Twice the audio source, +3 +6 or +10 dB SPL? I already know that for amplitude, it works this way: if you double the audio source, the dB variation is more or less +6dB.
But how does it work for the audio volume scale casually used, i.e. dB SPL? (please correct me if I'm wrong, I mean that scale: 20dB: quiet nature with no wind, 60dB: quietly speaking person, 110dB club, 140 dB: jet airplane, i.e. the scale we hear a lot in the media)
Question: Let's say a 80dB SPL vacuum cleaner is working. If we add a second vaccum cleaner (same model, same speed) near the first one, what is the total volume? 83 or 86 dB SPL?

Note: This topic has been widely discussed over internet (many results in search engine) and in nearly any audio-related forum, but as there is usually no votes / accepted answer on those forums, I don't know which answer to trust, thus the question here.
Note2: It might be linked to this question but not really a duplicate because both the questions are answers are much less accessible / more technical, and the "double volume, add +? dB" part of the qustion is mixed with lots of other technical considerations / not easy to grasp for a learner at first sight. Also this question here is specific to dB SPL, whereas the other one is not.
 A: The standard formula for calculating the Sound Pressure Level is
$$L_p=20\log_{10}\left(\frac{p}{p_0}\right)\text{dB}$$
where


*

*$p$ is the sound pressure being measured

*$p_0$ is the reference sound pressure (usually $20\mu\text{Pa}$ in air)


As you said, if the amplitude of the sound is doubled, then the dB variation is indeed $+6\text{dB}$, but this is not the same as adding a new sound source.
The formula for multiple incoherent sources is
$$L_\Sigma=10\log_{10}\left(\frac{{p_1}^2+{p_2}^2+...+{p_n}^2}{{p_0}^2}\right)\text{dB}$$

Now we can simply calculate the difference between doubling the sound pressure and adding another instance of the same sound pressure.
First the $80\text{dB}$ of the vacuum cleaner must be converted to $\text{Pa}$
$$80\text{dB}\rightarrow 2\text{Pa}$$
then we can run our calculations:


*

*Doubling the sound pressure


$$\Delta L_p=20\left(\log_{10}\left(\frac{4\text{Pa}}{2\mu\text{Pa}}\right)-\log_{10}\left(\frac{2\text{Pa}}{2\mu\text{Pa}}\right)\right)\text{dB}$$
$$\Delta L_p=6.02\text{dB}$$


*Adding another vacuum cleaner


$$\Delta L_\Sigma=10\left(\log_{10}\left(\frac{2\text{Pa}^2+2\text{Pa}^2}{2\mu\text{Pa}^2}\right)-\log_{10}\left(\frac{2\text{Pa}^2}{2\mu\text{Pa}}\right)\right)$$
$$\Delta L_\Sigma=3.01\text{dB}$$

TL:DR
The total sound pressure level if an additional identical vacuum cleaner is added is $83\text{dB, SPL}$.
A: Kieran's answer is correct: 83 dB. 
The key issues with this class of problems is whether the sources add in energy (+3 dB) or in amplitude (+6 dB). This depends on the correlation of the two sources. Correlated sources add in amplitude and uncorrelated ones add in energy.
Correlation only happens if both source are connected in some way to a single original excitation. That's not the case for two vacuum cleaners that each have their own engine. 
If you have two identical loudspeakers playing the same signal you will get +6dB at low frequencies, +3dB at high frequencies and something in between at mid frequencies. The amount of correlation is a complex function of wavelength, distance between the sources, directivity, room acoustics and listener/microphone location. 
That's the reason why there is often confusion of between +3dB and +6dB.
