A two-fold increase in intensity of a wave implies an increase of (Given: ${\log }_{10}2 = 0.3010$.)

$\mathrm{L} = 10 {\log }_{10}\frac{\mathrm{I}}{{\mathrm{I}}_0}\mathrm{decibel}$

${\mathrm{L}}^{\text{'}} = 10{\log }_{10}\frac{2\mathrm{I}}{{\mathrm{I}}_0}\mathrm{decibel}$

${\mathrm{L}}^{\text{'}}-\mathrm{L} = 10[{\log }_{10}\frac{2\mathrm{I}}{{\mathrm{I}}_0}-{\log }_{10}\frac{\mathrm{I}}{{\mathrm{I}}_0}]$

         = 10${\log }_{10}2 = 10\times 0.3010 = 3.01 \mathrm{dB}$