## College Physics (4th Edition)

When all three machines are running, the intensity level is $95.2~dB$ This intensity level is only $2.2~dB$ higher than the intensity level of the loudest machine.
We can find the intensity of the 85-dB machine: $\beta = 10~log\frac{I}{I_0}$ $85 = 10~log\frac{I}{I_0}$ $8.5 = log\frac{I}{I_0}$ $10^{8.5} = \frac{I}{I_0}$ $I = (10^{8.5})~I_0$ $I = (10^{8.5})~(1.0\times 10^{-12}~W/m^2)$ $I = 3.16\times 10^{-4}~W/m^2$ We can find the intensity of the 90-dB machine: $\beta = 10~log\frac{I}{I_0}$ $90 = 10~log\frac{I}{I_0}$ $9.0 = log\frac{I}{I_0}$ $10^{9.0} = \frac{I}{I_0}$ $I = (10^{9.0})~I_0$ $I = (10^{9.0})~(1.0\times 10^{-12}~W/m^2)$ $I = 1.0\times 10^{-3}~W/m^2$ We can find the intensity of the 93-dB machine: $\beta = 10~log\frac{I}{I_0}$ $93 = 10~log\frac{I}{I_0}$ $9.3 = log\frac{I}{I_0}$ $10^{9.3} = \frac{I}{I_0}$ $I = (10^{9.3})~I_0$ $I = (10^{9.3})~(1.0\times 10^{-12}~W/m^2)$ $I = 2.0\times 10^{-3}~W/m^2$ We can find the total intensity from all three machines: $I = (3.16\times 10^{-4}~W/m^2)+(1.0\times 10^{-3}~W/m^2)+(2.0\times 10^{-3}~W/m^2)$ $I = 3.316\times 10^{-3}~W/m^2$ We can find the intensity level: $\beta = 10~log\frac{I}{I_0}$ $\beta = 10~log\frac{3.316\times 10^{-3}~W/m^2}{1.0\times 10^{-12}~W/m^2}$ $\beta = 95.2~dB$ When all three machines are running, the intensity level is $95.2~dB$ This intensity level is only $2.2~dB$ higher than the intensity level of the loudest machine.