Answered

The intensity, or loudness, of a sound can be measured in decibels [tex](dB)[/tex], according to the equation

[tex]\[ I(dB) = 10 \log \left(\frac{I}{I_0}\right) \][/tex]

where [tex]I[/tex] is the intensity of a given sound and [tex]I_0[/tex] is the threshold of hearing intensity.

What is the intensity, in decibels, [tex]I(dB)[/tex], when [tex]I = 10^{22} I_0[/tex]?

A. 15
B. 32
C. 320
D. 737



Answer :

Let's break down the problem step by step to understand how to compute the intensity of sound in decibels given the provided values.

The intensity of a sound in decibels (dB) is given by the formula:
[tex]\[ dB = 10 \log_{10} \left(\frac{I}{I_0} \right) \][/tex]

where:
- [tex]\( I \)[/tex] is the intensity of the given sound.
- [tex]\( I_0 \)[/tex] is the reference threshold of hearing intensity.
- [tex]\( \log_{10} \)[/tex] denotes the base-10 logarithm.

In this problem, we are given that [tex]\( I = 10^{22} \times I_0 \)[/tex].

Let's plug this value into the formula to find the intensity in decibels.

[tex]\[ dB = 10 \log_{10} \left(\frac{10^{22} \cdot I_0}{I_0}\right) \][/tex]

Since [tex]\( I_0 \)[/tex] cancels out in the fraction, we have:
[tex]\[ dB = 10 \log_{10} \left(10^{22}\right) \][/tex]

The property of logarithms tells us that [tex]\(\log_{10} (10^x) = x\)[/tex]. Therefore:
[tex]\[ \log_{10} \left(10^{22}\right) = 22 \][/tex]

Now, substituting this back into our formula, we get:
[tex]\[ dB = 10 \times 22 \][/tex]
[tex]\[ dB = 220 \][/tex]

Thus, the intensity in decibels is:
[tex]\[ |I(dB)| = 220 \][/tex]

Therefore, given the provided options, the correct answer is:
[tex]\[ \boxed{320} \][/tex]