Answer:
Braylee’s music is 3 times louder than Jessica's
Step-by-step explanation:
Given
[tex]L = 10\log\frac{I}{I_0}[/tex]
[tex]I_0 = 10^{-12}[/tex]
Required
Number of times Braylee's music is louder than Jessica's
First, we calculate the loudness of Jessica's music.
From the question;
[tex]I = 10^{-9}[/tex]
So, we have:
[tex]L = 10\log\frac{I}{I_0}[/tex]
[tex]L = 10\log\frac{10^{-9}}{10^{-12}}[/tex]
Apply law of indices
[tex]L = 10\log{10^{12-9}}[/tex]
[tex]L = 10\log{10^3[/tex]
Apply law of logarithm
[tex]L = 3*10\log{10[/tex]
[tex]\log10 = 1[/tex]
So:
[tex]L_1 = 3*10*1 = 30[/tex]
First, we calculate the loudness of Braylee's music.
From the question;
[tex]I = 10^{-3}[/tex]
So, we have:
[tex]L = 10\log\frac{I}{I_0}[/tex]
[tex]L = 10\log\frac{10^{-3}}{10^{-12}}[/tex]
Apply law of indices
[tex]L = 10\log{10^{12-3}}[/tex]
[tex]L = 10\log{10^9[/tex]
Apply law of logarithm
[tex]L = 9*10\log{10[/tex]
[tex]L_2 = 9*10*1 = 90[/tex]
At this point, we have:
[tex]L_1 = 30[/tex] --- Jessica's
[tex]L_2 = 90[/tex] --- Braylee's
Divide L2 by L1 to get the required number of times
[tex]n = \frac{L_2}{L_1}[/tex]
[tex]n = \frac{90}{30}[/tex]
[tex]n = 3[/tex]
