Chapter 5 - Section 5.4 - Half-Life of a Radioisotope - Study Check 5.8 - Page 151: 1

17200 years.

Present activity $N=\frac{1}{8}N_{0}$ where $N_{0}$ is the original activity. Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,y}=1.21\times10^{-4}\,y^{-1}$ $\ln(\frac{N_{0}}{N})=kt$ $\implies \ln(\frac{N_{0}}{\frac{1}{8}N_{0}})=2.08=1.21\times10^{-4}\,y^{-1} \times t$ Or $t= \frac{2.08}{1.21\times10^{-4}\,y^{-1} }=17200\,y$