Question 1038963
you need a different variable to represent the intensity of the larger sound than the variable that represents the intensity of the smaller sound.


capital I for the larger sound and small i for the smaller sound would do it.


the formula is D = 10 * (log(I) - log(i))


since log(a) - log(b) = log(a/b), then D = 10 * (log(I) - log(i)) is equivalent to D = 10 * log(I/i).


the formula becomes D = 10 * log(I/i).


if the intensity of a sound is 100 times the intensity of a softer sound, then I can be replaced with 100 * i and the formula D = 10 * log(I/i) becomes D = 10 * log(100*i/i) which becomes D = 10 * log(100).


log(100) is equal to 2 because 10^2 = 100.


the formula becomes D = 10 * 2 which makes D = 20.


a sound 20 DB higher than another sound is 100 times as intense as that of the other sound.


the following reference supports this conclusion.


<a href = "http://www.allaboutcircuits.com/textbook/semiconductors/chpt-1/absolute-db-scales/" target = "_blank">http://www.allaboutcircuits.com/textbook/semiconductors/chpt-1/absolute-db-scales/</a>


the table in this article has 1 milliwatt of power as a reference.


you'll need to scroll down a little until you find it.


1 milliwatt is not shown, but it is assumed to be 0 db.


0 db is equal to 1 milliwatt which is equal to .001 watts.



100 milliwatts is shown as 20 db.


100 milliwatts is 100 times the intensity of 1 milliwatt of power.