How much intense is 80 dB sound in comparision to 40 dB ?

$\because \mathrm{B}_1=10 \log _{10}$\left(\frac{$\mathrm{I}$_1}{$\mathrm{I}$_0}\right)$and$\quad \mathrm{B}_2=10 \log _{10}\left(\frac{\mathrm{I}_2}{\mathrm{I}_0}\right)\because$Given,$\mathrm{B}_1=80 \mathrm{~dB}, \mathrm{~B}_2=40 \mathrm{~dB} \therefore \quad 80=10 \log _{10}\left(\frac{$\mathrm{I}$_1}{$\mathrm{I}$_0}\right)$or$\frac{$\mathrm{I}$_1}{$\mathrm{I}$_0}=10^8$and$40=10 \log _{10}\left(\frac{\mathrm{I}_2}{\mathrm{I}_0}\right) or \quad \frac{\mathrm{I}_2}{\mathrm{I}_0}=10^4 Hence, \frac{\mathrm{I}_1}{\mathrm{I}_2}=10^4