Posted by Joel on October 13, 2002 at 13:06:57:
In Reply to: log problem posted by Lisa on October 12, 2002 at 16:29:31:
: The noise level of one sound is 20 decibals higher than that of a second sound. Find the ratio of the intensity of the first sound to the second sound.
: Is this the way to do a problem like this.
: L = 10 (log I_1/I_2)= 10 log 100 = 20
Well, sort of, but since 100 is the answer, how did you jump to that so fast?
This is just simple algebra. You're starting with:
10*log(I_1/I_2) = 20dB and you want to solve for the ratio (I_1/I_2).
Just divide both sides by 10:
log(I_1/I_2) = 2
Now, since you are dealing with log to the base 10, this is equivalent to:
I_1/I_2 = 10^2
I_1/I_2 = 100
i.e. the louder one has 100 times the intensity (or power) of the other.
Post a Followup