how does Brandon's sound intensity level compare to ahmad's?

we know that
[tex]L=10log(\frac{I}{I0} )[/tex]
Find the value of L for Brandon
[tex]L=10log(\frac{10^{-10}}{10^{-12}} )[/tex]
[tex]L=10*2=20[/tex]
Find the value of L for Ahmad
[tex]L=10log(\frac{10^{-4}}{10^{-12}} )[/tex]
[tex]L=10*8=80[/tex]
therefore
the answer is
Brandon's sound intensity level is (1/4) the level of Ahmad