A quantity follows first-order decay when its instantaneous rate of decrease is proportional to how much of the quantity is currently present.

If $X(t)$ is the remaining amount and $k>0$ is a constant rate coefficient,

$$\boxed{\frac{dX}{dt}=-kX}.$$

The minus sign indicates decrease. The proportionality means that losing a fixed fraction per unit time matters, not losing a fixed absolute amount.

The function

$$\boxed{X(t)=X_0e^{-kt}}$$

has exactly this property. Differentiating verifies it:

$$\frac{dX}{dt}=-kX_0e^{-kt}=-kX.$$

Thus the fractional rate of change is constant:

$$\frac1X\frac{dX}{dt}=-k.$$

Half-life

The half-life $t_{1/2}$ is the time required for the quantity to fall to half its initial value:

$$X(t_{1/2})=\frac{X_0}{2}.$$

Substitute the exponential law:

$$\frac12=e^{-kt_{1/2}}.$$

Taking logarithms gives

$$-\ln2=-kt_{1/2},$$

so

$$\boxed{t_{1/2}=\frac{\ln2}{k}}.$$

Because this result contains no $X_0$, every halving takes the same amount of time. After $n$ half-lives,

$$X=X_0\left(\frac12\right)^n.$$

Example

If

$$k=0.0200,\mathrm{min^{-1}},$$

then

$$t_{1/2}=\frac{0.693}{0.0200}=34.7,\mathrm{min}.$$

After $69.4$ min, or two half-lives,

$$X=X_0\left(\frac12\right)^2=\frac{X_0}{4}.$$

A related characteristic timescale is

$$\tau=\frac1k,$$

for which $X(\tau)=X_0/e$. Half-life and $\tau$ are two ways to express the same exponential timescale:

$$t_{1/2}=\tau\ln2.$$

This mathematical pattern appears whenever a process has a constant first-order fractional removal rate, including radioactive decay, simple chemical first-order reactions and many relaxation or loss processes.