# Ln() & SI Units

• B
Const@ntine
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:

Mentor
2022 Award
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.

Const@ntine
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.
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.

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.

berkeman and fresh_42
Const@ntine
Well, thanks a lot for the help everyone! I appreciate it.

rumborak
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.

berkeman and anorlunda
Const@ntine
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.
Thanks for the info, I'll keep it in mind!