M2215.14 carbon-14 dating

[karush]

$\textsf{ The charcoal from a tree killed
in a vocano eruption }$
$\textsf{contained 62.8% percent of the carbon-14 found in living mater.}$
$\textsf{How old is the tree, to the nearest year? }$
$\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$
$$62.8=100e^{-0.00012160y}$$
$$y\approx 3826$$

 

[MarkFL]

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:

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

What do you get when you solve for $A$ (which is what we're asked to find)?

 

[karush]

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:

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

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 $a$?

 

[MarkFL]

$$62.8=100\left(\frac{1}{2}\right)^{a/5700}$$
$$a=3826$$

What I meant was to take:

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

And solve for $A$:

\(\displaystyle \frac{100}{P}=2^{\Large{\frac{A}{H}}}\)

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

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

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

why not use $y$ instead $a$?

Whatever you choose is fine. :)