Could we work with a multiplicative co-ordinate system with $1$ at the origin?

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I thought of a new kind of co-ordinate system which has $1$ at the origin. First, we draw two perpendicular lines intersecting at origin and write $1$ at the origin.

Then, we choose any positive $a>1$ and we write $a$ on the right of $1$ on the $x-axis$. We call the distance between $1$ and $a$ the unit distance. Now, we write the same number $a$ on $y-axis$ too at the same unit distance from $1$ in the upward direction.

Now, this is the process of writing the rest of the numbers on the co-ordinate axes: Any real number $k$ is written at a distance of $\log_ak$ from 1. Clearly the numbers, $a$, $a^2$, $a^3$ are at distances $1$unit, $2$units, $3$ units respectively from $1$ on the right of $x-axis$ and are also in the upward direction from $1$ on the $y-axis$. And the numbers, $\frac{1}{a}$, $\frac{1}{a^2}$, $\frac{1}{a^3}$ are at distances $1$ unit, $2$ units, $3$ units respectively from $1$ on the left of $x-axis$ and are in the downward direction from $1$ on the $y-axis$.

$0$ is located at the left-infinity on the $x-axis$ and also on the downward-infinity on the $y-axis$. There are no negative numbers on the co-ordinate axes.

Clearly, it differs from the normal co-ordinate system in the 'spacing' between the real numbers. In the normal co-ordinate system, the numbers $a$, $2a$, $3a$, etc are equidistant from each other, while in this co-ordinate system, the numbers $a$, $a^2$, $a^3$, etc are equidistant from each other.

The slope of the line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in this co-ordinate system will be: $$\frac{\log_ay_2-\log_ay_1}{\log_ax_2-\log_ax_1}=\frac{\log_a(\frac{y_2}{y_1})}{\log_a(\frac{x_2}{x_1})}=\log_{\left(\frac{x_2}{x_1}\right)}\left(\frac{y_2}{y_1}\right)$$

If we plot the points of $x^n$ in this co-ordinate system, it's graph will be a straight line of slope $n$. I guess that the graph of $e^x$ will be a parabola. Could this co-ordinate system be more convenient for some problems?