How are half lifes calculated for radioactive substances that decay over many years How are the half lives (or whole lives) calculated for slowly decaying radioactive substances such as tellurium? I must be having a fundamental misunderstanding of how we are able to predict the amount of time it will take for an isotope or radioactive material to decay fully including for ones that can remain unstable for practically eons.
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 $$\ce{T = \frac{0.693}{4.87·10^{-18}} = 1.423·10^{17} s = 4.51· 10^{9} years}$$