1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - simplification Date
Simplification of a complex exponential Dec 10, 2017
Simplification of complex expression Nov 9, 2017
Sqrt(xy)/y simplification? Feb 18, 2016
Complex numbers simplification Jan 24, 2016
Simplification question Apr 7, 2015