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