To solve for the loudness [tex]\( L \)[/tex] in decibels of a sound with intensity [tex]\( I = 10^{-7} \)[/tex] watts per square meter, we use the formula given:
[tex]\[ L = 10 \log \frac{I}{I_0} \][/tex]
where [tex]\( I_0 = 10^{-12} \)[/tex] watts per square meter is the reference intensity, the threshold of hearing.
Step-by-step solution:
1. Substitute the given values into the formula:
[tex]\[ L = 10 \log \frac{10^{-7}}{10^{-12}} \][/tex]
2. Simplify the fraction inside the logarithm:
[tex]\[ \frac{10^{-7}}{10^{-12}} = 10^{-7} \div 10^{-12} \][/tex]
Using the property of exponents [tex]\( a^{-m} / a^{-n} = a^{n-m} \)[/tex]:
[tex]\[ 10^{-7} \div 10^{-12} = 10^{(-7) - (-12)} = 10^{5} \][/tex]
Therefore:
[tex]\[ L = 10 \log (10^5) \][/tex]
3. Evaluate the logarithm:
[tex]\[ \log (10^5) = 5 \][/tex]
This is because the logarithm with base 10 of [tex]\( 10^5 \)[/tex] is just the exponent 5.
4. Multiply by 10 to find [tex]\( L \)[/tex]:
[tex]\[ L = 10 \times 5 = 50 \][/tex]
Thus, the approximate loudness of a dinner conversation with a sound intensity of [tex]\( 10^{-7} \)[/tex] watts per square meter is:
[tex]\[ L = 50 \text{ dB} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{50 \text{ dB}} \][/tex]
