To solve for the intensity of a sound in decibels when [tex]\( I = 10^8 \times I_0 \)[/tex], let's break down the problem step by step.
1. Understand the Equation:
The intensity in decibels ([tex]\( I_{dB} \)[/tex]) is given by the formula:
[tex]\[
I_{dB} = 10 \log_{10} \left( \frac{I}{I_0} \right)
\][/tex]
Here, [tex]\( I \)[/tex] is the intensity of the sound and [tex]\( I_0 \)[/tex] is the reference intensity (threshold of hearing).
2. Substitute Given Values:
We are given that [tex]\( I = 10^8 \times I_0 \)[/tex]. Substituting this into the formula, we get:
[tex]\[
I_{dB} = 10 \log_{10} \left( \frac{10^8 \times I_0}{I_0} \right)
\][/tex]
3. Simplify the Expression:
Since [tex]\( I_0 \)[/tex] is present in both the numerator and the denominator, they cancel each other out. This simplification leads to:
[tex]\[
I_{dB} = 10 \log_{10} (10^8)
\][/tex]
4. Apply the Logarithm Rule:
Using the property of logarithms, [tex]\( \log_{10} (10^8) = 8 \)[/tex]. Therefore, we can simplify further:
[tex]\[
I_{dB} = 10 \times 8
\][/tex]
5. Calculate the Decibel Intensity:
Finally, we multiply to get the answer:
[tex]\[
I_{dB} = 80
\][/tex]
Therefore, the intensity of the sound is 80 decibels. The correct answer is 80.
