shorthand practice that's become common through usage.