If the intensity is increased by a factor of `20`, by how many decibels is the intensity level increased.

Explanation

Let the initial intensity be `I` and the density level be `beta_(1)` and when the intensity is increased by `= 20` times, the intensity level increases to `beta_(2)`.
Then `beta_(1) = 10 log(I//I_(0))`
and `beta_(2) = 10 log (20I//I_(0))`
Thus, `beta_(2) - beta_(1) = 10 log (20I//I)`
`= 10 log 20`
`= 13 dB`.