An 87.0 dB sound has an intensity five times as great as an 80 dB sound. What difference in sound intensity level does a factor of 5 in intensity correspond to?

1. The sound intensity level difference is beta2 minus beta1 = 10 log10(I2 / I1). 2. For a factor of 5 in intensity, this is 10 log10(5) = 10 x 0.699 = 6.99 dB, which rounds to about 7 dB. 3. This matches option B, consistent with the comparison of an 87.0 dB sound to an 80 dB sound, a 7 dB gap. 4. Option C, about 3 dB, is instead the difference produced by a factor of 2 in intensity, not a factor of 5. _Source: OpenStax Physics (CC BY 4.0), Ch 14 "Sound", section 14.2 Sound Intensity and Sound Level_