Suppose that the minimum possible value is $a$, and the maximum possible value is $b$. We want to map the interval $[a,b]$ to the interval $[0,200]$.
There are a great many possibilities. One of the simplest is a linear mapping, where $x$ is mapped to $f(x)$, with $f(x)=px+q$.
We want $f(a)=0$, so $pa+q=0$. We want $f(b)=200$, so $pb+q=200$.
Now solve the system of equations $pa+q=0$, $\,pb+q=200$ for $p$ and $q$. We get
$$p=\frac{200}{b-a},\qquad q=-\frac{200a}{b-a},$$
and therefore
$$f(x)=\frac{200}{b-a}\left(x-a\right).$$
You may want to modify this so that the scaled values are integers. If so, then a sensible modification is to use, instead of $f(x)$, the function $g(x)$, where $g(x)$ is the nearest integer to $f(x)$.
Added: If you want to have only $N$ values, equally spaced with bottom at $0$ and top at $200$, then $N-1$ should divide $200$. Let $D=\frac{200}{N-1}$. Take the integer nearest to $\frac{f(x)}{D}$, and multiply the result by $D$. For example, if you want $26$ different values $0, 8, 16, 24, \dots, 200$, then $N=26$, so $N-1=25$ and $D=8$. We find the nearest integer to $\frac{f(x)}{8}$ and multiply the result by $8$.
Remark: One important real world example is when you want to convert the temperature $x$, in degrees Fahrenheit, to temperature in degrees Celsius. In the Fahrenheit scale, the freezing point of water is $32$ degrees, and the boiling point is $212$. In degrees Celsius it is $0$ and $100$. So we want to transform the interval $[32,212]$ to the interval $[0,100]$. the relevant $f(x)$ is equal to $\frac{100}{212-32}(x-32)$, which simplifies to $f(x)=\frac{5}{9}(x-32)$.
Suppose that $f(1) = 1, f(10) = 2, f(100) = 3.$ Let's suppose further that you measure position on your paper in centimeters, with the origin being at the origin of your graph.
If you plot $\log(x)$ vs $f(x)$, you'll plot points at $(0cm, 1cm), (1cm, 2cm),$ and $(2cm, 3cm)$.
If, on the other hand, you use the log paper's log-scale on the x-axis, let's suppose that the first "Decade" of the paper starts at the 0cm mark in the horizontal direction, and the leftmost vertical line of this decade is labelled "1", the second decade starts at the 1cm mark (and starts with "10"), and so on. Then for the point $f(10) = 2$, you'll go to $(1cm, 2cm)$; the other points you plot will be at $(0cm, 1cm)$ and $(2cm, 3cm)$.
In short, you'll draw the same three points.
The horizontal axis may be labelled $\log x$ in the sense that the physical distance from the vertical axis really is (up to a constant) the logarithm of the $x$-value that made you plot a point; the lines on the paper (and their labels "1", "10", "100") are just a way for you to easily exponentiate these distances to get the original value of $x$. In that sense, the paper is "taking the log" for you as you plot things: you have $x = 100$, you look for the vertical line labelled 100, and put a point there...and its distance from the $y$-axis turns out to be 2, which is $\log 100$.
Personally, I don't like it. I tend to label the thin vertical lines 1, 10, 100, etc., and label the axis $x$, but then I'm not an engineer or physicist. Maybe they know something I don't about graphs...
Best Answer
The specific function certainly has the formal property you seek, but it's probably not in wide use.
Non-Cartesian coordinate systems, such as log coordinates, are used to highlight mathematical features. For example, power laws become linear in suitable combinations of log and linear coordinates.
Depending on your intent for your graph, there might well be "better" choices of coordinates. For example, you could:
Embed the first quadrant $(0, \infty) \times (0, \infty)$ in three-dimensional space and perform point projection as if looking from a point "above" the third quadrant, toward the origin (diagram).
Construct a conformal map from $(0, \infty) \times (0, \infty)$ to the unit square $(0, 1) \times (0, 1)$ or the quarter-disk.
Offhand, I expect you'll get visually-nicer results by not using a map of the form $(x, y) \mapsto \bigl(f(x), f(y)\bigr)$, but if you do go that route, I'd suggest the hyperbolic tangent $$ f(x) = \tanh x = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}}. $$