A recording engineer works in a soundproofed room that is 47.5 dB quieter than the outside. If the sound intensity in the room is 1.40 x 10^-10 W/m2, what is the intensity outside?
-47.5=10log( 1.40 x 10^-10/Io
divid each side by ten, then the antilog of each side, and solve.
mary, I think we are in the same class
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To find the intensity outside, we can use the formula for the sound level difference in decibels:
ΔL = 10log(I/Io)
where ΔL is the difference in sound level in decibels, I is the sound intensity inside the room, and Io is the sound intensity outside the room (which we want to find).
Given that the sound level difference is -47.5 dB (quieter), and the sound intensity inside the room is 1.40 x 10^-10 W/m^2, we can substitute these values into the formula:
-47.5 = 10log(1.40 x 10^-10 / Io)
Dividing both sides by 10 gives:
-4.75 = log(1.40 x 10^-10 / Io)
To isolate Io, we need to take the antilog (inverse logarithm) of both sides. Since the logarithm is base 10, we can use the power of 10 function to undo the logarithm:
10^(-4.75) = 1.40 x 10^-10 / Io
Rearranging the equation to solve for Io gives:
Io = (1.40 x 10^-10) / (10^(-4.75))
Evaluating this expression:
Io ≈ 1.26 x 10^-6 W/m^2
Therefore, the intensity outside the room is approximately 1.26 x 10^-6 W/m^2.