Understanding why we multiply log(P/Po) by 10

[dervast]

Hi i want to ask sth for the db
We have denoted db as log10(P/Po)
I want to ask sth if a^x=theta=>x=log(a)theta
then the db is 10*a^x=10*log(a)theta Is that right?
I want to know why we multiply the log(P/Po) by 10? Why we didnt leave it alone?

 

[SGT]

Hi i want to ask sth for the db
We have denoted db as log10(P/Po)
I want to ask sth if a^x=theta=>x=log(a)theta
then the db is 10*a^x=10*log(a)theta Is that right?
I want to know why we multiply the log(P/Po) by 10? Why we didnt leave it alone?

Because $$log\frac{P}{P_0}$$, called Bel is too large a unit to be useful. So we use the deciBel, that is more manageable.

 

[dervast]

If bel is too large to be manageable then why we multiplt by 10?

 

[Averagesupernova]

Dervast, it is the same thing as microfarads vs farads. Most of the time we deal in signal ratios that make it more suitable to use decibels instead of bels. Same thing with farads and microfarads. However, if you choose NOT to multiply by 10 and work with bels it is up to you. It will work just fine.

 

[Ouabache]

If bel is too large to be manageable then why we multiplt by 10?

I think you are looking for an analytical reason.
Try applying dimensional analysis.
log (Pout/Pin) = y (Bel) equ (i)
how to convert to dB? Since a dB is 1/10 of a Bel, then 10db = 1 Bel
To convert equat (i) to db, multiply both sides by (10db/1Bel)
10 log (Pout/Pin) = y (Bel)(10db/1Bel) <------dimensions Bel/Bel = 1 leaving just db
10 log (Pout/Pin) = 10y (db) (ii)
right side of equation (ii) means we multiply the solution y in Bels by 10 to obtain a value in dB.