Calculating $E_b/N_0$ from given SNR ratio I have the following formula:
$$\frac{S}{N}=\frac{E_b \cdot R_b}{N_0 \cdot B }$$
Where $S/N$ is of course SNR in dB, $E_b$ is energy per information bit, $ R_b$ is information bit rate in [bit/s], $N_0$ is noise spectral density and $B$ is bandwitch in [Hz].
So let's say the $S/N$ ratio = $70\ \rm dB$

$R_b = 250\ \rm kb/s$
$B = 1\ \rm MHz$

So:
$$70\ \mathrm{dB}=\frac{E_b \cdot 250 000\ \mathrm{b/s}}{N_0 \cdot 1000000\ \mathrm{MHz}}$$
$$70\ \mathrm{dB}=\frac{E_b \cdot 1\ \mathrm{b/s}}{N_0 \cdot 4\ \mathrm{MHz}}$$
Now, of course, I can't simply multiply because this 4 is not on a logarithmic scale. But can I just logarithmize 4 like that? Won't there be a problem with units? How to do it to get $E_b/N_0$ ratio in decibels, because I keep doing something wrong.
HINT:
$$
\frac{S}{N}=\frac{E_b \cdot R_b}{N_0 \cdot B }
$$
With all units linear:
$$\implies \frac{E_b}{N_0} = \frac SN\cdot \frac BR_b\tag{linear}
$$
Now with $S/N$ in $[\rm dB]$ and $R_b$ and $B$ linear, you get $E_b/N_0$ in $[\rm dB]$ as follows:
$$\implies \frac{E_b}{N_0} = \frac SN +10\log_{10}\left(\frac BR_b\right)\tag{in [dB]}$$