Chapter 32 - Nuclear Physics and Nuclear Radiation - Problems and Conceptual Exercises - Page 1153: 36

$1.57\times 10^4y$

We can find how old the bones are as follows: $\lambda=\frac{\ln 2}{T_{\frac{1}{2}}}=\frac{\ln 2}{5730y}=1.210\times 10^{-4}y^{-1}$ We know that $0.150N_{\circ}=N_{\circ}e^{-\pi t}$ $\implies 0.150=e^{-\pi t}$ Taking the natural log and solving for t, we obtain: $t=\frac{\ln 0.150}{\pi}=-\frac{\ln 0.150}{1.210\times 10^{-4}y^{-1}}=1.57\times 10^4y$