Sound level when doubling the distance

So how does the sound intensity level change with a doubling of distance?

We know that the intensity will drop by $1/2^2$ and thus

$$L_I = 10 \log (1/2^2 \times I/I_0)$$

$$= 10\log(I/I_0) + 10\log(1/2^2)$$

$$= 10\log(I/I_0) + 10\log(2^{-2})$$

$$= 10\log(I/I_0) - 20\log(2)$$

$$= 10\log(I/I_0) - 20(.3)$$

$$= 10\log(I/I_0) -6 .$$

Doubling the distance from a source causes a decrease of 6dB in the sound level.