Question

The loudness, L, measured in decibels (Db), of a sound intensity, I, measured in watts per square meter, is defined as L = 10 log StartFraction I Over I 0 EndFraction, where I 0 = 10 Superscript negative 12 and is the least intense sound a human ear can hear. Jessica is listening to soft music at a sound intensity level of 10^-9 on her computer while she does her homework. Braylee is completing her homework while listening to very loud music at a sound intensity level of 10^-3 on her headphones. How many times louder is Braylee’s music than Jessica’s?

Answer 1

D.

Explanation:

Answer 2

Braylee’s music is 3 times louder than Jessica's

Explanation:

Given

`L = 10\log(I)/(I_0)`

`I_0 = 10^(-12)`

Required

Number of times Braylee's music is louder than Jessica's

First, we calculate the loudness of Jessica's music.

From the question;

`I = 10^(-9)`

So, we have:

`L = 10\log(I)/(I_0)`

`L = 10\log(10^(-9))/(10^(-12))`

Apply law of indices

`L = 10\log{10^(12-9)}`

`L = 10\log{10^3`

Apply law of logarithm

`L = 3*10\log{10`

`\log10 = 1`

So:

`L_1 = 3*10*1 = 30`

First, we calculate the loudness of Braylee's music.

From the question;

`I = 10^(-3)`

So, we have:

`L = 10\log(I)/(I_0)`

`L = 10\log(10^(-3))/(10^(-12))`

Apply law of indices

`L = 10\log{10^(12-3)}`

`L = 10\log{10^9`

Apply law of logarithm

`L = 9*10\log{10`

`L_2 = 9*10*1 = 90`

At this point, we have:

`L_1 = 30` --- Jessica's

`L_2 = 90` --- Braylee's

Divide L2 by L1 to get the required number of times

`n = (L_2)/(L_1)`

`n = (90)/(30)`

`n = 3`