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I'm working on species area relationships. Basically, the more area you have, the higher the species richness. It's often described by a power function:

$S=c A^z$

The contemporary way to use this power function is to transform S and A to log-scales so you end up with this:

$\log ⁡S = c + z \log A$

Where $c$ and $z$ are fitted constants, $S$ is species richness, and $A$ is area.

The value of $z$ is of my interest. I have been extracting $z$ values from papers to analyze for a project. However, I have run into papers with $z$ values from a semi-log version of this relationship.

$S = c + z \log A$

I am wondering if it's possibly to convert the semilog $z$ to a log-log $z$ without having any of the values of species richness or area? I have talked to my colleagues and they are not sure.

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I don't think you can, and I don't think you should.

The papers you read measured (or somehow obtained) a lot of $A$s and $S$s and then fitted $c$ and $z$ to do the best prediction using the model $S = c + z \log A$. If they have a reason to believe that is the model that needs to be used on their data, then why do you believe you should use your model? I think that on the fundamental level, one of you is making a mistake, and that's why I don't think you should do that.

Now as for why you can't do that. Yes, if you had their $A$s and $S$s you could fit $c$ and $z$ to do the best prediction for your model, but if you don't have neither $A$s nor $S$s there is nothing you can do. Because, knowing only their $c$ and $z$, you can draw a graph for $S$ and $A$. And now you want to approximate that graph by "completely" different graph from your model. It should be intuitively clear that your answer will depend on which region of their graph you want to approximate well.

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If I understand you correctly, in your model you look at the function $$S(A)=cA^z,$$ where $c>0$ and $z$ are some constants. Then you used the logarithm to get $$\log S(z) = \log c + z\log A.$$ In the paper you are reading they use $$\log S(A) = \bar{c} + z\log A$$ for some constant $\bar{c}>0$. The z in both papers are the same? Then, the relationship $$\bar{c}=\log c$$ holds and therefore $$e^{\bar{c}}=c.$$

I am assuming that $\log$ means the natural logarithm.

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  • $\begingroup$ Hi @Toni I forgot to mention these are log10. Does that make a difference for exp()? $\endgroup$
    – Leo Ohyama
    Apr 2, 2020 at 22:30

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