If the intensity of sound from one source is 10,000 times that of another, how many times more is the decibel level of the louder sound than the quieter one?

We are asked to find the increase in the loudness of a sound (in decibels) if the intensity of the source is increased by a factor of 10,000.
The loudness of a sound is commonly measured in decibels. The formula to convert from the intensity of a sound (often measured in watts per square meter) to decibels is L=10 + log (I/I_0) where I is the intensity of the sound in watts per square meter, I nought is set at 10^(-12) watts per square meter, and log is the common (base 10) logarithm.
To find the increase in decibels we can subtract:
L=[10+log(10000I/I_0)]-[10+log(I/I_0)]
= 10+log(10000I/I_0)-10-log(I/I_0)
Use the following properties of logarithms:
log(AB)=logA+logB, log(A/B)=logA-logB
to simplify further:
=10+log(10^5)+log(I/I_0)-10-log(I/I_0)
=log(10^5)+logI-logI_0-logI+logI_0
=log(10^5)
Now use another property of logarithms, log(A^n)=nlogA
=5log10
=5 (Since by definition the log base 10 of 10 is 1.)
Thus, a source with intensity 10,000 times greater than the original source will produce a sound which is 5 decibels louder.
Note that the original intensity did not matter. For example suppose the original intensity was 10^6 watts per square meter. Its loudness would be :
10+log(10^6/10^(-12))=10+log(10^(18))=10+18=28"dB"
Increasing the intensity by 10,000 watts per square meter to 10^(11) yields:
10+log(10^(11)/(10^(-12)))=10+log(10^(23))=10+23=33"dB"
The 10 we are adding is because we measure sound in decibels as opposed to bels, much like we measure food energy in Calories that are kilocalories.
33dB would be very quiet—50dB is close to the loudness of a quiet conversation.
http://mathworld.wolfram.com/Logarithm.html