## Calculus 8th Edition

$4.34\times10^{7}\frac{dB}{watt/m^{2}}$
Loudness of a sound with intensity $I, L=10log_{10}(\frac{I}{I_{0}})$ Given: $L = 50$ dB $10log_{10}(\frac{I}{I_{0}})=50$ dB $log_{10}(\frac{I}{I_{0}})=5$ dB or $I=(10)^{5}{I_{0}}$ ...(1) Using equation (1) and given value of intensity $I=10^{-7}watt/m^{2}$ The rate of change of loudness is $=\frac{dL}{dI}$ Therefore, $\frac{dL}{dI}=\frac{10}{ln10}\frac{1}{I}$ Hence, $\frac{dL}{dI}=\frac{10^{8}}{ln10 } watt/m^{2}\approx 4.34\times10^{7}\frac{dB}{watt/m^{2}}$