> To do so, we must solve (1+r2)^12 = (1+r), which requires logarithms

No, it doesn't. r_2 = (1 + r)^{1/12} - 1. Compound interest looks like (money) = (money_0)*r^t; you'd only need logarithms if you were trying to solve for time.

> Now, it might be natural to ask what about for continuous compounding?

I can't tell what you're getting at here. Once you've written down the equation (1 + r_2)^{12} = (1 + r), you've already provided a complete solution for continuous compounding. If the time you want to compound over is t, and Y is one year, then the solution is always given by (1 + r_2) = (1 + r)^{t/Y}. Nothing goes to infinity.