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Math Help - Decibels Rating

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    Decibels Rating

    Sound A measures 30 decibels and sound B is 5 times as loud as sound A. What is the decibel rating of sound B to the nearest integer?

    Alright so I know the formula for decibels is 10 log (I/I0). But how would I solve this problem? So would sound be be 150 decibels then since it is 5 times as loud as sound A? And what would I do with the decibels afterward to fit it in the equation. The equation seems to only allow inputs such as 10^-16 and such.

    Please help me with this. Thank you very much.
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  2. #2
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    Quote Originally Posted by krzyrice View Post
    Sound A measures 30 decibels and sound B is 5 times as loud as sound A. What is the decibel rating of sound B to the nearest integer?

    Alright so I know the formula for decibels is 10 log (I/I0). But how would I solve this problem? So would sound be be 150 decibels then since it is 5 times as loud as sound A? And what would I do with the decibels afterward to fit it in the equation. The equation seems to only allow inputs such as 10^-16 and such.

    Please help me with this. Thank you very much.
    decibels is a logarithm. the relationship is not linear.

    sound level in dB, L = 10\log\left(\frac{I}{I_0}\right) where I_0 = 10^{-12} watts per square meter ... the sound intensity at the threshold of human hearing.

    for a 30 dB sound level ...

    30 = 10\log\left(\frac{I}{I_0}\right)

    3 = \log\left(\frac{I}{I_0}\right)

    change to an exponential equation ...

    \frac{I}{I_0} = 10^3

    \frac{I}{10^{-12}} = 10^3

    I = 10^{-9} ... the intensity of a 30 dB sound source.


    now increase this intensity by a factor of 5 ...

    I = 5 \cdot 10^{-9}

    find the new sound level in dB ...

    L = 10\log\left(\frac{5 \cdot 10^{-9}}{10^{-12}}\right) \approx 37 dB


    ok ... that's the long way to do it. here is a shortcut.

    first, note that 30 = 10\log\left(\frac{I}{I_0}\right).


    increase the intensity by a factor of 5, and find the new sound level , L ...

    L = 10\log\left(\frac{5I}{I_0}\right)

    using the laws of logs ...

    L = 10\log(5) + 10\log\left(\frac{I}{I_0}\right)

    L = 10\log(5) + 30 \approx 37 db
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  3. #3
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    Thank you so much skeeter for helping me on this!

    Oh by the way is the sound intensity at the threshold of human hearing I0 = 10^-16? Well that is what my book says.
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