The factor by which the acoustic power is changed is 7943.28

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The discussion centers on calculating the change in acoustic power when the intensity level of sound decreases from 80 dB to 41 dB. The equation used is 10*log(I2/I1) = B1 - B2, leading to the conclusion that log(I2/I1) equals 3.9. By applying the anti-logarithm, the ratio of intensities I2 to I1 is determined to be 7943.28. This indicates that the acoustic power is reduced by a factor of 7943.28. The calculations confirm the accuracy of the method used.
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Homework Statement



The intensity level of a particular sound coming through a medium gets reduced from 80 dB to 41 dB. By what factor is the acoustic power that can pass through changed by?

B1=80
B2=41

Homework Equations



10*log(\frac{I2}{I1}) = B1-B2

The Attempt at a Solution



I was wondering if I applied the correct method.

log(\frac{I2}{I1}) = 3.9

Using anti-log I get:
\frac{I2}{I1} = 7943.28
 
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Looks right to me.
 

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