To find how much more intense the louder sound is compared to the softer sound, we need to analyze the relationship between decibel levels and sound intensity using the given decibel formula:
[tex]\[ D = 10 \log \left(\frac{I}{I_0}\right) \][/tex]
where [tex]\( D \)[/tex] is the decibel level, [tex]\( I \)[/tex] is the sound intensity, and [tex]\( I_0 \)[/tex] is the reference intensity.
Given that the decibel level of one sound is 5 more than that of another sound, let’s denote:
- [tex]\( D_{\text{higher}} \)[/tex] as the decibel level of the louder sound, and
- [tex]\( D_{\text{lower}} \)[/tex] as the decibel level of the softer sound.
We know:
[tex]\[ D_{\text{higher}} = D_{\text{lower}} + 5 \][/tex]
Using the decibel formula:
[tex]\[ D_{\text{higher}} = 10 \log \left(\frac{I_{\text{higher}}}{I_0}\right) \][/tex]
[tex]\[ D_{\text{lower}} = 10 \log \left(\frac{I_{\text{lower}}}{I_0}\right) \][/tex]
According to the given information:
[tex]\[ 10 \log \left(\frac{I_{\text{higher}}}{I_0}\right) = 10 \log \left(\frac{I_{\text{lower}}}{I_0}\right) + 5 \][/tex]
Let's isolate the logarithmic expressions:
[tex]\[ 10 \log \left(\frac{I_{\text{higher}}}{I_0}\right) - 10 \log \left(\frac{I_{\text{lower}}}{I_0}\right) = 5 \][/tex]
We can factor out the 10:
[tex]\[ 10 \left[ \log \left(\frac{I_{\text{higher}}}{I_0}\right) - \log \left(\frac{I_{\text{lower}}}{I_0}\right) \right] = 5 \][/tex]
Apply the properties of logarithms:
[tex]\[ 10 \log \left(\frac{I_{\text{higher}}}{I_{\text{lower}}}\right) = 5 \][/tex]
Divide both sides by 10:
[tex]\[ \log \left(\frac{I_{\text{higher}}}{I_{\text{lower}}}\right) = 0.5 \][/tex]
To get rid of the logarithm, rewrite the equation in exponential form:
[tex]\[ \frac{I_{\text{higher}}}{I_{\text{lower}}} = 10^{0.5} \][/tex]
Calculate the value:
[tex]\[ 10^{0.5} = \sqrt{10} \approx 3.16 \][/tex]
Thus, the louder sound is approximately [tex]\(\boxed{3.16}\)[/tex] times more intense than the softer sound.
