The decibel level [tex]L[/tex] of a sound is related to its intensity [tex]I[/tex] by [tex]L = 10 \log \left(\frac{I}{I_0}\right)[/tex]. If [tex]I_0[/tex] is [tex]10^{-12}[/tex], what is the intensity of a noise measured at 39 decibels? Express your answer in scientific notation, rounding to three significant digits if necessary.



Answer :

To solve for the intensity \(I\) of a noise measured at 39 decibels using the given formula, follow these steps:

1. Understand the given formula:
[tex]\[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \][/tex]

2. Identify the given values:
- Decibel level \( L = 39 \)
- Reference intensity \( I_0 = 10^{-12} \)

3. Rearrange the formula to solve for intensity \( I \):
[tex]\[ 39 = 10 \log_{10} \left( \frac{ I }{ 10^{-12} } \right) \][/tex]
Divide both sides by 10 to isolate the logarithm:
[tex]\[ 3.9 = \log_{10} \left( \frac{I}{10^{-12}} \right) \][/tex]

4. Solve for \( \frac{I}{I_0} \):
To remove the logarithm, rewrite the equation in its exponential form:
[tex]\[ 10^{3.9} = \frac{I}{10^{-12}} \][/tex]

5. Multiply both sides by \( 10^{-12} \) to solve for \( I \):
[tex]\[ I = 10^{-12} \times 10^{3.9} \][/tex]

6. Simplify the expression:
Combine the exponents:
[tex]\[ I = 10^{-12 + 3.9} \][/tex]
[tex]\[ I = 10^{-8.1} \][/tex]

7. Convert the result to scientific notation:
Using rounding to three significant digits:
[tex]\[ I \approx 7.94 \times 10^{-9} \][/tex]

Therefore, the intensity of the noise measured at 39 decibels is approximately [tex]\( 7.94 \times 10^{-9} \)[/tex] watts per square meter.