Verified Answer :
Answer:
Using logarithmic rule:
\(\log \frac{m}{n} = \log m -\log n\)
\(\log_b b^a = a\)
\(\log_b x = a\) then \(x = b^a\)
Given that,
Sound intensity of a typical vacuum cleaner = 80 dB
Sound intensity of burst of eardrum = 160 dB
Let, I₁ and I₂ b e the intensity of typical vacuum and intensity of eardrum.
Use the formula:
\(1 dB = 10\log_{10} (\frac{I}{I_0})\)
As per the statement
The sound intensity of a typical vacuum cleaner is 80 dB (decibels).
⇒ \(80 dB= 10\log_{10} (\frac{I_1}{10^{-12}})\)
Divide both sides by 10 we have;
\(8 dB= \log_{10} (\frac{I_1}{10^{-12}})\)
Apply the logarithmic rules:
then;
\(8 = \log_{10} I_1 - \log_{10} 10^{-12}\)
⇒\(8 = \log_{10} I_1 + 12\)
Subtract 12 from both sides we have;
\(-4= \log_{10} I_1\)
⇒\(10^{-4} = I_1\) ......[A}
It is also given that: Sound intensity of 160 dB will burst an eardrum.
⇒ \(160 dB= 10\log_{10} (\frac{I_2}{10^{-12}})\)
Divide both sides by 10 we have;
\(16 dB= \log_{10} (\frac{I_2}{10^{-12}})\)
Apply the logarithmic rules:
\(16 = \log_{10} I_2 - \log_{10} 10^{-12}\)
⇒\(16 = \log_{10} I_2 + 12\)
Subtract 12 from both sides we have;
⇒\(4 = \log_{10} I_2\)
⇒\(10^{4} = I_2\) ......[B]
We have to find how many times louder is the sound that will burst an eardrum than the sound of a vacuum cleaner.
\(\dfrac{I_2}{I_1} = \dfrac{10^4}{10^{-4}} = 10^8\)
⇒\(I_2 = 10^8 \times I_1\)
Therefore, \(10^8\) times louder is the sound that will burst an eardrum than the sound of a vacuum cleaner