We need daredevel7's mathematical skills here...
Someone say my name?
This is how I see it (I could be wrong, btw):
To me,
1 dB represents 10^0 = 1x
10 db represents 10^1 = 10x
20 db represents 10^2 = 100x
30 db represents 10^3 = 1000x
40 db represents 10^4 = 10000x
....
80 db represents 10^8 = 100000000x
100db represents 10^10 = 10000000000x
So.. PFH was correct when he said:
10 db = 10x. so 20db is 100x of 1db.
100db is 100x of 80db....
Because:
10^10 / 10^8 = 10^2 = 100x
(100dB) / (80dB)
Meaning 100 dB is definitely 100x bigger than 80dB
Now, about the 50% loss. That assumes that there is an "optimum" or a "maximum". Because you can't put a percentage of loss without a reference. So let's assume that 100dB is the "maximum".
100 dB is equal to 10^10.
So 50% of 10^10 is actually just 5^10.
Using my calculator to find the log of 5^10 using a base of 10... that equals to 7.6 ish. **
This means 50% loss from 100 dB is about 76 dB, which supports PFH's theory:
Then again, I can see it being 85dB being 50% less.... And logically that would make sense.
And I'd have to agree. 50% loss from 120 dB is actually 117 dB. So looks like 100dB is a better point of reference.
** If you want to know what the hell I am talking about with the log stuff, here is an explanation. For those who don't care, end of post!
If you use your calculator, you can use log to solve for X if you know Y for 10^X=Y. (This is assuming log is based on 10, but most are.)
For example, 10^2 = 100, right? So if I type in 100, then press log, my answer will be 2. Obviously, you don't need a calculator if you have something like 100, 10000, 100000000000000, you only have to count the number of zeros, but when you have a more complicated number, you will need your calculator for that. So, that's what I did for 5^10.