Rock Concert vs Whisper: Intensity Factor of 22026.4

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The intensity of sound at a rock concert, measured at 120 dB, is significantly louder than that of a whisper at 20 dB. The calculation shows that the intensity ratio between the two is 22026.4. This ratio indicates that the sound intensity at the concert is 22026.4 times greater than that of a whisper. The discussion involves using the decibel formula to derive this intensity factor. Understanding this ratio is crucial for comparing sound levels in different environments.
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Homework Statement



By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 dB and 20 dB respectively?


Homework Equations



The sound level of a sound wave in decibles is

\beta = 10 log (\frac{I}{I_0})

Also the ratio of intensities:

\beta_B - \beta_A = 10 log (\frac{I_B}{I_A})

The Attempt at a Solution



Using the latter equation:

100 = 10 log (\frac{I_B}{I_A})

Solving for the ratio I get 22026.4. Now how can I determine the factor one intensity is greater than the other? I'm not quite sure... :confused:
 
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roam said:

Homework Statement

\beta_B - \beta_A = 10 log (\frac{I_B}{I_A})

The Attempt at a Solution



Using the latter equation:

100 = 10 log (\frac{I_B}{I_A})

Solving for the ratio I get 22026.4. Now how can I determine the factor one intensity is greater than the other? I'm not quite sure... :confused:

The equation should be

100 = 10log_{10}(\frac{I_A}{I_B})

Or 10 = log_{10}(\frac{I_A}{I_B})

Now find the ratio of the intensities.
 
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