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Homework Help: Ln simplification

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm doing a physics lab and need to do the uncertainties, and the method I'm using is the natural log method, hich goes like this:
    (equation used was k*(x^a)*(y^b)*(z^c) )
    http://img297.imageshack.us/img297/3663/lnform.jpg [Broken]

    The equation I'm doing:

    http://img175.imageshack.us/img175/4214/40303282.jpg [Broken]

    I'm just wondering how it simplifies (the right most part of the first picture) so that I can take the partial dervs. (The Inside of ln is my actualy equation, calorimetry.)

    2. Relevant equations

    I don't know, otherwise I would apply them!

    3. The attempt at a solution

    It's a pretty straight foward problem, once I see it down once I think I can do it all the time in the future. The main problem I am having is dealing with the things in the denominator.

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 23, 2010 #2
    I would use ln(a/b) = ln(a) - ln(b)
  4. Feb 24, 2010 #3
    But what about the pluses? wouldn't that turn into ln(mhTh + mcTc) - ln(mc - mh), what then? I thought that each variable has to be on its own ( ln(mc), ln(Tc), ln(mh)...) ?
  5. Feb 24, 2010 #4


    Staff: Mentor

    There is no property of logs that allows you to break up the log of a sum or difference. IOW, log(a + b) != log(a) + log(b), and log(a - b) != log(a) - log(b).
  6. Feb 24, 2010 #5
    Ok, so ignore the logs and just start over, how would you get the uncertainty given this equation?
  7. Feb 24, 2010 #6


    Staff: Mentor

    The above is not an equation.
    From your following work, I'm assuming that the equation was f = k*(x^a)*(y^b)*(z^c).
    First thing to do is to take the natural log of both sides:
    ln f = ln(k*(x^a)*(y^b)*(z^c)) = lnk + a*lnx + b*lny + c*lnz

    Now take the total derivative on both sides.
    (1/f)df = 0 + a(1/x)dx + b(1/y)dy + c(1/z)dz

    The expressions in parentheses on the right side are the partials with respect to x, y, and z, respectively. It's now a simple matter to get df.
    Last edited by a moderator: May 4, 2017
  8. Feb 24, 2010 #7
    I know how to do that one, it was an example used by the instructor, I'm trying to do this one ( with the other side of the equation being = ln f) :

    The problem is the addition inside the ln, which prevents me from deconstructing it into a bunch of simple ln's ( ln(mh) + ln(Th) +....) since there are no rules for ln (a+b). Does that mean that the ln method doesn't apply, and I just have to take the total derivative from the outset (i.e. with just f=(mh*Th+mc*Tc)/(mh+mc) instead of ln f = ln ( (mh*Th+mc*Tc)/(mh+mc) )? Do you understand what I'm trying to say?

    Last edited by a moderator: May 4, 2017
  9. Feb 24, 2010 #8


    Staff: Mentor

    I think I understand what you're saying. Just take the total derivative right away.

    BTW, you should post problems like this in the Calculus & Beyond section, not the Precalc section.
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