Can you hear me now outside?

[sailordragonball]

A recording engineer works in a soundproofed room that is 45.0 dB quieter than the outside. If the sound intensity in the room is 1.30e-10 W/(m^2), what is the intensity outside?

I know this intensity formula ...

... I = (P/A) OR I = ( P / [(4)(pi)(r^2)] )

... and for dB ...

... beta = (10 dB)(log (I/Io)... I can't make sense of the 2 equations to know what to do next ... any ideas?

 

[sailordragonball]

Do I substitute 45 dB for the 10 in the dB equation?

 

[OlderDan]

The outside sound is the more intense sound. The intensity inside is a fraction of the intensity of the outside sound such that the ratio gives a 45 dB difference. What is the value of beta, including its sign, if you take the outside sound intensity as the reference? What is it if you take the inside intensity as the reference?

 

[sailordragonball]

I think I have to revisit working with logs ... it's been a while ... LOL - I'll get back to you though.

 

[sailordragonball]

I came up with this ...

... 45 = 10 log ( I / 1e-10 ) = 10 log ( 1.3e-12 / 1e-10 ) ...

... does that sound right? LOL

 

[OlderDan]

I came up with this ...

... 45 = 10 log ( I / 1e-10 ) = 10 log ( 1.3e-12 / 1e-10 ) ...

... does that sound right? LOL

No it doesn't. The 45 is in the right place, but where did 1e-10 come from? With a +45 on the left, the sound intensity in the room should be the reference I_o and the intensity outside the I. Waht is the inverse function of the log functiuon?