# Why do things decay in terms of base $e$?

For instance, why do radioactive elements follow some $$e^{-x}$$ relationship? Why not $$\pi^{-y}$$ or some other numerical constant as the base? Is there some fundamental reason for Euler's number being the base of an exponential relationship for physical phenomena like radioactive decay, scale heights in atmospheres, etc.?

• But you can always change the base of an exponential function. That is, $e^x = 2^{x\log_2(e)}$. Nov 17, 2022 at 17:33
• For radioactivity, we typically don't use $e$ as the base. Base 2 is much more common ("half-life"). Nov 17, 2022 at 17:35
• @honeste_vivere "Half life" comes from the decay constant that naturally appears in base 2. Nov 17, 2022 at 17:41
• @honeste_vivere It is an exponential relationship. We could just as well use $e$ as the base. I don't know why it's customary to use 2 for radioactivity. Nov 17, 2022 at 17:42
• @John We use both for radioactivity, base 2 for the half-life, base e for the mean lifetime. en.wikipedia.org/wiki/Exponential_decay#Mean_lifetime Eg en.wikipedia.org/wiki/Free_neutron_decay Nov 17, 2022 at 17:47

In some sense, all bases are equivalent. If something decays as $$\sim e^{-t/\tau}$$, where $$\tau$$ is a decay constant, then it also decays as $$\sim b^{-t/\tau_b}$$ for any base $$b$$, where $$\tau_b = \tau / \log_b e=\tau \ln b$$ (which you can verify using properties of logs).

The reason we like to use base $$e$$ in physics is because it arises naturally in calculus. For example, $$y(t) = y_0 e^{-t/\tau}$$ is the solution to the differential equation $$$$\frac{dy(t)}{dt} = -\frac{y(t)}{\tau}$$$$ with the initial condition $$y(0)=y_0$$. In other bases, you would have an extra conversion factor appearing in the equation.

In fact in other fields where doing calculus isn't as important, it is more common to use other bases that are considered more "intuitive", like $$10$$ or $$2$$.

There's nothing special about $$e$$ except that it's convenient in the sense that $$e^t$$ is its own derivative. The description of exponential decay arises from the model that the rate of change is proportional to the amount of stuff remaining, i.e., $$\frac{dN}{dt} = -\gamma N\,,$$ which is solved the most succinctly and easily by $$N(t) = N_0e^{-\gamma t}\,.$$ However, we can change the base of the exponential to be anything we like, and we often do to get things like half-lives. That is, $$N(t) = e^{-\gamma t} = 2^{\log_2(e^{-\gamma t})} =2^{-(\gamma\log_2(e))t} \,.$$

$$\pi^x = e^{ln(\pi)x} = e^y$$

You can always write it as a power of e

• And of course $\log_a b = 1 / \log_b a$ Nov 17, 2022 at 17:58