Answer
The general solution to the problem $Q' = -rQ$ is : $Q = Q(t) = ce^{-rt}$, where $c$ is a constant depending on the initial amount of material, $r$ is its decay-rate, $t$ is the number of years.
If $\tau$ is the half-life of the material, expressed in years as well, the equation $Q(\tau) = 1/2 Q(0)$ yields $r\tau = ln(2)$.
Work Step by Step
First, we find a general solution to the problem $Q' = -rQ$, where $Q' = dQ/dt$ and $r$ is the decay-rate constant.
If we suppose $Q \neq 0$, we can rewrite the equation as $Q'/Q = -r$.
Then, we notice that $Q'/Q = (ln(\lvert Q \rvert))' = -r$.
By integrating both sides of the equality relatively to $t$, we get that $ln(\lvert Q \rvert) = -rt + C$, where $C$ is the integration constant.
By applying the exponential function to both sides of the equality, we get that $\lvert Q \rvert = e^Ce^{-rt}$.
Since the exponential function e$^x$ is always positive, either $Q$ is always positive, or always negative, so we get $Q = e^Ce^{-rt}$ or $Q = -e^Ce^{-rt}$. We rewrite that as $Q = Q(t) = ce^{-rt}$, where $c$ is a constant depending on the initial condition (here, it would be the initial amount of material).
Let's say that $t$ corresponds to the number of years and $\tau$ is the half-life of our material, expressed in years as well (it doesn't matter much; we just have to make sure that the units of $t$ and $\tau$ are the same), i.e. $\tau$ is the number of years it takes for an amount of this material to decay to one-half its original value. In mathematical words, we have the equation :
$Q(\tau) = ce^{-r \tau} = 1/2 Q(0) = 1/2ce^{-0r} = 1/2c \iff e^{-r\tau} = 1/2$.
By applying the natural logarithm to both sides of the equation, we get $-r\tau = ln(1/2) = -ln(2) \iff r \tau = ln(2)$.
