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I'm working on this problem.

The radioactive element carbon-14 has a half-life of about 5,750 years. The percentage of carbon-14 present in the remains of animal bones can be used to determine age. How old is an animal bone that has lost 40% of its carbon-14?

I'm using the formula $y = Ce^kt$ to model the decay.

To make my life easier, I consider $5,750$ years to be 1 time unit.

So knowing this is a half life, for a single time unit I can say

$$50 = 100 e^k$$

Then

$$\frac{1}{2}=e^k$$

Then log of both sides

$$\ln\frac{1}{2}=\ln e^k$$

$$\ln\frac{1}{2}= k \cdot \ln e$$

So

$$k = \ln\frac{1}{2}$$

Good so far.

Next we need to determine the $t$ after Carbon14 has decayed to $60\%$.

So

$$60 = 100 e^kt$$

which is

$$\frac{3}{5}= e^kt$$

Then I take the log of both sides

$$ \ln \frac{3}{5}= \ln e^kt$$

then

$$\ln\frac{3}{5} = kt\cdot \ln e$$

simplifies to

$$ \ln\frac{3}{5} = kt$$

Plugging in $k$ we get

$$t =\frac{ \ln\frac{3}{5}}{\ln\frac{1}{2}}$$

Recall that t is a unit where each unit is $5,750$ years, so the final answer is

$$5750\cdot \left(\frac{\ln\frac{3}{5}}{\ln\frac{1}{2}}\right)$$

My answer is $\approx4,237.55$ years.

The answer key says the $4,257$ years. http://www.sosmath.com/cyberexam/precalc/EA3001/EA3001.html

Where did I go wrong?

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    $\begingroup$ It seems to be that site has a mistake. What you did is correct. $\endgroup$ – DonAntonio Dec 14 '13 at 20:06
  • $\begingroup$ Minor comment: It would be more conventional to use $e^{-kt}$. $\endgroup$ – André Nicolas Dec 14 '13 at 20:11
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I personally think you did it right. The structure and execution is correct. I checked my answer over and over again and it is correct. I think the answer key is wrong.

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