Help Figuring Out Decibels: What Is 2% of 90 dB?

Hey. I have a question that I can't get a good answer to. What is 2% of 90dB? I know dB is a logarithmic scale, so could anyone help me figure this out? Thanks.

 

Are you asking how loud a sound is going to be if it were to be 50 times quieter?

I might be completely off, but here is my thinking. X db is 10^(x/10). Log base 10 of 0.02 is -1.69897. So 10^(-1.69897) = 10^(x/10). Thus x = -17 db, so 2% of 90dB is 90 - 17 = 73 dB.

I think the above is right, as 1% = 100 times less power = -20 dB => 70 dB. And doubling power (going from 1% to 2%) is known to mean an increase of 3 dB.
http://www.animations.physics.unsw.edu.au/jw/dB.htm
and the first table in https://en.wikipedia.org/wiki/Decibel

 

Interesting math process. Thanks for the help!

 

THAT hurt more than my ears ringing...