The half-life for radioactive decay of C-14 is 5730 years. An archaeological artifact containing wood had only 80% of the C-14 found in a living tree. The age of the sample is

Radioactive decay follows first order kinetics. therefore, 

Decay constant $\left(\lambda \right)=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730}$

Given, $R_0=100$        $\therefore R = 80$

and $t=\frac{2.303}{\lambda }\log \frac{{\left[R\right]}_0}{\left[R\right]}=\frac{2.303}{\left({\displaystyle \frac{0.693}{5730}}\right)}\log \frac{100}{80}$

$=\frac{2.303\times 5730}{0.693}\times 0.0969=1845$ years