1. **Problem Statement:** A fossilized leaf contains 70% of its normal amount of carbon-14. We need to find how old the fossil is, given the half-life of carbon-14 is 5730 years. 2. **Formula Used:** The decay of carbon-14 follows the exponential decay formula: $$N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T}}$$ where: - $N(t)$ is the amount of carbon-14 remaining at time $t$, - $N_0$ is the original amount of carbon-14, - $T$ is the half-life (5730 years), - $t$ is the time elapsed (age of the fossil). 3. **Set up the equation:** Given $\frac{N(t)}{N_0} = 0.70$, substitute into the formula: $$0.70 = \left(\frac{1}{2}\right)^{\frac{t}{5730}}$$ 4. **Solve for $t$:** Take the natural logarithm of both sides: $$\ln(0.70) = \ln\left(\left(\frac{1}{2}\right)^{\frac{t}{5730}}\right)$$ Using logarithm power rule: $$\ln(0.70) = \frac{t}{5730} \times \ln\left(\frac{1}{2}\right)$$ 5. **Isolate $t$:** $$t = 5730 \times \frac{\ln(0.70)}{\ln(\frac{1}{2})}$$ 6. **Calculate the value:** $$\ln(0.70) \approx -0.3567$$ $$\ln(\frac{1}{2}) = \ln(0.5) \approx -0.6931$$ So, $$t = 5730 \times \frac{-0.3567}{-0.6931} = 5730 \times 0.5145 \approx 2947 \text{ years}$$ 7. **Interpretation:** The fossil is approximately 2947 years old.