The logarithm ("common logarithm") of x is the number n such that 10^n=x (10 to the n-th power is x), e.g. log(1000)=3 because 10^3=1000.
This logarithm is used for the (deci)bel scale: every time you multiply the intensity of the sound source by 10, you add 1 bel, i.e. 10 decibels.
Now, the math: ASSUMING that these two billion cats are in the exact same place (huge assumption...), we have log(2000000000)=9.3 (approx.), so if a cat meows at 65dB, two billion cats in the same place would meow at 65+10*9.3=158dB. If a single cat meows at 75dB, we get 168dB for the 2bn cats.
If you stand 1 meter away from a .30-06 rifle being fired, that's 171dB for you (about twice the 168dB intensity, or 20 times the 158dB intensity). And you have to take into account that the cats, of course, are not in the same place, sound does not propagate instantly (it's even pretty slow), and its intensity decreases with the distance, so... Not that impressive, actually.
Every time you add 10 decibels, sound pressure level multiplies tenfold. So 70 dB is ten times the sound pressure of 60 dB. And, as ajhedges already stated, the maximum achievable soundpressure level (at room temperature and sea level air pressure) is 194 dB. Mind you, that is about 250,000 times as loud as a jet plane taking off.
It gets weirder. An increase of 10 dB, which is a tenfold increase in SPL, is PERCEIVED as just a doubling of sound VOLUME. In other words: the human hearing is ALSO logarithmic when it comes to perceived volume. Then there's weighting (putting frequency response into the equation), different reference levels (decibels express a ratio, not an absolute number), etc. and it's easy to lose track completely.
The sound would multiply together. Its not a meow with the power of all the cats in the world, its all the cats in the world meowing. It would be unnoticeable. Unless you work at petco.
This logarithm is used for the (deci)bel scale: every time you multiply the intensity of the sound source by 10, you add 1 bel, i.e. 10 decibels.
Now, the math: ASSUMING that these two billion cats are in the exact same place (huge assumption...), we have log(2000000000)=9.3 (approx.), so if a cat meows at 65dB, two billion cats in the same place would meow at 65+10*9.3=158dB. If a single cat meows at 75dB, we get 168dB for the 2bn cats.
If you stand 1 meter away from a .30-06 rifle being fired, that's 171dB for you (about twice the 168dB intensity, or 20 times the 158dB intensity). And you have to take into account that the cats, of course, are not in the same place, sound does not propagate instantly (it's even pretty slow), and its intensity decreases with the distance, so... Not that impressive, actually.
That’s not how any of this works!
Maximum volume is 194dB