1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    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!

B Ln() & SI Units

  1. May 3, 2017 #1
    I tried with Google but I couldn't find anything, so here goes: When I "use ln on a quantity" (I don't really know how to phrase it in english, as we just have a verb for it), say, I have n = 0.00149 kg/m*s, and I put it into the ln, so now I have ln(0.00149 kg/m*s) what happens to the SI Units? The result of 0.00149 ~-6.508, but I'm not sure on the kg/m*s. It never came up during HS so I now have to fill a board with the ln of various values of n, and I'm not sure what to do with SI.

    Any help is appreciated!
    Last edited: May 3, 2017
  2. jcsd
  3. May 3, 2017 #2


    User Avatar
    2017 Award

    Staff: Mentor

    The unwritten sign between ##0.00149## and ##\frac{kg}{m \cdot s}## is a multiplication. So ##\ln n \approx -6.51 + \ln kg - \ln m - \ln s## which can hardly be interpreted and thus raises the question: what do you want to express and what's the goal? What should ##\ln n## stand for? If it is only a scaling for some plot, then the units remain as they are, as only the graphic representation of the magnitude of ##n## changes, not the quantity.
  4. May 3, 2017 #3
    Yeah, the first thing that popped to my mind was the classic ln(a*b) = lna + lnb as well.

    In my case n is the viscosity index of a liquid (alcoholic, specifically). It's not used in any formula or anything, we just have to fill this board (it's for Lab), and for each n, we need the ln. I was just curious whether there was some "rule" about such cases.
  5. May 3, 2017 #4


    User Avatar
    Science Advisor

    Normally, you would want to convert to some dimensionless value before you take the logarithm. This could be done by dividing by some arbitrary constant, which you could call n0.
  6. May 3, 2017 #5
    Well, thanks a lot for the help everyone! I appreciate it.
  7. May 3, 2017 #6
    FYI, this is actually one way of double checking the validity of your calculations. If you suddenly find yourself taking the square root of for example 5kg, that very often shows something went wrong somewhere.
  8. May 5, 2017 #7
    Thanks for the info, I'll keep it in mind!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted