Sure, let's solve this step by step.
First, let's recall the formula which defines loudness [tex]\( L \)[/tex] in decibels (dB):
[tex]\[ L = 10 \log_{10} \left( \frac{I}{I_0} \right) \][/tex]
where:
- [tex]\( I \)[/tex] is the sound intensity we want to measure,
- [tex]\( I_0 \)[/tex] is a reference intensity, specifically the least intense sound a human ear can hear, usually taken as [tex]\( 10^{-12} \)[/tex] watts per square meter.
In this problem, we are given:
- The sound intensity of the rock concert, [tex]\( I = 10^{-1} \)[/tex] watts per square meter.
- The reference intensity, [tex]\( I_0 = 10^{-12} \)[/tex] watts per square meter.
Let's plug these values into the formula:
[tex]\[ L = 10 \log_{10} \left( \frac{10^{-1}}{10^{-12}} \right) \][/tex]
First, calculate the ratio inside the logarithm:
[tex]\[ \frac{10^{-1}}{10^{-12}} \][/tex]
When dividing powers of 10, we subtract the exponents:
[tex]\[ 10^{-1} / 10^{-12} = 10^{-1 - (-12)} = 10^{-1 + 12} = 10^{11} \][/tex]
So, now we have:
[tex]\[ L = 10 \log_{10} (10^{11}) \][/tex]
Recall that [tex]\( \log_{10} (10^x) = x \)[/tex]. Thus:
[tex]\[ \log_{10} (10^{11}) = 11 \][/tex]
So the equation for loudness simplifies to:
[tex]\[ L = 10 \times 11 \][/tex]
[tex]\[ L = 110 \, \text{dB} \][/tex]
Hence, the approximate loudness of the rock concert is:
[tex]\[ \boxed{110 \, \text{dB}} \][/tex]
Therefore, the correct answer choice is:
[tex]\[ 110 \text{dB} \][/tex]
