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Math Help - Mass Bias Calculation

  1. #1
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    Question Mass Bias Calculation

    Hi I need some help just reorganizing this equation for mass bias:

    Robs = Rtrue(1+e)^m

    I need the equation rearranged for "e" the mass bias factor. It seems to get tricky with logarithms and I am not sure if I have it correct or not

    Any help would be much appreciated
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  2. #2
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    Well, how can we tell you if you are right if you don't present anything.

    Logarithms are not tricky. Stop thinking that.

    R_{obs} = R_{true}(1+e)^{m}

    log(R_{obs}) = log(R_{true}(1+e)^{m})

    log(R_{obs}) = log(R_{true}) + log((1+e)^{m})

    log(R_{obs}) = log(R_{true}) + m \cdot log(1+e)

    log(R_{obs}) - log(R_{true}) = m \cdot log(1+e)

    log(R_{obs}/R_{true}) = m \cdot log(1+e)

    \frac{1}{m}log\left(\frac{R_{obs}}{R_{true}}\right  ) = log(1+e)

    You're not going to make me do ALL the work, are you?
    Last edited by TKHunny; January 7th 2010 at 08:46 PM.
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  3. #3
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    My apologies, whether you believe me or not I actually got to that part of the equation. What I can't figure out is how to get rid of the logarithms now. I'll show you what I have as my temporary solution:

    Robs = Rtrue(1+e)^m
    Robs/Rtrue = (1+e)^m
    log (Robs/Rtrue) = log(1+e)^m
    log (Robs/Rtrue) = (m)log(1+e)
    log (Robs/Rtrue)/m = log(1+e)

    Now this is where my confidence ends, to get the "e" out of the logarithm, I believe you need to put the logarithm to the base of 10:

    so log(1+e) becomes 10^log(1+e), which of course is just 1+e but then you need to do the same to the other side which is where I start to be not so sure:

    10^log(Robs/Rtrue)/m = 10^log(1+e)
    e = 10^log(Robs/Rtrue)/m - 1
    e = ((Robs/Rtrue)^m-1) - 1

    I am really not sure that's right but I can't seem to figure out what to do with the other side once you raise it to the 10.

    If you can help me out with this I would seriously appreciate it, thanks. And thank you for the last part it has been a long time since I did logarithms so I am glad I actually did it correctly.
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  4. #4
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    \frac{1}{m}log\left(\frac{R_{obs}}{R_{true}}\right  ) = log(1+e)


    log\left(\left[\frac{R_{obs}}{R_{true}}\right]^{\frac{1}{m}}\right) = log(1+e)


    \left[\frac{R_{obs}}{R_{true}}\right]^{\frac{1}{m}} = 1+e

    Notice, please, how this demonstration did NOT require me to disclose the base of the logarithms I used.

    log(a^b) = b*log(a) Is works both ways.

    log(a*b) = log(a) + log(b) It works both ways.
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