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Math Help - Log Applications

  1. #1
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    Log Applications

    Hey I have 3 questions to ask and I've done both of them, i just want to know if I did it right or wrong.

    1. On September 26, 2001, an earthquake in North Bay measured 5.0 on the Richter scale. What is the magnitude of an earthquake 3 times as intense as North Bay's earthquake?


    My work:

    The answer I got was 5.477


    2. How much louder is a mp3 player (100dB) to a crying baby (80dB)?

    My work:

    The answer I got was approx 10 (9.69)


    3. A liquid has a pH of 3.46. Find the hydrogen ion concentration [H+].


    My Work

    The answer I got was 10^-3.64 which is .000229

    *

    Basically the work for this process I did:

    PH = -logH^+
    H^+ = 10^-3.64
    H^+ = .000229


    So are these answers right? For the last one did I show the correct process? Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by agent2421 View Post
    Hey I have 3 questions to ask and I've done both of them, i just want to know if I did it right or wrong.

    1. On September 26, 2001, an earthquake in North Bay measured 5.0 on the Richter scale. What is the magnitude of an earthquake 3 times as intense as North Bay's earthquake?


    My work:

    The answer I got was 5.477


    2. How much louder is a mp3 player (100dB) to a crying baby (80dB)?

    My work:

    The answer I got was approx 10 (9.69)


    3. A liquid has a pH of 3.46. Find the hydrogen ion concentration [H+].


    My Work

    The answer I got was 10^-3.64 which is .000229

    *

    Basically the work for this process I did:

    PH = -logH^+
    H^+ = 10^-3.64
    H^+ = .000229


    So are these answers right? For the last one did I show the correct process? Thanks
    1. is correct but you don't tell us how you got there.

    2. is incorrect, but again you don't tell us your reasoning.

    3. looks OK

    CB
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  3. #3
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    Thanks... yeah it was too much work to type up the whole thing so I didn't show it.... Can you tell me how to do #2 though? If you want I cant post w/e work I had, just I thought it'd be more simple like this.
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  4. #4
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    can anyone help me out for the 2nd question please?
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by agent2421 View Post
    can anyone help me out for the 2nd question please?
    A difference of 20\text{dB} in sound level is a difference of 100 in intensity, sound level:

    SL=10\log_{10}(I) + K

    where K is related to the reference sound intensity.

    CB
    Last edited by CaptainBlack; November 1st 2009 at 09:19 AM.
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  6. #6
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    why do you do 10log (Base 10) though? From the opening question where do we get that info from? I know the diff between 80 DB and 100 DB is 20 dB but I'm not sure why you have 10 log (base 10) (I) + K....
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by agent2421 View Post
    why do you do 10log (Base 10) though? From the opening question where do we get that info from? I know the diff between 80 DB and 100 DB is 20 dB but I'm not sure why you have 10 log (base 10) (I) + K....

    To do these you are supposed to know something about the Richter scale, the relationship between a sound level in dB and intensity and the relationship between pH and concentration of H+ ions.

    A sound level S is related to the intensity by:

    S=10\;\log_{10}\left[\frac{I}{I_{ref}}\right]=10\;\log_{10}(I) - 10\;\log_{10}(I_{ref})

    where I_{ref} is the reference intensity for the sound level type in question (the reference levels used in air and underwater are different).

    Now if two sound levels differ by 20 dB we have:

    20=S_1-S_2=10\;\log_{10}(I_1)-10\;\log_{10}(I_2)=10\;\log_{10}\left[\frac{I_1}{I_2}\right]

    so:

    2=\log_{10}\left[\frac{I_1}{I_2}\right]

    or:

    10^2=\frac{I_1}{I_2}

    But the definition of sound level in dB in terms of intensities will be (should) in your notes or text book somewhere.

    (how much louder means how much greater in the sound intensity in the two cases)

    CB
    Last edited by CaptainBlack; November 1st 2009 at 09:28 AM.
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