A sample of a radioactive isotope initially weighs 200 g. After 100 years, 18.5 g remains. What is the half-life of the isotope?

The formula for radioactive decay is $N_t = N_0 \left(\frac{1}{2}\right)^n$, where $N_t$ is the final mass, $N_0$ is the initial mass, and $n$ is the number of half-lives.

$$18.5 \text{ g} = 200 \text{ g} \left(\frac{1}{2}\right)^n$$ $$\frac{18.5}{200} = \left(\frac{1}{2}\right)^n$$ $$0.0925 = (0.5)^n$$

To solve for $n$, take the logarithm of both sides:

$$\log(0.0925) = n \log(0.5)$$ $$n = \frac{\log(0.0925)}{\log(0.5)}$$ $$n \approx \frac{-1.0339}{-0.3010}$$ $$n \approx 3.4349$$

The total time ($T$) is related to the number of half-lives ($n$) and the half-life ($t_{1/2}$) by the formula $T = n \times t_{1/2}$.

$$100 \text{ years} = 3.4349 \times t_{1/2}$$ $$t_{1/2} = \frac{100 \text{ years}}{3.4349}$$ $$t_{1/2} \approx \boxed{29.11 \text{ years}}$$