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Question 969599: The decibel level of a sound with intensity x watts per square centimeter is given by the formula D= 10 log(10^16x).
a) Normal conversation is about 60 decibels. What is the sound intensity of normal conversation?
b) A loud car is 85 decibels. How many times more intense is the loud car than normal conversation?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Let's start with 60 db
60=10 log(10^16x)
Divide both sides by 10
6=log (10^16x)
raise each to the power of 10. That removes the log. Furthermore, 10 ^log x=x. e^ln x=x. They are opposites just like multiplication and division are. Therefore, 10^{log 10^16x}=10^16x
10^6=10^16x
Divide by 10^16.
(10^6/10^16)=x
x= 10^(-10) because when you divide, you subtract exponents. 6-16=-10
x=10^(-10) watts/sq meter.
For 85 decibels
85=10 log (10^16x)
Divide by 10
8.5=log (10^16x)
10^8.5=10^16x. Don't try to get a specific number for 10^8.5 yet.
Divide by 10^16, both sides
x=10^(-7.5)
Still don't get a specific number. Leave it as is for now.
The louder is divided by the smaller, so we divide 10^(-7.5)/ 10^(-10)
Moving the negative exponent to the top makes it positive, so (-7.5+10)=2.5
The louder is 10^2.5 times louder than the smaller. That answer is not unreasonable in that form, but most would want a specific ratio, not a power of 10. NOW you can raise 10^2.5 to get 316.2, which is how much louder 85 dB is than 60 dB.
Notice that decibels from 60 to 85 will be 10^[(85-60)/10] times louder.
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