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Question
the loudness, l, measured in decibels (db), of a sound intensity, i, measured in watts per square meter, is defined as $l = 10\log\frac{i}{i_0}$, where $i_0 = 10^{-12}$ and is the least intense sound a human ear can hear. brandon is trying to take a nap, and he can barely hear his neighbor mowing the lawn. the sound intensity level that brandon can hear is $10^{-10}$. ahmad, brandons neighbor that lives across the street, is mowing the lawn, and the sound intensity level of the mower is $10^{-4}$. how does brandons sound intensity level compare to ahmads mower?
brandons sound intensity is $\frac{1}{4}$ the level of ahmads mower.
brandons sound intensity is $\frac{1}{8}$ the level of ahmads mower.
brandons sound intensity is 20 times the level of ahmads mower.
brandons sound intensity is 80 times the level of ahmads mower.
Step1: Identify the sound - intensity levels
Brandon's sound intensity $I_{B}=10^{- 10}$, Ahmad's mower sound intensity $I_{A}=10^{-4}$.
Step2: Find the ratio of Brandon's sound intensity to Ahmad's
Calculate the ratio $\frac{I_{B}}{I_{A}}=\frac{10^{-10}}{10^{-4}}$.
Using the rule of exponents $\frac{a^{m}}{a^{n}}=a^{m - n}$, we have $\frac{10^{-10}}{10^{-4}}=10^{-10-(-4)}=10^{-6}=\frac{1}{10^{6}}$. This is incorrect. Let's find the ratio the other way around $\frac{I_{A}}{I_{B}}=\frac{10^{-4}}{10^{-10}}$.
Using the rule $\frac{a^{m}}{a^{n}}=a^{m - n}$, we get $\frac{10^{-4}}{10^{-10}}=10^{-4-(-10)}=10^{6}$. Or $\frac{I_{B}}{I_{A}}=\frac{10^{-10}}{10^{-4}}=\frac{1}{10^{-4 + 10}}=\frac{1}{10^{6}}$.
We can also calculate the decibel levels for each and then compare.
For Brandon: $L_{B}=10\log\frac{I_{B}}{I_{0}}=10\log\frac{10^{-10}}{10^{-12}}=10\log(10^{-10+12})=10\log(10^{2}) = 20$ dB.
For Ahmad: $L_{A}=10\log\frac{I_{A}}{I_{0}}=10\log\frac{10^{-4}}{10^{-12}}=10\log(10^{-4 + 12})=10\log(10^{8})=80$ dB.
The ratio of Brandon's decibel - level to Ahmad's is $\frac{L_{B}}{L_{A}}=\frac{20}{80}=\frac{1}{4}$.
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Brandon's sound intensity is $\frac{1}{4}$ the level of Ahmad's mower.