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A sample collected from cave paintings on an archeological site in France shows that only
2% of the carbon-14 still remains. How old is the sample? Round to the nearest year.
Use the model for radiocarbon dating:
Q(t)=Q_(0)e^(-0.000121 t) where
Q_(0) is the original quantity of carbon-14

A sample collected from cave paintings on an archeological site in France shows that only $2 \%$ of the carbon $-14$ still remains. How old is the sample? Round to the nearest year. $\newline$ Use the model for radiocarbon dating: $Q(t)=Q_{0} e^{-0.000121 t}$ where $Q_{0}$ is the original quantity of carbon $-14$

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Reflections of functions

Full solution

Q. A sample collected from cave paintings on an archeological site in France shows that only

2 \%

of the carbon

-14

still remains. How old is the sample? Round to the nearest year.

\newline

Use the model for radiocarbon dating:

Q(t)=Q_{0} e^{-0.000121 t}

where

Q_{0}

is the original quantity of carbon

-14

Given Information: We know that $2\%$ of the original carbon $-14$ remains, so $\frac{Q(t)}{Q_{0}} = 0.02$ . We'll use the radiocarbon dating model $Q(t) = Q_{0}e^{-0.000121t}$ to find $t$ .
Substitution into Model: First, we substitute $\frac{Q(t)}{Q_{0}} = 0.02$ into the model to get $0.02 = e^{-0.000121t}$ .
Solving for $t$ : Now, we need to solve for $t$ . We take the natural logarithm of both sides to get $\ln(0.02) = \ln(e^{-0.000121t})$ .
Taking Natural Logarithm: Using the property of logarithms, we simplify to $\ln(0.02) = -0.000121t$ .
Simplification: Divide both sides by $-0.000121$ to isolate $t$ , so $t = \frac{\ln(0.02)}{-0.000121}$ .
Calculating t: Now we calculate t using a calculator: $t \approx \ln(0.02) / -0.000121 \approx 37708.3$ years.

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