I have a measure that uses radiation dose (M.U. $Gray$) and $\log(Volume)$. The measure is $[\frac{Dose}{\log(Volume)}]$ that is $[\frac{D}{\log(l^3)}]$ with $D$ as radiation dose (M.U. unit is Gray) and $l$ is length (M.U. in international measure units is meter). What is the correct measure unit for this kind of measure after the following steps $[\frac{D}{\log(l^3)}]\Leftrightarrow [\frac{D}{3\log(l)}]\Rightarrow [\frac{D}{\log(l)}]$ that is, what is the transformation of logarithm of length $\log(l)$?
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3$\begingroup$ Logarithms only make sense on dimensionless quantities. Usually there is an implied unit, in this case $\text{m}^3$. $\endgroup$– DanielSankCommented Nov 3, 2015 at 0:13
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$\begingroup$ Possible duplicates: physics.stackexchange.com/q/13060/2451 , physics.stackexchange.com/q/109995/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Nov 3, 2015 at 0:42
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2 Answers
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You cannot meaningfully evaluate the logarithm of a dimensionful number. The reason for this should become clear if you attempt to series expand a logarithm with a dimensionful argument.
The usual way to get around this is to write something like:
$$\frac{\frac{D}{1\,{\rm Gray}}}{\log_{10}(\frac{V}{{1\,\rm cm}^3})} = 1.23$$
Some people prefer to omit the $1$'s, but this is a matter of style. Either way, it makes the dimensions of all relevant quantities clear.
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You have to take the log of a dimensionless quantity. I assume you have a model for the relationhship between $D$ and $V$ that looks like $D=m \log V + b$, but since you can only take the log of a dimensionless quantity, this should really be $D=m \log \left(V/V_0\right) + b$. Typically $V_0$ is just $1$ in some choice of units, so for example, $V_0 = 1 m^3$ or $V_0 = 1 cm^3$. Notice that a change in $V_0$ just shifts the log by a constant so it will not affect the value of $m$, only the value of $b$. Thus your value for $m$ does not care about the units of $V$, and it will only contain Grays. So the units for $m$ is Gray.
However, this is not quite the whole story. The value for $m$ will depend on what the base of the logarithm is: if you take the natural log, you will get a different number for $m$ than if you take the base ten log. Therefore it makes sense to indicate the choice that was made when you give a value for $m$. So you could say that $m$ is $5$ Gray/decade of volume, or $1.5$ Gray/octave of volume. But technically, the number of decades is dimensionless because it is just a number, so you do get that the units for $m$ are Grays.
