Dimensions in the logarithmic form of the Arrhenius equation

Question

The logarithmic form of the Arrhenius equation is:

$\displaystyle\ln k=\ln A-\frac{E_a}{RT}$

Here $k$ and $A$ have dimensions whereas $\displaystyle\frac{E_a}{RT}$ is dimensionless. In other words, $k$ and $A$ have dimensions whereas $\ln k$ and $\ln A$ are dimensionless. How can this be?

Possible duplicate of What is the logarithm of a kilometer? Is it a dimensionless number? — John Rennie, Jul 21 '19 at 05:16
Hi Thomas. See the question I've linked for a discussion on this subject. — John Rennie, Jul 21 '19 at 05:17
How can this be? Because $\ln{k}-\ln{A}=\ln{(k/A)}$ and this is the logarithm of a dimensionless quantity. — G. Smith, Jul 21 '19 at 05:25

score 3 · Accepted Answer · answered Jul 21 '19 at 10:27

As pointed out by @G. Smith in the comments, your logarithms are all dimensionless function of dimensionless quantities.
The initial equation is in fact consistent only if $k$ and $A$ share the same unit (which is, in this case, $s^{-1}$); they can thus both be rewritten as $k = \bar k u$ and $A = \bar A u$, where $ u = 1 s^{-1}$ and $ \bar k$,$\bar A$ are dimensionless quantities.
Therefore, using the properties of logarithms, $$ \log k - \log A = \log \bar k - \log \bar A$$ and your logarithms are dimensionless.

Dimensions in the logarithmic form of the Arrhenius equation

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