Prove that `DeltaL=10log_(10)[(I_(1))/(I_(0))]` decibel, from Weber-Fechner's law.

Explanation

According to Weber-Fechner's law, ''loudness (L) is proportional to the logarithm of the actual intensity (I) measured with an accurate non-human instrument''. It is meant that
`LproplnI`
`L=klnI`
The difference between two loudnesses, `L_(1)andL_(0)` measures the relative loudness between two precisely measured intensities and is called as sound intensity level. Mathematically, sound intensity level is
`DeltaL=L_(1)-L_(0)=klnI_(1)-klnI_(0)`
`=kln[(I_(1))/(I_(0))]`
If k=1, then sound intensity level is measured in bel, in honour of Alexander Graham Bell.
`:.DeltaL=ln[(I_(1))/(I_(0))]`bel
However, this is practically a bigger unit, so we use a convenient smaller unit, called decibel. Thus, decibel `=(1)/(10)bel`. Hence, by multiplying and dividing by 10 we get
`DeltaL=10(ln[(I_(1))/(I_(0))])(1)/(10)bel`
`DeltaL=10ln[(I_(1))/(I_(0))]` decibel with k=10
For practical purposes, we use logarithm to base 10 instead of natural logarithm,
`DeltaL=10log_(10)[(I_(1))/(I_(0))]` decibel.