Testing filter code w/ Octave
I think the solution is to use $\log_{10}(y)$ instead of $\log(y)$.  log(y) in octave is the natural $\log$, but you want a log base $10$, which is log10(y).

Confirming this:

$20\log_{10}(0.5) = -6 \textrm{ dB/octave}$

$20\ln(0.5) = \frac{-13.86}{20} \textrm{ nepers/octave}$

I am using $\textrm{ln}$ for natural log to avoid any confusion.