1. **Problem statement:** We want to find the time $t$ in years until only 5% of the original amount of strontium-90 remains in the bones. 2. **Formula:** The exponential decay formula based on half-life is: $$A(t) = A_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}$$ where: - $A(t)$ is the amount remaining after time $t$, - $A_0$ is the original amount, - $h$ is the half-life (30 years for strontium-90), - $t$ is the time elapsed. 3. **Given:** - $A(t) = 0.05 A_0$ (5% remains), - $h = 30$ years. 4. **Set up the equation:** $$0.05 A_0 = A_0 \left(\frac{1}{2}\right)^{\frac{t}{30}}$$ 5. **Divide both sides by $A_0$ (assuming $A_0 \neq 0$):** $$0.05 = \left(\frac{1}{2}\right)^{\frac{t}{30}}$$ 6. **Take the natural logarithm of both sides:** $$\ln(0.05) = \ln\left(\left(\frac{1}{2}\right)^{\frac{t}{30}}\right) = \frac{t}{30} \ln\left(\frac{1}{2}\right)$$ 7. **Solve for $t$:** $$t = 30 \times \frac{\ln(0.05)}{\ln(\frac{1}{2})}$$ 8. **Calculate the values:** - $\ln(0.05) \approx -2.9957$ - $\ln(\frac{1}{2}) = \ln(0.5) \approx -0.6931$ 9. **Substitute:** $$t = 30 \times \frac{-2.9957}{-0.6931} = 30 \times 4.3219 = 129.66$$ 10. **Interpretation:** It will take approximately 130 years for only 5% of the original amount of strontium-90 to remain in the bones. **Final answer:** $$\boxed{130 \text{ years}}$$