f(0.5) = c

you can solve the equation

b = (1 - 2 * c)/Math.pow(c, 2)

which gives

a = Math.log((1 - 2 * c)/Math.pow(c, 2) + 1)

and the final formula:

(Math.exp(Math.log((1 - 2 * c)/Math.pow(c, 2) + 1) * x) - 1) / ((1 - 2 * c)/Math.pow(c, 2))

I advise strongly to precalculate a and b for a given c so you can use the simpler and faster formula:

(Math.exp(a * x) - 1) / b.

log16 = Math.log(16) // same as the magic constant used above

var scaledValue = (Math.exp(log16 * x -1) /15 * MAX_WIDTH);
var unscaledValue = log16 * (Math.log(scaledValue * 15 /MAX_WIDTH + 1))

unscaledValue then fuzzy-equals x.


$x = 1;
$min = 5;
$max = 1000;
print f($x, $min, $max);

function f($x, $min, $max) {
return ((($max - $min) * (exp($x) - 1) / (exp(1) - 1)) + $min);
}


function step2euro(step) {
if (step < 1) return 1;
if (step < 101) return step * 100;
if (step < 201) return (step – 100) * 1000;
if (step < 301) return (step – 200) * 10000;
if (step < 401) return (step – 300) * 100000;
return (step – 400) * 1000000;
}

is it possible to make a range slider for big values?

for example a price selector:

1 euro to 100 mio euro
1 – 10’000 = hundred steps
10’000 – 100’000 = thousend steps
100’000 – 1 mio = 10k steps
1 mio – 10 mio = 100k steps
10 mio – 100 mio = 1m steps

Couldn’t you please show some live example with this formula in use? thanks!