Confused about = 10 logI/I0

[MedPR]

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I get the basic relationship (I think), but I'm confused about how it could be presented on the MCAT.

For instance, they could give you I in terms of I0. So, when I=1000 I0, then 10log(1000/1) = 10*log1000 = 10*3 = 30decibels.

However, if they give you some off the wall number for Intensity, like 50000, then you have (5*10^4)/(10^-12) = 5*10^16. 10log(5*10^16) = 10*(log5+log(10^16)) = 10*(16+.7) = 167decibels.

This is just arithmetic though, not MCAT type stuff.

 

[chiddler]

i'm not complaining. this makes it easier lol

lets see....think out of the box...

what if you change the reference intensity, how does that affect anything? what does increasing intensity have on dB? (seen this many times, though) Or those annoying "what if distance is doubled, intensity = ?". I had to practice those a lot, and have probably forgotten it now.

 

[milski]

Most related problems I've seen are something in the lines of 1 noise maker makes X db of noise, how many noise makers do you need to add to make Y db of noise. Or how much closer/further do you need to be to the source to make it so much louder/quieter.