A member of an aircraft maintenance crew wears protective earplugs that reduce the sound intensity by a factor of 355. When a jet aircraft is taking off, the sound intensity level experienced by the crew member is 72 dB. What sound intensity level would the crew member experience if he removed the protective earplugs?

Answer 1

what is log 355? 2.55

multiply that by 10, add it to 72 db

intensity= 10 log (power)=10log(original)+10log355

Answer 2

A factor of 355 in intensity is 25.5 dB more than 72 dB. That makes 97.5 dB

Here's how I got that: Take 10 times the log(base10) of 355.