Problem
The loudness level of a sound can be expressed by comparing the sound's intensity to the intensity of a sound barely audible to the human ear. The formula $D=10\left(\log I-\log I_{0}\right)$ describes the loudness level of a sound, $D$, in decibels, where $\mathrm{I}$ is the intensity of the sound, in watts per meter ${ }^{2}$, and $\mathrm{I}_{0}$ is the intensity of a sound barely audible to the human ear. Use this information to answer parts $(a)$ and $(b)$ below. a. Express the formula so that the expression in parentheses is written as a single logarithm. \[ D=10 \log \left(\frac{I}{I_{0}}\right) \] b. Use the form of the formula from part (a) to answer this question. If a sound has an intensity $1 / 1,000,000$ th of the intensity of a louder sound, how much smaller on the decibel scale is the loudness level of the less intense sound? decibels softer (Type a whole number.)
Solution
Step 1 ：Use the properties of logarithms to combine the two logarithms into one. The property we will use is \(\log a - \log b = \log \left(\frac{a}{b}\right)\). So, the formula becomes \(D=10 \log \left(\frac{I}{I_{0}}\right)\).
Step 2 ：Substitute the given intensity ratio into the formula to find the difference in decibel levels. The intensity ratio is \(\frac{I}{I_{0}} = \frac{1}{1,000,000}\).
Step 3 ：Calculate the difference in decibel levels using the formula \(D=10 \log \left(\frac{I}{I_{0}}\right)\).
Step 4 ：The loudness level of the less intense sound is \(\boxed{60}\) decibels smaller.
