SOLUTION: The decible level of sound is given by D=10 log (I/10^-12, where I is the sound intensity measured in watts per square meter. Find the decible level of a whisper at an intensity of

Algebra ->  College  -> Linear Algebra -> SOLUTION: The decible level of sound is given by D=10 log (I/10^-12, where I is the sound intensity measured in watts per square meter. Find the decible level of a whisper at an intensity of      Log On


   

Question 72681: The decible level of sound is given by D=10 log (I/10^-12, where I is the sound intensity measured in watts per square meter. Find the decible level of a whisper at an intensity of 5.4 ? 10-10 watts per meter.
Does anyone understand this? If so please break it down for me. Thanks to anyone who can help.
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Since nobody else has taken a shot at this, I'll give it a go ...
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dB+=+10%2Alog%28%28I%2F10%5E%28-12%29%29%29
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is the equation that governs the relationship between sound intensity and decibels.
.
I think what you meant for the given sound intensity was 5.4%2A10%5E%28-10%29
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The units of your given intensity and the units of the intensity in the decibel equation
are consistent, so we can just substitute 5.4%2A10%5E%28-10%29 for I in the decibel
equation without having to make any conversion in units. With this substitution the decibel
equation becomes:
.
dB+=+10%2Alog%28%285.4%2A10%5E%28-10%29%2F10%5E%28-12%29%29%29
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We can now divide the 10%5E%28-12%29 of the denominator into the 10%5E%28-10%29 of the
denominator. Recall that when you divide, you subtract exponents. The subtraction
of the exponents involves %28-10%29-%28-12%29 and this reduces to -10%2B12+=2. Therefore,
the decibel equation becomes:
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dB+=+10%2Alog%28%285.4%2A10%5E%282%29%29%29
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The rules of logarithms say that if you have the log of a product, you can split this into
the sum of the logs of each of the terms in the product. Applying this rule gives us:
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dB+=+10%2A%28log%28%285.4%29%29+%2B+log%28%2810%5E%282%29%29%29%29
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The rule of exponents for logarithms says that an exponent comes out as the multiplier
of the log. So log+%28%2810%5E%282%29%29%29 becomes 2%2Alog+%28%2810%29%29. But log%28%2810%29%29 is simply 1,
and therefore, log%28%2810%5E%282%29%29%29 reduces just to 2. Substitute this back into the
decibel equation and you get:
.
dB+=+10%2A%28log%28%285.4%29%29+%2B+2%29
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Finding the log%28%285.4%29%29 is just a calculator problem. It is 0.732393759 or just round it
to 0.732 which is probably close enough for your needs.
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Substituting this value makes the equation:
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dB+=+10%2A%280.732+%2B+2%29+=+10%2A2.732+=+27.32+
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So the answer appears to be 27.32 dB
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Hope you could follow this through and that it helps you to understand this power ratio
equation.