Unify principles of two formulae

[Twinbee]

What's the most mathematical elegant way of converting the numbers on the left (x) to the numbers on the right (y). Note that the numbers don't have to be exactly the same, as long as the property of how the numbers act is left intact - so that numbers (x) closer to 1 act as the formula 1/(1-x) = y, while numbers (x) closer to 0 act as the formula that twice as many zeros equals half the amount of y:

In other words, I want to unify these formula to act as one 'master formula'. So there you go, convert x to y:

0.00000000000000000000000000000001 --> 0.0125
0.0000000000000001 --> 0.025
0.00000001 --> 0.05
0.0001 --> 0.1
0.01 --> 0.2
0.1 --> 0.4
0.5 --> 0.75
0.9 --> 4
0.99 --> 40
0.999 --> 400
0.9999 --> 4000
0.99999 --> 40000

 

[Twinbee]

Just managed to solve it. For those who don't want to see the answer, it's below.
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1/logBase2(1/x) = y

 

[HallsofIvy]

Any "mathematically elegant" way of solving such a problem would start by specifying exactly what "unify principles" means and what "act as the formula that twice as many zeros equals half the amount of y" means. I think once you had done that, the formula would be apparent.

 

[Twinbee]

"Mathematically elegant" in this context would be perhaps the simplest/smallest way of converting between the two.

The solution I found seems to be the best bet as you can see.

I'll rephrase "act as the formula that twice as many zeros equals half the amount of y".

to:

"so that twice as many decimal places in x translates to a geometric halving in y". If that still doesn't make it clear, then you should see the pattern after a quick scan of the numbers in my first post.

===========================

EDIT:

a: oops, you didn't ask me what "mathematically elegant" was anyway.

b: I'll be honest, and say that in place of "numbers (x) closer to 0 act as the formula that twice as many zeros equals half the amount of y", I originally had 1/logBase2(1/x) = y. It only ocurred to me afterwards that this formula works for the whole range anyway, and is thus the solution.