I understand how to square a number, but how do you square a speed? 
Speed’s just a number (albeit with an attached unit, e.g. m/s or mph). You square it the same way you square any number, by multiplying it by itself.
ETA: Example: You can square 5 mph (I’m going to use mi/h instead of mph here, and you’ll see why): (5 mi/h)(5 mi/h) = 25 mi²/h²
c[sup]2[/sup] = (300,000km/s)[sup]2[/sup] = 300,000km[sup]2[/sup]/s[sup]2[/sup].
As long as you keep your units straight, it’s all groovy.
ETA: Check it out: Wolfram Alpha is smart enough to keep units straight as well.
Don’t you mean 90,000,000,000 km²/s²?
You’re right; I forgot to square the 300,000 part.
This is what happens to your brain when your last physics class was 11 years ago.
To the OP: It’s not entirely unreasonable to ask this question, because, as you realize, speed is not a number, though ratios between speeds are numbers.
However, nonetheless, one can make sense of squaring a speed, or indeed of multiplying one speed by another more generally; it’s just that the result won’t be a speed, but will rather be a speed[sup]2[/sup]. What’s a speed[sup]2[/sup], you ask?
Well, you can think of a speed as a distance/time, which you can think of as a linear function from lengths of time to distances; when we say “5 MPH”, we mean the function which sends one hour to five miles, which sends two hours to ten miles, which sends 1.5 hours to 7.5 miles, and so on. The key property is that multiplying the input to such a function by k causes the output to multiply by k.
A speed[sup]2[/sup], on the other hand, is a distance[sup]2[/sup]/time[sup]2[/sup], which is to say an (area/time)/time, which you can think of as a function that takes two lengths of time as input, and produces an area as output, with the property that multiplying either input to such a function by k causes the output to multiply by k. If f and g are speeds, then their product f * g is a speed[sup]2[/sup]; specifically, (f * g)(t1, t2) = the area f(t1) * g(t2) [multiplying two lengths gives the area of a corresponding rectangle]. So, for example, squaring 5 MPH gives us a function which sends (1 hour, 1 hour) to 25 square meters, which sends (1 hour, 4 hours) to 100 square meters, which sends (2 hours, 3 hours) to 150 square meters, and so on. Note that nothing here actually depends on what units we express the results in; if we used kilometers per second, we would still end up with the same function when we squared that same speed, even though the particular numbers used to express it would change. The process of multiplying two speeds is perfectly well defined and well behaved without selecting any particular system of units.
Mathy discussion:
In general, you can think of dimensions (length, time, area, speed, etc.) as describing (usually, but not necessarily, one-dimensional) various abstract vector spaces/modules. Multiplying dimensions corresponds to taking tensor products of such spaces; division (e.g., speed as length/time) corresponds to taking the space of linear maps from the divisor to the dividend. When these spaces are one-dimensional, the selection of a reference non-zero element within them (i.e., a choice of units) allows you to identify the space with real numbers; however, the tensor and function-space constructions for multiplication and division work perfectly fine even purely abstractly (allowing us to make sense of them without having to make an arbitrary selection of units).
Er, in editing, words floated around to places they don’t fit. Reword this as:
In general, you can think of dimensions (length, time, area, speed, etc.) as describing various abstract (usually, but not necessarily, one-dimensional) vector spaces/modules.
While I understood squaring speeds well enough, it’s taking their logs that I don’t get. The log of 3e8 m/s is about 8.5 m/s. The units go right through the log function unchanged. But what about the log of a squared speed:
2log(3E8)(m/s)^2 = log(3E8^2)(m/s)^2 = log(3E8^2 (m/s)^2) = log((3E8 m/s)^2) = 2log(3E8 m/s) = 2log(3E8)(m/s)
so
2log(3E8)(m/s)^2 = 2log(3E8)(m/s)
so
(m/s) = 1
Never did get this. I mean, I think it’s just me, and the universe still works right.
Units don’t go through a logarithm unchanged. Why would you think that?
More to the point, let’s reiterate: speeds aren’t numbers. Only ratios between speeds are numbers. Speeds themselves live in an abstract vector space; you can add them to each other, you can multiply them by numbers to get other speeds, you can divide them by each other to get numbers, but they aren’t numbers themselves. And you can’t in any nontrivial sense take their logarithm (though there is a trivial sense in which you can take their logarithm, the sense in which you just write things about their logarithms using addition to mean the corresponding things about the speeds themselves using multiplication; that’s all a logarithm is, an isomorphism from a multiplication-structure to an addition-structure).
