# Exponential growth and decay (Part 11): Half-life

In this series of posts, I provide a deeper look at common applications of exponential functions that arise in an Algebra II or Precalculus class. In the previous posts in this series, I considered financial applications. In today’s post, I’ll discuss radioactive decay and the half-life formula.

One way of writing the formula for how a radioactive substance (carbon-14, uranium-235, brain cells) decays is $A(t) = A_0 e^{-kt}$

In yesterday’s post, I showed how this formula is a natural consequence of a certain differential equation. Of course, students in Algebra II or Precalculus (or high school chemistry) are usually not ready to understand this derivation using calculus. Instead, they are typically given the final formula and are expected to use this formula to solve problems. Still, I think it’s important for the teacher of Algebra II or Precalculus to be aware of how the origins of this formula, as it only requires mathematics that’s only a year or two away in these students’ mathematical development.

There is another correct way to write this formula in terms of half-life. Sadly, my experience is that many students are familiar with these two different formulas but are not aware of how the two formulas are connected. As we’ll see, while yesterday’s post using differential equations is inaccessible to Algebra II and Precalculus students, the derivation below is entirely elementary and can be understood by good students in these courses.

Let $h$ be the half-life of the substance. By definition, this means that $A(h) = \displaystyle \frac{1}{2} A_0$

Substituting into the above formula, we find $\displaystyle \frac{1}{2} A_0 = A_0 e^{-kh}$

Let’s now solve for the constant $k$ in terms of $h$: $\displaystyle \frac{1}{2} = e^{-kh}$ $\displaystyle \ln \left( \frac{1}{2} \right) = - k h$ $\displaystyle -\frac{1}{h} \ln \left( \frac{1}{2} \right) = k$

Let’s now substitute this back into the original formula: $A = A_0 e^{-kt}$ $A = A_0 e^{ -\left[ -\frac{1}{h} \ln \left( \frac{1}{2} \right) \right] t}$ $A = A_0 e^{\ln \left( \frac{1}{2} \right) \cdot \frac{t}{h}}$ $A = A_0 \left[e^{\ln \left( \frac{1}{2} \right)} \right]^{t/h}$ $A = A_0 \displaystyle \left( \frac{1}{2} \right)^{t/h}$

This is the usual half-life formula, where the previous base of $e$ has been replaced by a new base of $\displaystyle \frac{1}{2}$ . For most applications, a base of $e$ is preferred. However, for historical reasons, the base of $\displaystyle \frac{1}{2}$ is preferred for problems involving radioactive decay. For example, $A(2h) = A_0 \displaystyle \left( \frac{1}{2} \right)^{2h/h}$ $A(2h) = A_0 \displaystyle \left( \frac{1}{2} \right)^{2}$ $A(2h) = \displaystyle \frac{1}{4} A_0$

In other words, after two half-life periods, only one-fourth (half of a half) of the substance remains.