To solve this problem we need to apply the corresponding sound intensity measured from the logarithmic scale. Since in the range of intensities that the human ear can detect without pain there are large differences in the number of figures used on a linear scale, it is usual to use a logarithmic scale. The unit most used in the logarithmic scale is the decibel yes described as
[tex]\beta_{dB} = 10log_{10} \frac{I}{I_0}[/tex]
Where,
I = Acoustic intensity in linear scale
[tex]I_0[/tex] = Hearing threshold
The value in decibels is 17dB, then
[tex]17dB = 10log_{10} \frac{I}{I_0}[/tex]
Using properties of logarithms we have,
[tex]\frac{17}{10} = log_{10} \frac{I}{I_0}[/tex]
[tex]log_{10} \frac{I}{I_0} = 1.7[/tex]
[tex]\frac{I}{I_0} = 10^{1.7}[/tex]
[tex]\frac{I}{I_0} = 50.12 W/m^2[/tex]
Therefore the factor that the intensity of the sound was [tex]50.12W/m^2[/tex]
