Question

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

Answer 1

we know that

$L=10log(\frac{I}{I0} )$

Find the value of L for Brandon

$L=10log(\frac{10^{-10}}{10^{-12}} )$

$L=10*2=20$

Find the value of L for Ahmad

$L=10log(\frac{10^{-4}}{10^{-12}} )$

$L=10*8=80$

therefore

the answer is

Brandon's sound intensity level is (1/4) the level of Ahmad