# Units of a log of a physical quantity

So I have never actually found a good answer or even a good resource which discusses this so I appeal to experts here at stack exchange because this problem came up again today. What happens to the units of a physical quantity after I take its (natural) logarithm. Suppose I am working with some measured data and the units are Volts. Then I want to plot the time series on a log-scale, only the ordinate is on the log scale, not the abscissa. So the x-axis is definitely in time (seconds let's say) but what are the units on the y-axis? Will it be Volts or log(Volts) or something? If I square the quantities, then the units are squared too so what if I take the log? A rationale in addition to the answer will be appreciated as well.

I guess whatever the answer, the same goes for taking the exponential or sine of the data too, right?

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The units remain the same, you are just scaling the axes. As an analogy, plotting a quantity on a polar chart doesn't change the quantities, it just 'warps' the display in some useful way. However, some quantities are 'naturally' expressed as logs (dB, for example), but these are always dimensional quantities (sometimes implicitly referenced to a known quantity). – copper.hat Nov 16 '12 at 2:45
If $x = 0.5$ is measured in some units, say, seconds, then taking the log actually means $\ln (0.5s/1s) = \ln (0.5)$. Hope this helps. – glebovg Nov 16 '12 at 2:51
Prince Ali, you wrote "same goes for taking the exponential or sine of the data too, right?" - what about taking the square as you mention - the square of length is area, which is not length. – alancalvitti Nov 16 '12 at 3:22

Overall, the argument $x$ of $\ln(x)$ must be unitless, and a log transformed quantity must be unitless. If $x = 0.5$ is measured in some units, say, seconds, then taking the log actually means $\ln(0.5s/1s) = \ln(0.5)$. See this for more information about other transcendental functions. Hope this helps.
The paper seems interesting, but I can't read it (not willing to pay $35 for it at the moment). Can you give a 25 words or less summary of their main argument? I read the abstract but they don't cover the rationale there. – Robert Dodier Mar 11 '14 at 20:04 @RobertDodier I found the aforementioned article on Matta's (one of the authors) website. – glebovg Mar 13 '14 at 0:14 @glebovg Thanks for the link. I read the paper, but I have to admit I'm doubtful about their conclusions. They don't have any kind of an argument as to exactly why transcendental functions must have dimensionless arguments; simply repeating the conclusion over and over isn't an argument. Stuff like log(10 meters) = log(10) + log(meters) doesn't make sense, they claim, but that's not a mathematical attitude, right? The mathematical approach would be to find a consistent interpretation of apparently-nonsensical expressions, or prove there cannot be one. Simply giving up isn't proof of anything. – Robert Dodier Mar 13 '14 at 1:21 Logarithm of a quantity really only makes sense if the quantity is dimensionless, and then the result is also a dimensionless number. So what you really plot is not$\log(y)$but$\log(y/y_0)$where$y_0$is some reference quantity in the same units as$y$(in this case$y_0 = $1 Volt). Similarly for$\exp$and$\sin$. - In the expression$\ln{x}$,$x$must be unitless. This is because the log function is a series with x raised to differing powers. For instance: $$-\sum_{k=1}^{\infty}{\frac{(-1)^k(-1+x)^k}{k}}$$ for $$|-1 + x| < 1$$ Let's say$x$had units of meters (for example). Then the first term in the series would have units of meters, the second term units of square meters, 3rd term in cubic meters, etc. You can't add quantities with differing powers of units, thus$x$must be unit-less. The same argument applies for$|-1+x|\not<1$. - These are all incorrect. We teach students at an early age to multiply physical quantities with units as follows:$ 2\,\mathrm{meters} * 2\,\mathrm{meters} = 4 \,\mathrm{meters}^2 $. So, we syntactically multiple both the numeric value ad the units. So, same also applies to exponents.$ (2\,\mathrm{meters})^3 = 2^3\,\mathrm{meters}^3 $. Right? Since logarithm is the inverse of the exponent, it MUST work for units also. So,$\log_{10}(270\,\mathrm{meters}) = \log_{10}(270) \log_{10}(\mathrm{meters}) \approx 2.4314 \log_{10}(\mathrm{meters})$This then works when the inverse is applied$ 10^{2.4314 \log_{10}(\mathrm{meters})} = 270\,\mathrm{meters}$To do anything else with the representation of log-units would make the exponent not work as the inverse function and we all know that 3 meters cubed is 27 cubic meters. - You reasoning is flawed. Different units can be multiplied/divided but they cannot be added/subtracted. For example, you can divide meters by seconds and get meters per seconds but you cannot add seconds+meters. That is why taking the log of a unit doesn't make sense because log(1+x) = x-x^2/2+x^3/3... So how can you add sec-sec^2+sec^3...? – Fixed Point Jul 2 '14 at 7:04 -1 - This reasoning does not work -$(a b)^c$distributes to$a^c b^c$, but$log(a*b)$does not distribute to$\log(a) \log(b)\$ – Eric Oct 30 '14 at 15:45