The intensity level of a sound is reported in unit decibel(dB). how does IL change if we increase a sound intensity by a factor of 10?

a)remains the same
b) increases by 1dB to 2dB
c)it increases by 2dB to 20dB
d)it increases by 20dB to 200dB
e) it decreases

IL=10(log I- logIo)

..i have posted this before and the person who answered got the same thing I had been getting...that it increases by a factor of ten...this is a textbook question so the answer must be there and our conclusion it possible that they mean the sound intensity, after taking the log of it, is multiplied by 10..and in that case e would be the correct answer?

any help is greatly appreciated.

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  1. Say intensity 1 = Io
    then intensity 2 = 10 Io

    dB1 = 0 dB
    dB2 = 10 (log 10 Io - log Io)
    which is 10 log 10 = 10
    10-0 = 10
    We all agree.

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  2. I bet the meant the PRESSURE increases by a factor of ten. Then it is 20 log for dB.

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