A smoke detector holds 0.52 mg of americium-241, with a half-life of 432 years…

Step 1. Half-lives passed. $\dfrac{100\ \text{years}}{432\ \text{years}} = 0.2315$.

Step 2. Mass left. $m = 0.52\ \text{mg} \times 0.5^{0.2315} = 0.443\ \text{mg}$.

Step 3. The 10 per cent point. $0.5^{t/432} = 0.10$. Take logarithms: $\dfrac{t}{432\ \text{years}} \times \log 0.5 = \log 0.10$, so $\dfrac{t}{432\ \text{years}} = \dfrac{-1}{-0.3010} = 3.32$.

Step 4. The time. $t = 3.32 \times 432\ \text{years} = 1435\ \text{years}$.