My first guess would be to create a linear interpolation between the points and the grid.
For example, 200, we have that between `100` and `500`, assuming that we have a linear equation to describe this particular piece
$$ y = mx + c $$ we can determine the parameters as follows $$ y(100) = m \cdot x_{100} + C\\\ y(500) = m\cdot x_{500} + C $$ solving for `m` and `c` we find $$ m = \frac{y(500) - y(100)}{x_{500} - x_{100}} = \frac{500-100}{x_{500} - x_{100}}\\\ $$ now we can figure out what `x` has to be to fit to that particular scale $$ \frac{500-100}{x_{500} - x_{100}} = \frac{200-100}{x_{200} - x_{100}} $$ and re-arrange. I have used the fact that the gradient would be constant.
If you have some other scale in mind - i.e. a nonlinear curve to fit through the range then you can apply interpolation to obtain a curve fit, $y=f(x)$ and solve for the inverse of that equation for `x`. Alternatively, you can try to convert to a log scale and see if that achieves what you want.