College Algebra, Chapter 5, 5.5, Section 5.5, Problem 40

A power mower has the noise of 106 dB. The noise level at a rock concert was measured at 120 dB. Determine the ratio of the intensity of the rock music to that of the power mower.



Recall that the formula for Sound Intensity in dB is represented as

$\displaystyle B = 10 \log \frac{I}{I_0}$

where

$B$ = intensity level in dB

$I$ = Sound intensity

$I_0 = 10^{-12} W/m^2$ reference intensity

If we let $I_1$ be the intensity of the rock concert and $I_2$ be the sound intensity of the power mower, then we need to find $\displaystyle \frac{I_1}{I_2}$, so

@ rock concert


$
\begin{equation}
\begin{aligned}

120 =& 10 \log \frac{I_1}{I_0}
\\
\\
12 =& \log \frac{I_1}{I_0}
\\
\\
10^{12} =& \frac{I_1}{I_0}
\\
\\
I_0 =& \frac{I_0}{10^{12}}
\qquad \text{Equation 1}

\end{aligned}
\end{equation}
$


@ power mower


$
\begin{equation}
\begin{aligned}

106 =& 10 \log \frac{I_2}{I_0}
\\
\\
10.6 =& \log \frac{I_2}{I_0}
\\
\\
10^{10.6} =& \frac{I_2}{I_0}
\\
\\
I_0 =& \frac{I_2}{10^{10.6}}
\qquad \text{Equation 2}

\end{aligned}
\end{equation}
$


By using equations 1 and 2


$
\begin{equation}
\begin{aligned}

\frac{I_1}{10^{12}} =& \frac{I_2}{10^{10.6}}
&& \text{Multiply each side by $10^{12}$ and divide both sides by } I_2
\\
\\
\frac{I_1}{I_2} =& \frac{10^{12}}{10^{10.6}} = 25.12
&&

\end{aligned}
\end{equation}
$