Radioactive Decay Equations and Some Related Confusion on Discrete vs. Continuous Growth/Decay, Continuously Compounding Interest, etc.

Question

UPDATE: I've gotten some great answers regarding the first part of my question, related to radioactive decay specifically - thanks! However, I'd still very much appreciate a response to the second part of my post, regarding discrete vs. continuous compound interest and other such stuff I'm still confused about. Really everything starting with "More specifically, this new discovery..." is still uncertain to me.

---

While there are a number of questions on this site already regarding radioactive decay, my question has not, I believe, been asked here.

Essentially, I am confused by the fact that Wikipedia (and other sources) lists three formulas for decay that are said to be equivalent but look quite different from one another:

Wikipedia screenshot here

Previously, I had always used the simple half-life equation with 1/2 being raised to a power (the first equation on the above-linked list). That made sense to me.

But now I see from my research that there are other equations for decay involving the number e! And I am consequently confused about the relationship between the half-life equation I've always used and these other equations with e in them.

More specifically, this new discovery is prompting me to question some of my old understandings about the simple half-life equation. First, I'd always thought that the simple half-life equation graph created a smooth curve for even non-integer numbers of half lives. But if continuity is already achieved WITHOUT using e in the equation, what is so special about e? I always thought there was something unique about e that applied to continuous growth (think continuously compounding interest) ... whereas other growth wouldn't be continuous. I know I'm missing something here but I can't quite put it into words.

More broadly, this has really started to make me wonder about the difference between "regular" compound interest and continuously compounded interest. Does "regular" compounding create a smooth curve, or does it technically just create a set of discrete points? The formula for ordinary compound interest - let's say compounding annually for simplicity: [final amount = P(1+r)^n] - is clearly a continuous formula without "gaps" in the graph, since I never see it limited to integer numbers of years (n). But compounding would appear intuitively to be discrete (except in the special case of continuous compounding). E.g., you don't make any extra interest between January and May if the compounding date is December 31, right?

I wonder if one of my sources of confusion could be the difference (if there is one) between continuous compounding and a continuous function...?

Note that I am only in Pre-Calculus, so while I understand the general concept of a derivative as the instantaneous rate of change, I don't know too much about them (and I know basically NOTHING about integrals). So if possible please avoid resting your answers on Calculus topics I haven't learned yet.

Anyway, I apologize for the length. But I really hope someone here has a way of explaining this that can resonate with me. I really like to fully understand math concepts and find being confused really uncomfortable. Thank you so much. :)

Answer 1

$\frac{1}{2}=e^{-ln2}$. Therefore $\frac{ln2}{t_{1/2}}=\frac{1}{\tau}={\lambda}$ shows that they are all essentially the same in defining radioactive decay.

Answer 2

These are all the same equation with the time constant defined differently to give the same result. $$N(t)=N_0\left(\frac 12\right)^{\frac t{t_{1/2}}}\\N(t)=N_0e^{-\frac t\tau}\\N(t)=N_0e^{-\lambda t}$$

The second two are related by $\frac 1\tau=\lambda$ where $\tau$ is the mean lifetime and $\lambda$ is the decay rate. For the first, we can use the fact that $e^{-\log 2}=\frac 12$ to write it as $$N(t)=N_0\left(e^{-\log 2}\right)^{\frac t{t_{1/2}}}=N_0e^{-\frac{t \log 2}{t_{1/2}}}$$ and with $\frac {t_{1/2}}{\log 2}=\tau$ we recover the second. There is nothing magic about using $e$ or $\frac 12$ or something else as the base of the exponent. You just need to adjust the constant correctly. Using $e$ is natural because it comes out of solving the differential equation $$\frac {dN(t)}{dt}=-\lambda N(t)$$ This is also the motivation for calling $\lambda$ the decay rate.