Question

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

Answer 1

we know that

`L=10log((I)/(I0) )`

Find the value of L for Brandon

`L=10log((10^(-10))/(10^(-12)) )`

`L=10*2=20`

Find the value of L for Ahmad

`L=10log((10^(-4))/(10^(-12)) )`

`L=10*8=80`

therefore

the answer is

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