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?
 A: 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 $|-1+x|\not<1$.
A: 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.
A: 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 $\sin$.
