Where does this decibel equation come from, why a factor of 20?
Sound power (\$P\$) is proportional to the sound intensity (\$I\$):
$$
P = A\ I
$$
where \$A\$ is the surface area.
And sound intensity (\$I\$) is proportional to the square of sound pressure:
$$
I = \frac{p^2}{z_0}
$$
where \$p\$ is the sound pressure level (SPL) and \$z_0\$ is the specific acoustic impedance (think of it as something like a characteristic impedance). Just like in \$\mathrm{P = V^2/R}\$.
Putting all together yields us
$$
dB = 10 \log (\frac{P}{P_0}) = 20 \log (\frac{p}{p_0})
$$

EDIT / ADDITIONS / PROOFS
This will be too mathematical, so be warned!
We all know that the work (\$W\$) is defined by the product of the force (\$F\$), and the displacement caused by that force (\$dx\$):
$$
W = \int \mathbf F \ dx = \int \mathbf F \ \mathbf v\ dt
$$
and thus, the power is
$$
P = dW/dt = \mathbf F \ \mathbf v
$$
So, sound power (\$P_s\$) is calculated in the same manner:
$$
P_s = \mathbf {F_s \ v_p}
$$
where \$\mathbf {F_s}\$ is the sound force vector and \$\mathbf {v_p}\$ is the particle velocity vector (or, particle flow rate). The particles here are the particles inside the medium the force is applied (e.g. air molecules).
Since we know that the pressure is defined as the applied force per unit area, we obtain
$$
\mathbf {F_s} = p \ \mathbf A   \\
\therefore P_s = p \ \mathbf A \ \mathbf{v_p} = \mathbf{A\ I}
$$
where \$p\$ is the sound pressure level (SPL), \$\mathbf A\$ is the surface area vector. Note that the term \$p \ \mathbf{v_p}\$ is replaced with \$\mathbf I\$, which as defined as Sound Intensity (i.e. the sound power applied to a surface area).
When the force \$\mathbf {F_s}\$ is applied to an area \$\mathbf A\$ in a medium (results in the acoustic pressure, \$p\$), the particles in that medium will flow with a rate (flow rate, \$\mathbf v\$) but present opposition to that pressure. This opposition is called acoustic impedance, \$z\$. Just like the impedance in electricity: The opposition to electron flow caused by the voltage (yes, voltage is a force!).
$$
p = \mathbf {z \ v}
$$
The final results can be obtained by using the absolute values of the vectors above. Also, remember that the particles' velocities are caused by the sound force itself, so the absolute values (i.e. magnitudes) instead of the vectors can be used.
So,
$$
p = z \ v \Rightarrow v = p / z \\
I = p \ v \Rightarrow I = p (p / z) = p^2/z \\
P = A \ I = A \ p^2 / z
$$
Finally,
$$
\mathrm{
dB = 10 \log (P/P_0) =} 10 \log (\frac{A\ p^2/z} {A\ p_0^2/z_0})
$$
Where \$P_0\$ is the reference sound power and \$p_0\$ is the reference SPL. Now \$z = z_0\$ here for the same medium (characteristic sound impedance); just like the R in V²/R (same load).
Putting all together yields
$$
\mathrm{
dB = 10 \log (P/P_0)} = 10 \log (\frac{A\ p^2/z} {A\ p_0^2/z_0}) = 20 \log (\frac{p} {p_0})
$$
I hope these are enough and satisfactory.