# Convert ln(mm^2) to linear cm^2 [closed]

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? • 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.
– Community Bot
Sep 27 at 0:16
• 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.
– Rose
Sep 27 at 0:31
• This question has been asked previously on Physics SE.
– Jam
Sep 27 at 19:28

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 $$A_c = A_m / 100 = \frac{e^x}{100}$$.