Levi Porter said:
Are quantum bits just a form of a ternary numeral system?
Not really, for one because upon measurement a qubit will always have the value 0 or 1. For example, if i have 4 bits then i can store [itex]2^{4}[/itex] states, and pass any of them to you and you'd be able to recognize which state i had passed you (0100, 0011, etc). On the other hand if i have 4 qubits, then any of the [itex]3^{4}[/itex] quantum states i send you will always collapse to one of [itex]2^{4}[/itex] states, so although we can interpret a qubit as a base-3 digit, we can't use it as base-3 digit for storing and passing information.
In quantum computation we also use more information for each qubit than can be stored ina base-3 digit. For example rather than having the three states:
0, 1, 01
In quantum computation you might have:
[x] 1, [y] 0
Where [itex]x^{2}[/itex] gives the probability of the qubit being 1 and [itex]y^{2}[/itex] gives the probability of it being 0.
Therefore a qubit can have any number of states defined by [x y], where [tex]x^{2} + y^{2} = 1[/tex].
In operation, a quantum machine isn't equivalent to a base-3 machine either (which is actually equivalent to a base-2 machine, as they both can be modeled by a turing machine). When a quantum computer with n qubits performs an operation, it's performing an operation on 2^5 states (the number of states that the system of 5 qubits is in a superposition of). An operation changes the probabilities of each possible state, the one with the highest probability being the most likely to be revealed upon measurement.
This is something that a system of conventional bits would need years to model for even some small number of qubits.
Scott Aaronson has in his blog a pretty good explanation of Peter Shor's quantum factoring algorithm using simple terms:
http://scottaaronson.com/blog/?p=208
I particularly like one of the explanations given in one of the comments:
http://scottaaronson.com/blog/?p=208#comment-10026