$A$ sample is obtained from sea water. $11.9$ $C^{14}$ atoms dissociate from $1 \ g$ of carbon. $15.3$ $C^{14}$ atoms dissociate from a live sample. How many years old is the sample? $(t_{1/2} = 5730 \ \text{years})$

(N/A) The decay of $C^{14}$ follows first-order kinetics. The rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} \ \text{yr}^{-1}$.
Using the first-order integrated rate equation: $t = \frac{2.303}{k} \log \left( \frac{N_0}{N_t} \right)$.
Here,$N_0 = 15.3$ and $N_t = 11.9$.
$t = \frac{2.303 \times 5730}{0.693} \log \left( \frac{15.3}{11.9} \right)$.
$t \approx 19033.8 \times \log(1.2857) \approx 19033.8 \times 0.1091 \approx 2077 \ \text{years}$.
Rounding to the nearest value,the age is approximately $2079 \ \text{years}$.