Select the correct answer.

Using your knowledge of exponential and logarithmic functions and properties, what is the intensity of a fire alarm that has a sound level of 120 decibels?

[tex]\[ \left(\beta = 10 \log \left(\frac{I}{I_0}\right)\right) \][/tex]

A. [tex]\(1.0 \times 10^{-12}\)[/tex] watts/m²

B. [tex]\(1.0 \times 10^0\)[/tex] watts/m²

C. 12 watts/m²

D. [tex]\(1.10 \times 10^2\)[/tex] watts/m²



Answer :

To determine the intensity [tex]\( I \)[/tex] of a sound given its level [tex]\( \beta \)[/tex] in decibels (dB), we can use the decibel formula for intensity:

[tex]\[ \beta = 10 \log \left(\frac{I}{I_0}\right) \][/tex]

where [tex]\( I_0 \)[/tex] is the reference intensity, often taken to be [tex]\( 1.0 \times 10^{-12} \)[/tex] watts/m².

Given:
[tex]\[ \beta = 120 \text{ dB} \][/tex]
[tex]\[ I_0 = 1.0 \times 10^{-12} \text{ watts/m}^2 \][/tex]

We need to solve for [tex]\( I \)[/tex]. Start by isolating [tex]\( \frac{I}{I_0} \)[/tex] in the equation:

[tex]\[ 120 = 10 \log \left(\frac{I}{I_0}\right) \][/tex]

Divide both sides by 10:

[tex]\[ 12 = \log \left(\frac{I}{I_0}\right) \][/tex]

To solve for [tex]\( \frac{I}{I_0} \)[/tex], remember that if:

[tex]\[ \log x = y \][/tex]

then:

[tex]\[ x = 10^y \][/tex]

So,

[tex]\[ \frac{I}{I_0} = 10^{12} \][/tex]

Now, multiply both sides by [tex]\( I_0 \)[/tex]:

[tex]\[ I = I_0 \times 10^{12} \][/tex]

Substitute [tex]\( I_0 \)[/tex] with [tex]\( 1.0 \times 10^{-12} \)[/tex]:

[tex]\[ I = (1.0 \times 10^{-12}) \times 10^{12} \][/tex]

[tex]\[ I = 1.0 \times 10^{0} \][/tex]

[tex]\[ I = 1.0 \text{ watts/m}^2 \][/tex]

Therefore, the correct answer is:

B. [tex]\( 1.0 \times 10^0 \text{ watts/m}^2 \)[/tex]