Nope: log(3E8 m/s) = 8.5 + log(m) - log(s).
ETA: to the extent that it makes sense to define this. In physical laws, one can always find a way to need only the logarithms of unitless numbers.
(Totally pedantic ass-covering nitpick on myself: I should say, when I say speeds aren’t numbers, I mean they aren’t real numbers (which are perhaps better called “scaling numbers” or something like that); they don’t have the same structure. There’s no distinguished speed called 1, there’s no distinguished way to multiply speeds into another speed, and so on. But you could still call them “numbers”, if you like; they have some arithmetic structure and anyway, you can call anything a number, if it suits your fancy. The important thing is that speeds are not the same kind of numbers as the familiar 1, 2, 3, …, and they don’t have all the same structure)
There is a joke about a cabin somewhere in here, but I’m just not clever enough to pull it off…
I think the OP is referring to Einstien’s famous mass energy equivalence equation E=mc^2
In this case, ‘c’ represents a number between 0 and 1 where 1 is the speed of light.
Tell me more.
I’m an electrical engineer and I was just always taught that you can’t take logarithms of non-dimensionless numbers. Nor exponentials and trig functions for that matter.
Eh? The speed of light in a vacuum is a constant. c is always the same thing. it can’t be a number between zero or one. If the speed of light is 1 then c is 1. The speed of light is only 1 if you’re using weird units, like lightyears-per-year.
And indeed you generally* can’t, in any nontrivial sense. Pasta said so as well. The trivial sense is the purely formal sense in which logarithm means “Everything that I used to write with multiplications, I’ll write with additions now instead, in logarithm-world. But I’ll never do anything else with the logarithms other than this”. This is the sense in which you can write log(1000 m/s) = log(1000) + log(m) - log(s). But there’s nothing more to it than that; log(m) isn’t a distance, it’s a logarithmic distance. And there’s nothing you can do with logarithmic distances other than manipulate the corresponding actual distances. So there’s no point to this trivial notion of logarithm; it’s just a way of saying “I’ll use + notation instead of * notation now…”.
*: For certain particular kinds of things other than dimensionless numbers, you can still take the logarithm. But not speeds…
Sure there is; c. A speed is just a ratio of two things that can be measured in the same units. It’s just a quirk of the way we look at the world that we usually use different units to measure lengths and times.
Meanwhile, if I might digress, I’ve a hunch that there’s another question floating around in the OP’s mind: Namely, what units is the equation E = mc^2 expressed in? And the answer to that is, you can pick any units you want for m and c, and that’ll determine what units you use for E. You might, for instance, choose to measure m in kilograms, and c in meters per second, in which case you’ll get E in kg m^2/s^2. That’s a unit of energy that shows up a lot, in practice, so it’s given its own name: The Joule. Alternately, you could use grams for your mass, and centimeters per second for c, in which case you’ll get energy in g cm^2/s^2. That’s also a unit of energy that gets used often, so it also has its own name: The erg. Or, if you want to be eccentric about it, you could measure mass in carats, and c in cubits per coon’s age. That’d give you an energy in units of carats cubits^2/(coon’s age)^2. Now, nobody ever actually uses those units, so that one doesn’t have its own name, but it’s still a perfectly valid energy unit.
Heh, good point, fair enough.
There is a natural picture of the world in which length and time are different abstract dimensions, and there is no distinguished isomorphism between the two. Of course, it turns out there is a particular isomorphism between the two which pops up all over the place in physics, so for many purposes, you may not want to adopt that picture as one’s mathematical model; you would want a model which gives more respect to the structure relating/identifying them. So… in general, given two unrelated abstract dimensions, there is no distinguished rate of change of the one with respect to the other called 1. But you may not want to think of length and time as unrelated dimensions.
Though I also suppose the notion of addition of speeds one is often most interested in relativistically is not the notion of pointwise addition of the corresponding linear functions from lengths of time to distance. So perhaps quite a lot of things about the particular abstract vector space model I was describing for speed are less useful from a relativistic point of view. Well, there you go then. But it’s a fine way to work in the Newtonian picture. 
Hey guys, how do I explain to people only slight less ignorant than myself, that ‘c’ represents the ‘square root of the speed of light’? At least in Einstein’s equation, which is the only way they’ve ever heard of it.