Questions: To quantify the intensity of sound, the decibel scale was developed. The formula for loudness L on the decibel scale is L=10 log (1/L0), where 10 is the intensity of sound just below the threshold of hearing, which is approximately 10^-16 watt per square centimeter. Find the decibel reading for a sound with intensity 32,000 times 10. The decibel reading is approximately . (Round to the nearest whole number as needed.)

Solution Steps

The problem asks us to find the decibel reading for a sound with a given intensity. The formula for loudness \( L \) on the decibel scale is:

\[ L = 10 \log \left( \frac{I}{I_0} \right) \]

where \( I \) is the intensity of the sound, and \( I_0 \) is the reference intensity, which is \( 10^{-16} \) watt per square centimeter.

The problem states that the intensity of the sound is 32,000 times 10. Therefore, the intensity \( I \) is:

\[ I = 32,000 \times 10 = 320,000 \]

Substitute the given intensity \( I = 320,000 \) and the reference intensity \( I_0 = 10^{-16} \) into the formula:

\[ L = 10 \log \left( \frac{320,000}{10^{-16}} \right) \]

Calculate the ratio inside the logarithm:

\[ \frac{320,000}{10^{-16}} = 320,000 \times 10^{16} = 3.2 \times 10^{21} \]

Now, calculate the logarithm:

\[ \log(3.2 \times 10^{21}) = \log(3.2) + \log(10^{21}) \]

Using the properties of logarithms, we have:

\[ \log(3.2) \approx 0.5051 \quad \text{and} \quad \log(10^{21}) = 21 \]

Thus:

\[ \log(3.2 \times 10^{21}) = 0.5051 + 21 = 21.5051 \]

Substitute back into the formula for \( L \):

\[ L = 10 \times 21.5051 = 215.051 \]

Round the decibel level to the nearest whole number:

\[ L \approx 215 \]

Final Answer