Chapter 5 - Section 5.6 - Nuclear Fission and Fusion - Challenge Questions - Page 163: 5.78

$1.146\times10^{4}$ years.

$A_{0}=40\,\text{counts per min}$ $A=10\,\text{counts per min}$ Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,y}=1.209\times10^{-4}\,y^{-1}$ $\ln(\frac{A_{0}}{A})=kt$ $\implies \ln(\frac{40}{10})=1.386=1.21\times10^{-4}\,y^{-1}(t)$ Then, $t=\frac{1.386}{1.209\times10^{-4}\,y^{-1}}=1.146\times10^{4}\,y$ The final answer in one significant figure is 10000 years.