MHB How Old Is the Tree Based on Carbon-14 Dating?

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The discussion revolves around determining the age of a tree using carbon-14 dating, specifically from charcoal found in a volcanic eruption. The charcoal contains 62.8% of the original carbon-14, leading to the calculation of the tree's age using the half-life of carbon-14, which is 5700 years. The formula used is P=100(1/2)^(A/H), where P is the percentage of carbon-14 remaining, A is the age, and H is the half-life. After solving the equation, the calculated age of the tree is approximately 3826 years. The conversation also touches on the choice of variable names in the calculations but ultimately emphasizes the same result.
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$\textsf{ The charcoal from a tree killed
in a vocano eruption }$
$\textsf{contained 62.8% percent of the carbon-14 found in living mater.}$
$\textsf{How old is the tree, to the nearest year? }$
$\textsf{Use $5700$ years for the half-life of carbon-14} $

$$1=2e^{k\cdot 5700}$$
$$k=-0.00012160$$

presume this is how we find $ k$
$$62.8=100e^{-0.00012160y}$$
$$y\approx 3826$$
 
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If $A$ is the age of the tree (in years), and $P$ is the percentage of the original amount of carbon-14 still present and $H$ is the half-life of carbon-14 (in years), then I would write:

$$P=100\left(\frac{1}{2}\right)^{\Large{\frac{A}{H}}}$$

What do you get when you solve for $A$ (which is what we're asked to find)?
 
MarkFL said:
If $A$ is the age of the tree (in years), and $P$ is the percentage of the original amount of carbon-14 still present and $H$ is the half-life of carbon-14 (in years), then I would write:

$$P=100\left(\frac{1}{2}\right)^{\Large{\frac{A}{H}}}$$

What do you get when you solve for $A$ (which is what we're asked to find)?
$$62.8=100\left(\frac{1}{2}\right)^{a/5700}$$
$$a=3826$$

why not use $y$ instead $a$?
 
karush said:
$$62.8=100\left(\frac{1}{2}\right)^{a/5700}$$
$$a=3826$$

What I meant was to take:

$$P=100\left(\frac{1}{2}\right)^{\Large{\frac{A}{H}}}$$

And solve for $A$:

$$\frac{100}{P}=2^{\Large{\frac{A}{H}}}$$

$$A=H\log_{2}\left(\frac{100}{P}\right)$$

Now plug in the given data (we have a general formula now for other problems):

$$A=5700\log_{2}\left(\frac{100}{62.8}\right)\approx3826$$

karush said:
why not use $y$ instead $a$?

Whatever you choose is fine. :)
 
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