A wooden artifact from a Chinese temple has a \(^{14} \mathrm{C}\) cactivity of 38.0 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 \(^{14} \mathrm{C}\) decay, 5715 yr, determine the age of the artifact.

We are given the following values: - \(N_t\): the remaining activity count of the artifact, which is 38.0 counts per minute. - \(N_0\): the initial activity count of the standard, which is 58.2 counts per minute. - \(T\): the half-life of the \(\mathrm{^{14}C}\) decay, which is 5715 years. We need to find the decay time \(t\) to determine the age of the artifact.

Now we set up the decay formula: \(N_t = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{T}}\)

Plug in the given values to the decay formula: \(38.0 = 58.2 \cdot \left(\frac{1}{2}\right)^{\frac{t}{5715}}\)

We need to solve the equation for \(t\): 1. Divide both sides by 58.2: \(\frac{38.0}{58.2} = \left(\frac{1}{2}\right)^{\frac{t}{5715}}\) 2. Take the natural logarithm of both sides: \(ln\left(\frac{38.0}{58.2}\right) = ln\left(\left(\frac{1}{2}\right)^{\frac{t}{5715}}\right)\) 3. Use the property of logarithms that states \(ln(a^b) = b \cdot ln(a)\): \(ln\left(\frac{38.0}{58.2}\right) = \frac{t}{5715} \cdot ln\left(\frac{1}{2}\right)\) 4. Multiply both sides by the reciprocal of \(ln\left(\frac{1}{2}\right)\) to isolate \(t\): \(t = 5715\frac{ln\left(\frac{38.0}{58.2}\right)}{ln\left(\frac{1}{2}\right)}\) 5. Calculate the value of \(t\): \(t \approx 3616.4\) Therefore, the age of the artifact is approximately 3616.4 years.