**$\overline{){\mathbf{t}}_{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}\mathbf{}\mathbf{=}\frac{\mathbf{ln}\mathbf{}\mathbf{2}}{\mathbf{k}}}\phantom{\rule{0ex}{0ex}}\mathbf{k}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{ln}\mathbf{}\mathbf{2}}{{\mathbf{t}}_{{\displaystyle \raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}}}\phantom{\rule{0ex}{0ex}}\mathbf{k}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{0}\mathbf{.}\mathbf{693}}{\mathbf{5715}\mathbf{}\mathbf{yr}}$**

**k = 1.2126x10 ^{-4} yr^{-1}**

A wooden artifact from a Chinese temple has a ^{14}C activity of 37.6 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age.

From the half-life for ^{14}C decay, 5715 yr, determine the age of the artifact.

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