At a distance of 5.0 m from a point source, the sound level is 40...

The intensity of sound from a point source varies inversely with the square of the distance:

$I \propto \dfrac{1}{r^2}$

The ratio of intensities at distances $r_1 = 5.0 \text{ m} and$r_2 = 50 \text{ m}$is:<br><br>$\dfrac{I_2}{I_1} = \left(\dfrac{r_1}{r_2}\right)^2 = \left(\dfrac{5.0}{50}\right)^2 = 10^{-2}$<br><br>The sound level in decibels is given by$\beta = 10 \log_{10} \left(\dfrac{I}{I_0}\right). The difference in sound levels is:

$\beta_2 - \beta_1 = 10 \log_{10} \left(\dfrac{I_2}{I_1}\right)$

Substituting the given values:

$\beta_2 - 40 = 10 \log_{10}(10^{-2})$

$\beta_2 - 40 = -20$

$\beta_2 = 20 \text{ dB}$