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The volume control on a surround-sound amplifier is adjusted so the sound intensity level at the listening position increases from 23 to 61 dB. What is the ratio of the final sound intensity to the original sound intensity?
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The decibel scale is defined as
beta = 10*log_10(I/I0)

Apply subscripts 1 and 2 to correspond to the conditions given:
beta1 = 10*log_10(I1/I0)
beta2 = 10*log_10(I2/I0)

Subtract (you will see why), beta2 - beta1:
beta2 - beta1 = 10*(log_10(I2/I0) - log_10(I1/I0))

Combining the logarithm, and you get:
beta2 - beta1 = 10*log_10(I2/I1)

Solve for the ratio of I's:
(beta2 - beta1)/10 = log_10(I2/I1)

Exponentiate both sides with base 10:
10^((beta2 - beta1)/10) = I2/I1

Thus:
I2/I1 = 10^((beta2 - beta1)/10)

Data:
beta2:=61 dB; beta1:=23 dB;

Result:
I2/I1 = 6309.57
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