the loudness level of a sound can be expressed by comparing the sounds intensity to the intensity of a sound…

the loudness level of a sound can be expressed by comparing the sounds intensity to the intensity of a sound barely audible to the human ear. the formula d = 10(log i - log i₀) describes the loudness level of a sound, d, in decibels, where i is the intensity of the sound, in watts per meter², and i₀ is the intensity of a sound barely audible to the human ear. use this information to answer parts (a) and (b) below.\na. express the formula so that the expression in parentheses is written as a single - logarithm.\nd = 10 log(\\frac{i}{i₀}) (use parentheses to clearly denote the argument of each function.)\nb. use the form of the formula from part (a) to answer this question. if a sound has an intensity 10,000 times the intensity of a softer sound, how much larger on the decibel scale is the loudness level of the more intense sound?\ndecibels louder\n(type a whole number.)

the loudness level of a sound can be expressed by comparing the sounds intensity to the intensity of a sound barely audible to the human ear. the formula d = 10(log i - log i₀) describes the loudness level of a sound, d, in decibels, where i is the intensity of the sound, in watts per meter², and i₀ is the intensity of a sound barely audible to the human ear. use this information to answer parts (a) and (b) below.\na. express the formula so that the expression in parentheses is written as a single - logarithm.\nd = 10 log(\\frac{i}{i₀}) (use parentheses to clearly denote the argument of each function.)\nb. use the form of the formula from part (a) to answer this question. if a sound has an intensity 10,000 times the intensity of a softer sound, how much larger on the decibel scale is the loudness level of the more intense sound?\ndecibels louder\n(type a whole number.)

Answer

Explanation:

Step1: Recall logarithm property

We know that $\log a-\log b=\log\frac{a}{b}$. Given $D = 10(\log I-\log I_0)$, by this property, we can rewrite it as $D = 10\log\left(\frac{I}{I_0}\right)$.

Step2: Solve part (b)

Let the intensity of the softer sound be $I_1$ and its loud - ness level be $D_1 = 10\log\left(\frac{I_1}{I_0}\right)$. The intensity of the more intense sound is $I_2=10000I_1$, and its loud - ness level is $D_2 = 10\log\left(\frac{I_2}{I_0}\right)=10\log\left(\frac{10000I_1}{I_0}\right)$. Using the property $\log(ab)=\log a+\log b$, we have $D_2 = 10\left(\log10000+\log\left(\frac{I_1}{I_0}\right)\right)$. Since $\log10000 = \log10^4=4$, then $D_2=10\times4 + 10\log\left(\frac{I_1}{I_0}\right)$. The difference $\Delta D=D_2 - D_1$. Substituting the expressions for $D_1$ and $D_2$: [ \begin{align*} \Delta D&=(10\times4 + 10\log\left(\frac{I_1}{I_0}\right))-10\log\left(\frac{I_1}{I_0}\right)\ &=40 \end{align*} ]

Answer:

40