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Math Help - Log problem

  1. #1
    Junior Member
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    Log problem

    I have  L=10*log(x/I)

    Its a problem about sound levels where I = 10^-12 W/m^2 and L is counted in decibels (db)

    The problem, what is x if L = 65 db

    I know i should take the log of 65 somehow and i think * it by the log of I, but am not really sure

    also in the picture the I has a lowewed 0 on it but i dont know how to right that or even what that means i have only seen raised ^
    Last edited by JBswe; October 5th 2007 at 06:40 AM. Reason: add
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by JBswe View Post
    I have  L=10*log(x/I)

    Its a problem about sound levels where I = 10^-12 W/m^2 and L is counted in decibels (db)

    The problem, what is x if L = 65 db

    I know i should take the log of 65 somehow and i think * it by the log of I, but am not really sure

    also in the picture the I is lowered to 0 but i dont know how to right that or even what that means i have only seen raised ^
    <br />
65=10 \log_{10}(x) - 10 \log_{10}(I) = 10 \log_{10}(x) + 120<br />

    so:

    <br />
-55=10 \log_{10}(x)<br />

    or:

    <br />
x=10^{-5.5} \approx 3.16 \times~10^{-6} \ \rm{W/m^2}<br />

    RonL
    Last edited by CaptainBlack; October 5th 2007 at 07:17 AM.
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  3. #3
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    Hello, JBswe!

    Give: .  L\:=\:10\cdot\log\left(\frac{x}{I_o}\right) . where: I_o \:=\:10^{-12}

    If L = 65, find x.
    The "sub-zero" usually means an initial value.
    So that I_o is some constant, not a variable.


    We have: . 10\cdot\log\left(\frac{x}{10^{-12}}\right) \;=\;65

    Divide by 10: . \log\left(\frac{x}{10^{-12}}\right) \;=\;6.5

    Exponentiate both sides: . \frac{x}{10^{-12}} \;=\;10^{6.5}

    Multiply by 10^{-12}\!:\;\;x \;=\;10^{-12}\cdot10^{6.5}

    Therefore: . x \;=\;10^{-5.5} \;\approx\;0.000003162

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