# Unit cancellation in integration with natural logarithm?

 P: 87 I was recently doing some problems with Entropy, and came to the startling conclusion of incorrect units for most of my answers! One problem, for example--was to calculate the change in entropy of a given sample as it gained heat per temperature. $$\Delta S=\int^{f}_{i}\frac{dQ_{r}}{T}$$ The dQr is assumed to be relatively reversible for the problem, and you end up solving the problem by recognizing Q as mc(delta)t. $$\Delta S=mc\int^{f}_{i}\frac{dT}{T}$$ Integrating yields mc(ln(Tf)-ln(Ti)), and then you recognize the properties of the natural logarithm and divide Tf by Ti, take the natural log, and finish the problem. The point where I get screwed up logically is that the units don't come out correctly if you don't treat the natural logarithm this way, and simply compute through subtraction. I know its a property of the natural logarithm to divide, and I've gotten stuck on problems like this in the past (simple rocket fuel propulsion examples, etc). My main question is--does this happen every time a natural logarithm is involved after integration? Are there cases where the units don't cancel like this? I know its a silly question, but it has been bugging me all night.
 Sci Advisor HW Helper Thanks P: 26,148 Hi Cvan! Yeah … never really occurred to me before … when we write: $$\int^{f}_{i}\frac{dT}{T}\,=\,\left[logT\right[^{f}_{i}\,,$$ the dT and T have the same units, so dT/T is dimensionless (a scalar), but the logT looks as if it has the same dimensions as T. I suppose the answer is that log (to any base) can only act on dimensionless numbers, so technically one ought to write log(T/c), where c is a constant with the same dimensions as T! For sanity-threatening conundrums like this, I only have two pieces of advice …a) panic! b) don't panic!… they both work … !
 Mentor P: 12,069 Unit cancellation in integration with natural logarithm? It may help to write the difference-in-logs as $$\ln(\frac{T_f}{1 K})-ln(\frac{T_i}{1 K})$$ This gives pure-number arguments for the logs. Also, it does not matter what units you use in the denominators, as long as they're the same, since $$\ln(\frac{T_f}{1 \ Anything})-ln(\frac{T_i}{1 \ Anything}) = \ln\left[(\frac{T_f}{1 \ Anything}) \cdot (\frac{1 \ Anything}{T_i})\right] = \ln(\frac{T_f}{T_i})$$ independent of what "1 Anything" actually is.