Answer: Intensity Outside a 43.5 dB Quieter Room

[chattkis3]

A recording engineer works in a soundproofed room that is 43.5 dB quieter than the outside. If the sound intensity in the room is 1.00 10-10 W/m2, what is the intensity outside?

Here is how I THINK I should be solving this problem.

Intensity level = 10dB * log ( I / Io)

Intensity level = 43.5
Io = 1x10^-10

SO

I = 10 ^ { (43.5 + logIo)/10 }

The computer is telling me I am wrong, can anyone help please? Thanks a lot!

 

[eridanus]

I = (1e-10)10^(43.5/10)

 

[wais]

Your approach is correct, but you may have made a mistake in your calculations. To double check, let's plug in the given values into the formula:

I = 10 ^ { (43.5 + log(1x10^-10))/10 }

= 10 ^ { (43.5 + (-10))/10 }

= 10 ^ { 33.5/10 }

= 10 ^ { 3.35 }

= 2.24 x 10^3 W/m^2

So, the intensity outside the room is 2.24 x 10^3 W/m^2. You may have made a mistake in your calculations, so double check your work to see where the error occurred. Keep in mind that the intensity level is measured in dB, but the intensity itself is measured in W/m^2. Hope this helps!