How do I add measurement units to a power-function relationship like $y = 0.17x^{0.52}$? I have the equation $y = 0.17x^{0.52}$, where the units on $x$ are $\rm cm$ and the units on $y$ are seconds. I can't figure out how to put units on the constants so that they work out and I get the correct unit on my $y$ value.
 A: My preferred way to notate this is
$$
y = 0.17\:\mathrm{s} \times (x/\mathrm{cm})^{0.52}
$$
but if you wanted you could also do
$$
y = (0.17\:\mathrm{s} \:\mathrm{cm}^{-0.52}) \ x^{0.52}
$$
if you wanted to put all of the units in a single place. Or heck, you could even do
$$
\frac{y}{\mathrm{s}} = 0.17 \left(\frac{x}{\mathrm{cm}}\right)^{0.52}
$$
or
$$
\frac{y}{1\:\mathrm{s}} = 0.17 \left(\frac{x}{1\:\mathrm{cm}}\right)^{0.52}
$$
if you wanted.
A: I would suggest:
$$y = (0.17\ \mathrm{s}) \left[(1\ \mathrm{cm}^{-1}) x\right]^{0.52}$$
What this does is it shows explicitly that before you're passing into the "troublesome" non-polynomial functions like the fractional power, you are making explicit there are "hidden" dimensional proportionality constants that are being used to clear the units and also what they are.
It's kind of like the hold dilemma of whether Newton's second law should be
$$F = ma$$
versus
$$F = kma$$.
The latter is technically the most general form, because there is no need for the unit of force to be simply related to the units of mass and acceleration: a very sensible real-life example of this would be if you are using US Customary or British Imperial units (though not necessarily a nice system, but they exist) and you measured $F$ in pounds of force, $m$ in pounds of mass and $a$ in, say, feet per second or miles per hour per second. Then there will be a $k$ that is not equal to 1 in front, because the unit of pound force does not accelerate a pound mass by 1 $\mathrm{ft/s^2}$ or 1 $\mathrm{mph/s^2}$ - yet all four units are entirely legitimate. The suppression of the initial $k$ is effectively a slight degree of "nondimensionalization" attained by choosing systems like SI or CGS where that the units of force are consistent with those of mass and acceleration in just the right way to make that possible that allows us to assume it equals $1$ without fault. (This can go even further when it comes to systems of "natural units" like Planck units and is their whole simplificatory basis.)
In the second form, the proportionality constant is shown explicit, and thus it is important to use when it matters. Likewise, in the initial example, you need proportionality constants, and the standard mathematical way to write the constant when $y \propto x$ is $y = kx$, thus
$$y = 0.17x^{0.52}$$
becomes
$$y = k_1 (k_2 x)^{0.52}$$
then fill in $k_1$ and $k_2$ appropriately.
