To solve the problem, we need to find the fraction of the sound intensity of the first game to the sound intensity of the second game. Below is a detailed step-by-step solution:
1. Understanding the given data:
- The loudest sound measured at the first game is [tex]\(L_1 = 112 \text{ dB}\)[/tex].
- The loudest sound measured at the second game is [tex]\(L_2 = 118 \text{ dB}\)[/tex].
- [tex]\(L = 10 \log \left(\frac{I}{I_0}\right)\)[/tex]
where [tex]\(I\)[/tex] is the sound intensity in watts per square meter, and [tex]\(I_0 = 10^{-12} \text{ watts/m}^2\)[/tex] is the reference sound intensity.
2. Calculate the sound intensity for the first game:
Using the formula [tex]\(L = 10 \log \left(\frac{I}{I_0}\right)\)[/tex], we can isolate [tex]\(I\)[/tex]:
[tex]\[
L_1 = 10 \log \left(\frac{I_1}{10^{-12}}\right)
\][/tex]
[tex]\[
112 = 10 \log \left(\frac{I_1}{10^{-12}}\right)
\][/tex]
Divide both sides by 10:
[tex]\[
11.2 = \log \left(\frac{I_1}{10^{-12}}\right)
\][/tex]
To solve for [tex]\(I_1\)[/tex], we exponentiate both sides with base 10:
[tex]\[
10^{11.2} = \frac{I_1}{10^{-12}}
\][/tex]
Multiply both sides by [tex]\(10^{-12}\)[/tex]:
[tex]\[
I_1 = 10^{-12} \times 10^{11.2} = 10^{-0.8} \times 10^{12} = 10^{11.2 - 12} = 10^{-0.8} \times 10^{11.2}
\][/tex]
Therefore,
[tex]\[
I_1 \approx 158489319246.11 \text{ watts/m}^2
\][/tex]
3. Calculate the sound intensity for the second game:
Similarly, for the second game's loudness:
[tex]\[
L_2 = 10 \log \left(\frac{I_2}{10^{-12}}\right)
\][/tex]
[tex]\[
118 = 10 \log \left(\frac{I_2}{10^{-12}}\right)
\][/tex]
Divide both sides by 10:
[tex]\[
11.8 = \log \left(\frac{I_2}{10^{-12}}\right)
\][/tex]
To solve for [tex]\(I_2\)[/tex], we exponentiate both sides with base 10:
[tex]\[
10^{11.8} = \frac{I_2}{10^{-12}}
\][/tex]
Multiply both sides by [tex]\(10^{-12}\)[/tex]:
[tex]\[
I_2 = 10^{-12} \times 10^{11.8} = 10^{-0.2} \times 10^{11.8} = 10^{11.8 - 12} = 10^{-0.2} \times 10^{11.8}
\][/tex]
Therefore,
[tex]\[
I_2 \approx 630957344480.19 \text{ watts/m}^2
\][/tex]
4. Calculate the fraction of sound intensity:
[tex]\[
\text{Fraction of sound intensity} = \frac{I_1}{I_2}
\][/tex]
Substituting the values we found:
[tex]\[
\frac{158489319246.11}{630957344480.19} \approx 0.251
\][/tex]
Hence, the fraction of sound intensity of the first game to the second game is approximately 0.25. Therefore, the answer is:
[tex]\[
\boxed{0.25}
\][/tex]
