SI units and prefixed points

Question. A liter is smaller than a cubic meter. But a milli-liter is larger than a cubic milli-meter — the prefix changes which one is larger. Is there a prefix which makes them equal? In other words, can you find a prefix "pre-" such that one pre-liter is equal to one cubic pre-meter?

I found this puzzle difficult to even set up at first, because I'm still learning how to think about unit conversions. I solved it eventually, though, and found some interesting algebraic insights. The general solution follows.

General solution.

In the general setup, you have units A (e.g. liters) and B (e.g. meters).
The units may have different dimensions (e.g. volume and length), so they need dimensional exponents to make them commensurate. Call the dimensional exponents r and s; they ensure that A^r is commensurate with B^s. (e.g. liters L¹ are commensurate with cubic meters m³).
With commensurate units, there's now some constant conversion factor. Say that $a$ units of A^r equal $b$ units of B^s . (For example, there are 1000 liters per one cubic meter.) The conversion equation is:

$$a \text{A}^r = b \text{B}^s$$

We're looking for a prefix which makes the two quantities equal. We can think of prefix in terms of powers of 10, as follows:

exponent $k$	number $10^k$	named prefix
-9	0.000000001	nano
-6	0.000001	micro
-3	0.001	milli
-2	0.01	centi
-1	0.1	deci
0	1	(prefix-less)
1	10	deca
2	100	hecto
3	1000	kilo
6	1000000	mega
9	1000000000	giga

For some problems, the prefix we're looking for might not exist in this table. For example, it might be a fractional number instead of a whole number.

When we don't have a name for a prefix $k$, we can just use $10^k$ as a placeholder name. For example, a 10³-meter is a kilometer, and a 10⁴-meter is ten kilometers.

The proper setup for the prefixed-point problem is to find the prefix (exponent) $k$ which makes the following two quantities equal:

$$(10^k \text{A})^r = (10^k \text{B})^s$$

Note that the dimensional exponents are placed on the outside so that, for example, we have "cubic micrometers" rather than "micro cubic-meters"
Shuffling around terms, we have:

$$10^{k(r-s)} = \text{B}^s / \text{A}^r$$

And based on the conversion equation, we know that the right-hand ratio of one B^s to one A^r is $a/b$:

$$10^{k(r-s)} = a/b$$

Taking the logarithm of both sides:

$$k(r-s) = \log_{10}(a) - \log_{10}(b)$$

or

$$k = \frac{\log_{10}(a) - \log_{10}(b)}{r-s}.$$

In slogan form, if you like, the formula for finding $k$ is "The difference between the logs of the conversion coefficients, divided by the difference between the dimensional exponents."
For example, in the original problem we had A (liters), B (meters), dimensional exponents r=1 and s=3, conversion coefficients a=1000, b=1. Hence the prefixed point is:

$$k = \frac{\log_{10}(1000) - \log_{10}(1)}{1 - 3} = \frac{3-0}{-2} = \;-\frac{3}{2}$$

This is a prefix of -1.5; situated halfway between the "deci" (-1) and "centi" (-2) prefixes. You can confirm that one 10^-1.5-liter is equal to one cubic 10^-1.5-meter. As a prefix name, I suggest sesqui, as in sesquimeter and sesquiliter. (Or a variant of sesqui- to indicate the negative sign.)
When you apply a prefix to the units A and B, the difference in dimensionality causes them to scale at different rates. This property is what allows for there to be a prefixed point at all: because of that difference, the prefix k multiplies both sides by a different amount. As k becomes large, the unit with larger dimensionality gets much larger. As k becomes small, the unit with larger dimensionality gets much smaller. And they meet somewhere in between.

To see this disparate scaling, note that a liter is only 1000¹ (one thousand) times larger than a milliliter, but a cubic meter is 1000³ (one billion) times larger than a cubic millimeter.

If the units A and B have different dimensional exponents, you can always find a unique solution for $k$. On the other hand, if the units A and B have the same dimensional exponents (for example, inches and meters), then they will remain at a fixed ratio regardless of what prefix is added. If they were different without prefixes, they will be different after scaling by prefix. There will be no solution for $k$, and correspondingly the formula contains division by zero.
Actually even if two different units have the same dimensional exponents, we can find a solution for $k$ as long as we allow infinities. We can define the omega prefix (±∞) which makes the two units equal. The omega prefix corresponds to 10^+∞ (or 10^-∞, or both, log(0/0), whatever). For example, one omega-mL is equal to one omega-L. This is because any two infinite piles of commensurate stuff are equal. Similarly any two empty piles of commensurate stuff are equal.
If we want to have a different prefix for positive and negative infinities, we can borrow Greek letters and refer to them as, respectively, omega- and omicro- (for example omega-meter and omicro-meter).