When computing power spectral density via Fourier transform and Parseval's theorem, one gets units of $V^2/Hz$ (e.g. Wiki). As a popular next step, the following unit conversion is widely used: $V^2/Hz \to dB/Hz$, and it's done by taking the famous $10.\log10(\cdot)$.
Question: while $\mathrm{dB}$ is a log of a ratio of two $\mathrm{V}^2$'s (second $\mathrm{V}^2$ is a fictive one) and I get where $\mathrm{dB}$ comes from, I'm struggling to understand why $1/\mathrm{Hz}$ remains unchanged after taking the logarithm?
Alex
P.S. Some obscure explanation is in the last paragraph here (and I'm afraid the word 'loosely' conceals all the real stuff here):
Since the argument of the logarithm is in units of $\mathrm{Hz}^{-1}$, this spectral measure can loosely be said to be in units of `dB/Hz'