Carbon 14 dating

Task

Carbon 14 is a common form of carbon which decays over time. The amount of Carbon 14 contained in a preserved plant is modeled by the equation $$ f(t) = 10e^{-ct}. $$ Time in this equation is measured in years from the moment when the plant dies ($t = 0$) and the amount of Carbon 14 remaining in the preserved plant is measured in micrograms (a microgram is one millionth of a gram). So when $t = 0$ the plant contains 10 micrograms of Carbon 14.

1. The half-life of Carbon 14, that is the amount of time it takes for half of the Carbon 14 to decay, is approximately 5730 years. Use this information to find the constant $c$.
2. If there is currently one microgram of Carbon 14 remaining in the preserved plant, approximately when did the plant die?

IM Commentary

The task requires the student to use logarithms to solve an exponential equation in the realistic context of carbon dating, important in archaeology and geology, among other places. Students should be guided to recognize the use of the natural logarithm when the exponential function has the given base of $e$, as in this problem. Note that the purpose of this task is algebraic in nature -- closely related tasks exist which approach similar problems from numerical or graphical stances.

The standards do not prescribe that students use or know with log identities, which form the basis for the "take the logarithm of both sides" approach.  The two solutions provided differ slightly in their approach in this regard.

Solutions

Solution: 1

1. Since the half life of Carbon 14 is 5730 years, this means that after 5730 years there will only be 5 micrograms of Carbon 14 left in the preserved plant: $$ f(5730) = 10e^{-5730c} = 5. $$ To solve for $c$, notice that $c$ is in the exponent and so we need to take a logarithm to isolate $c$. Since the base of the exponent is $e$, the natural logarithm is appropriate. Dividing by 10 first (to isolate the exponential expression) and taking the logarithm of both sides gives $$ -5730c = \ln{e^{-5730c}} = \ln{\frac{1}{2}} = -\ln{2}. $$ Solving for $c$ gives $$ c = \frac{\ln{2}}{5730} $$ which is approximately equal to $0.000121$.
2. To find when $f(t) = 1$ we use the value of $c$ from part a and are left with the equation $$ 10e^{-\frac{\ln{2}}{5730}t} = 1. $$ Dividing by 10 and taking the natural logarithm on both sides, as above, gives $$ -\frac{\ln{2}}{5730}t = \ln{e^\left(-{\frac{\ln{2}}{5730}t}\right)} = \ln{\frac{1}{10}} $$ or $$ t = -5730\frac{\ln\frac{1}{10}}{\ln{2}}. $$ Thus, evaluating on a calculator, gives a value of approximately 19,035 years since the plant has died. Note that if the approximate value $0.000121$ is used in place of $\frac{\ln{2}}{5730}$ then an approximate value of 19,030 years is found instead.

In either case, it is more appropriate to report the time since the plant has died as approximately 19,000 years since these measurements are never completely precise.

Solution: Alt version for sol'n 1

1. Since the half life of Carbon 14 is 5730 years, this means that after 5730 years there will only be 5 micrograms of Carbon 14 left in the fossilized plant: $$ f(5730) = 10e^{-5730c} = 5. $$ To solve for $c$, notice that $c$ is in the exponent, so one approach is to isolate the exponential expression and use the definition of the natural logarithm. Dividing both sides by 10 gives $$ {e^{-5730c}} = {\frac{1}{2}} . $$ By the definition of the natural log, $$ -5730c = {\ln{\frac12}} $$ To find $c$, we simply divide both sides by -5730: $$ c = \frac{\ln{2^{-1}}}{-5730}=\frac{-\ln2}{-5730}=\frac{\ln2}{5730} $$ which is approximately equal to $0.000121$.
2. To find when $f(t) = 1$, we use the value of $c$ from part a which gives us the equation $$ 10e^{-\frac{\ln{2}}{5730}t} = 1. $$ Dividing both sides by 10 and applying the definition of natural log gives $$ -{\frac{\ln{2}}{5730}t} = \ln{\frac{1}{10}} $$ If we solve for $t$, we get: $$ t = -5730\frac{\ln\frac{1}{10}}{\ln{2}}=5730\frac{\ln{10}}{\ln{2}}. $$

   If we evaluate this expression on a calculator, we get a value of approximately 19,035 years since the plant has died. Note that if the approximate value $0.000121$ is used in place of $\frac{\ln{2}}{5730}$ then an approximate value of 19,030 years is found instead.

   In either case, it is more appropriate to report the time since the plant has died as approximately 19,000 years since these measurements are never completely precise.