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Suppose you have some physical quantity $x$ of dimension $l$. We all know that the dimension of $x^2$, for example, will be $l^2$, and that of $\dfrac{1}{x}$ is $l^{-1}$. However, what will be the dimensions of $ln(x)$ and $e^x$ ?

I think I've never encountered such a case before, wherever I found $ln(x)$ or $e^x$, $x$ had no dimensions. I don't know if we can ever have a case where $x$ possesses a dimension.

Supposing this case exists, I may be able to prove that $ln(x)$ has no dimensions, as $d(ln(x))=\dfrac{dx}{x}$ which is a dimensionless quantity for any $x$, but I can prove nothing for exponentials, and I'm not even sure that this is a valid argument. However, there is a different problem with exponentials:

$e^x.e^y=e^{x+y}$, how can you sum $x$ and $y$ if they have different dimensions ?

Tofi
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    Often in the case of $e^x$, the derivation shows it's really $e^{kx}$, where the units of $k$ and $x$ cancel out and $kx$ is then dimensionless. But if $k$ is unity, then you won't see that. – Gert Apr 27 '16 at 14:03

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