I'm working on a meta-analysis of data extracted from scientific literature. One paper reported their data as $ln(mm^2)$ (see panel F in the figure below from https://doi.org/10.1007/s11258-009-9695-z). I need to convert this to raw $cm^2$ as they were originally measured.

How should I do this conversion?

enter image description here

  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Sep 27 at 0:16
  • $\begingroup$ Thanks for showing me where things are unclear. I've tried to edit my question and include the original graph to make it more clear. $\endgroup$
    – Rose
    Sep 27 at 0:31
  • $\begingroup$ This question has been asked previously on Physics SE. $\endgroup$
    – Jam
    Sep 27 at 19:28

2 Answers 2


It does not make sense to take $\ln$ of a dimension; i.e. the argument of $\ln$ should be dimensionless. That said, if we are considering the values $y = \ln \left( \frac{x}{1 \text{ mm}^2} \right)$, then note that $x = (1 \text{ mm}^2) e^y$. And since $1 \text{ mm}^2 = 10^{-2} \text{ cm}^2$, then $x = (10^{-2} \text{ cm}^2) e^y$.


Let $A_m$ be the area in $mm^2$ and $A_c$ be the area in $cm^2$. We have $A_c = A_m / 100$.

We want a formula for $A_c$ given $x = ln(A_m)$. We have $e^x= A_m$, so $A_c = A_m / 100 = \frac{e^x}{100}$.


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