The intensity, or loudness, of a sound can be measured in decibels ($dB$), according to the equation $I(dB)=10 \log \left[\frac{I}{I_0}\right]$, where $I$ is the intensity of a given sound and $I_0$ is the threshold of hearing intensity. What is the intensity, in decibels, [$I(dB)$], when $I=10^{32}\left(I_0\right)$?
The intensity level of a sound in decibels, denoted as \(I(dB)\), is calculated using the formula:
\[I(dB) = 10 \log_{10} \left( \frac{I}{I_0} \right)\]
where \(I\) represents the intensity of the sound in question, and \(I_0\) is the reference intensity known as the threshold of hearing. We are given the ratio of the sound intensity to the reference intensity as \(\frac{I}{I_0} = 10^{32}\). Our objective is to determine the sound level in decibels for this specific ratio.
Substitute the given ratio \(\frac{I}{I_0} = 10^{32}\) directly into the decibel formula:
\[I(dB) = 10 \log_{10} \left( 10^{32} \right)\]
Using the logarithmic property \(\log_b(b^x) = x\), we can simplify the expression \(\log_{10}(10^{32})\). Since the base of the logarithm is 10 and the exponent is 32, this term evaluates to exactly 32.
Now, multiply this result by the factor of 10 in our formula to find the final intensity in decibels:
\[I(dB) = 10 \times 32 = 320\]
Comparing this result with the provided options:
The calculated value corresponds to option (a).
In audio engineering and acoustics, decibels are used to measure the power or intensity of sound. Because the range of human hearing covers a vast scale, logarithmic scales like the decibel system make it easier to compare extremely quiet sounds to incredibly loud ones. For example, comparing the sound of a whisper to the sound of a jet engine involves a difference in power that is billions of times larger; using decibels converts these massive ratios into manageable numbers, helping engineers design safe sound systems and architects create better acoustic environments in buildings.