How are half lifes calculated for radioactive substances that decay over many years
Suppose you have a sample of $\ce{N}$ atoms of a radioactive material. As it is radioactive, you may state that a small fraction k is decomposed by second. This fraction does not depend on the total number of atoms. If this sample is placed in a Geiger counter, each decomposition of its atoms produces a ray that crosses the inner volume of the Geiger counter, and it will be detected and counted. In a given time, the number of rays is equal to the number of disintegrations. If dN/dt is the number of atoms disintegrated per second (or per minute), the factor k can be calculated by dividing the number of rays detected every second in the counter, by the total number of atoms N of the radioactive material. k = $\ce{\frac {dN}{dt}·\frac{1}{N}}$
Once k is known, the half-life T can be derived from the integrated law, if $\ce{N_o}$ is the original number to atoms at t = $\ce{0}$  $$\ce{N/N_o = e^{-kt}}$$ The half-life is the time T after which $\ce{N = N_o/2}$, which gives $$\ce{1/2 = e^{-kT}}$$ with the following result : $$\ce{T = \frac{ln 2}{k}} = 0.693/k $$
So the period may be obtained from the knowledge of k. It is not necessary to wait up to the radioactive material has decreased to one half.
For example, if $0.0238$ g, or $\ce{6.02·10^{19}}$ atoms U-$238$ is inserted in a Geiger counter, $292$ rays will be detected per second. From this measurements, the constant k can be calculated : $$k\ce{ = \frac{292 s^{-1}}{6.02·10^{19}} = 4.87·10^{-18} s^{-1}}$$ and the half-life is $$\ce{T = \frac{0.693}{4.87·10^{-18}} = 1.423·10^{17} s = 4.51· 10^{9} years}$$