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A method I tried in scaling this range is this common:

$(100 - 0) * (x-\min)/(\max - \min)$

If we take $1\times 10^{-9}$ which is somehow in the middle it has to be mapped to $50$ (right?) but it gives me $0.000999$. Anything I'm missing here? Or there is some kind of special methods for such ranges?

To give you more info, these are pressure readings, examples can be: $2.7\times 10^{-10}$, $5.27\times 10^{-8}$, etc. with max two digits precision.

lioness99a
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3bdalla
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    $10^9$ is not in the middle of $[10^{-12},10^{-6}]$. The middle point of that interval is in fact $$\frac{10^{-6}+10^{-12}}{2}=\frac{10^{-12}\cdot 1000001}{2}\approx 5\cdot 10^{-7}=500\cdot 10^{-9}$$ The thing you are missing is that the average of two numbers is completely different fom the average of their logarithms. –  Mar 01 '17 at 11:07
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    As for "special methods", it depends on what you want to accomplish wih these "pressure readings". For instance, you can keep "two digits precision", but (5.27E-8)-(2.7E-10) = 5.243E-8, which, well, kind of negates the purpose of calling two digits "precision". –  Mar 01 '17 at 11:15
  • In short words I want to map any value in $[1\times 10^{-12}, 1\times 10^{-6}]$ to a value in the range [0, 100] – 3bdalla Mar 01 '17 at 12:24

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