The Ratio intensities of two sounds

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Homework Statement



I need to find the ration of the intensities of two sounds if one is 12 dB higher than the other ?



Homework Equations



I am not sure


The Attempt at a Solution



I took the inverse log of 12 and multiple the answer with ( 1 * 10^-12) to find I = 1.584 * 10^-11

I know that the ration = (the higher values/the smaller value)

but I am not sure about what I am doing
 
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The decibel scale is a logarithmic one, and a relative one (that is, unlike temperature or length, you define any intensity to be 0 dB and then measure everything relative to that specific intensity). If you understand this question, you have understood the concept of it.

So let's start easier: do you know what the increase in level is (measured in dB) if the intensity doubles?
 
You pretty much had the answer already. You found the intensity of a sound that corresponds to a 12dB sound level, which was 1.58 x 10^-11. Now as you said, the ratio of the intensities would be the intensity you found over the intensity of the original sound, in this case the 0dB sound, or one with an intensity of 1.0 x 10^-12. So now just calculate this ratio.
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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