I am very confused how ham operators deal with all this, because most of the information in internet is controversial.
I have coaxial cable with loss of 34dB at 100 meters at 2400mhz. People say that if you send 1 mW of power (0dBm) trough this cable @ 2400mhz at the other end of it you will get -34dBm. Is this true ? (because of 0dBm - 34dB = -34dBm)
If this is true, then why people say that a source signal of 20dBm will be attenuated with the same calculation ( 20dBm - 34dB = -14dBm) when 20dBm is actually 100 times more watts of power than 0dBm ? My logic is that if you have 100 times more transmitted power you have 100 times the distance in order to get the same -34dB... 100 times the distance is 100meters * 100 = 10km !!? Obviously that's too much distance to believe myself but that's more logical than what many other say that -14dBm will attenuate to -34dBm after couple more meters of additional cable length (because of 34dB * 1.59 = 54dB. 1.59 is 1.59 times the distance of the cable length, which is 159 meters. So 20dBm - 54dB = -34dBm again at 159 meters cable length, which is obviously not 10 km!). So to recap - 0dBm goes 100 meters and 20dBm goes 159 meters. Makes no sense...
Similar, (but not identical as this scenario is wireless) I also see some guys say that the difference between 31dBm and 20dBm is only one brick wall, because brick wall attenuates 11dB for example. Again the difference between 31dBm and 20dBm is 12 times the power. If 20dBm can penetrate 1 brick wall and have 8dBm remainder, why wouldn't 31dBm penetrate 12 times thicker brick wall and have the same remainder of 8dBm ?
Please tell what is right and what is wrong in these statements ?
P.S. Let's don't talk about that WiFi is bidirectional, we're talking about TX only / signal delivery only scenario and how it propagates. I really hope for simple answers than complicated math/physics calculations