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Thread: Log problem

  1. #1
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    Log problem

    I have $\displaystyle L=10*log(x/I)$

    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; Oct 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 $\displaystyle L=10*log(x/I)$

    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 ^
    $\displaystyle
    65=10 \log_{10}(x) - 10 \log_{10}(I) = 10 \log_{10}(x) + 120
    $

    so:

    $\displaystyle
    -55=10 \log_{10}(x)
    $

    or:

    $\displaystyle
    x=10^{-5.5} \approx 3.16 \times~10^{-6} \ \rm{W/m^2}
    $

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

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

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


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

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

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

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

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

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