The Math!

(that's an exclamtion, not a factorial sign)

Exponential Decay Formula created by Willard Libby
A = Ao2^(-t/k)
A = present amount of the radioactive
isotope
Ao = original amount of the radioactive
isotope, measured in the same units as A
t = time it takes to reduce the original
amount of the isotope to present amount
k = half-life of the isotope, measured in
the same units as t
The reasoning behind this formula can easily be
seen. First, the Ao is there because the amount
original would be the amount when t is 0, or the
y-intercept which A0 is in this equation:
For t=0 A = Ao2^(-0/k)
A = Ao2^0
A = Ao
Thus (0,A0) is the y-intercept in the equation.
The reasoning behind the 2 is also rather clear
A = Ao2-t/k
and
A = Ao(2^-1)^(t/k)
are equivalent equations.
Thus r, or the multiplier is actually the
reciprocal of 2, or 1/2. The reasoning behind 1/2 is,
probably easy to see, but will be explained below.
The reason behind the t/k is because when you
multiply by one half you’re doing it every
half-life or the time gone by divided by the
half-life. t/k gives me that number. We hope this
brief explanation has helped better explain this
formula.

Please note : Another formula exists which is Y=Yo e^(Kt). This formula is the equation for exponentialgrowth. The reason for the difference is that they
actually are the same--only the value of k is
different. To see this, imagine Ao2^(-t/k) =Yo e^(Kt) is
true for every value of t. That’s what it means to
be equivalent expressions. In particular they must
be equivalent when t=0, e^0=1, and 2^0=1, thus Yo=Ao
(For t=0 Y=Yo*1 and A=Ao 1). Thus they can be
divided out. Now we have e^(Kt)=2^(-t/k). Take the log
of both sides (you can choose base 2 or e depending
on which ratio for k or K you want). We will use
log base e, or ln (natural logarithm) because we
want an equation comparing the rate of decay (k) to
the value of K in the exponential growth formula.
We get Kt=(-t/k) ln 2 : remember (ln ex=x). And then by
dividing t from both sides we get K=(-ln 2)/k.
Thus with the relationship between K and k, the
expressions are equivalent. The one using e is
more convenient if you’re doing calculus and is
often used. However the exponential decay formula
is more convenient for using half-lives. Thus we
will use the exponential decay formula.
One might ask what ever happened to F(x)=ab^x.
The truth is these are all forms of that equation.
For instance Ao2-t/k has a= Ao, b=2, and x=-t/k.

Now to solve for t (using the exponential decay
formula):
 

the 7th equation may cause an overflow error on
the TI-86 (and other calculators). Number 6 may
be more useful to you.

To show the usefulness of carbon dating we’ve
included the following story. Similar events have
occurred before and without radiocarbon dating one
would never have discovered the truth. To use
radiocarbon dating in the following story just
remember t = [k[log (Ao/A)]]/log 2.
A = present amount of the radioactive
isotope
Ao = original amount of the radioactive
isotope, measured in the same units as A
t = time it takes to reduce the original
amount of the isotope to present amount
k = half-life of the isotope, measured in
the same units as t

It is worth noting that we have used a common log
in solving the equation for t, but any base could
have been used. Also there are other ways to solve
for t, all of them using logarithms.
 

The following story is true
A newspaper commented that:
The body of a man wearing the traditional
clothes of the snow plains nomads was
found at the bottom of one of the many
peat bogs that remain from the last
glacial retreat. He had a stone ax buried
in the back of his skull. The ax was made
in the ancient nomad style, stone chipped
to a sharp edge and bound by leather
strips to a forked branch. Because of the
victim’s dress and the unusual murder
weapon, the police are assuming this
homicide involves a clan dispute among the
wandering deep snow people, the reindeer
herders, who still maintain their ancient
territorial ways.

The acid in the bog had tanned the man’s
body preserving it well. Although the
man’s skin is wrinkled and pulled tightly
over his bones, his features are still
distinguishable and his clothes and
internal organs are still intact and
available for police analysis. The
withered condition of the body has
convinced the police that the homicide
happened at least 10 years ago. Forensic
evidence shows that the man was killed
elsewhere, dragged to the bog and then
thrown, presumably to hide the crime.
Anyone having pertinent information or
knowing of any of the reindeer herders who
have disappeared, are urged to contact
police.

A follow up comment stated:
After months of investigation no new
evidence of information was tuned up.
Eventually the police requested a C14
radio-active dating test done on the
victim’s body and clothes. To their
astonishment the test found that for every
100,000,000 C12 atoms present in the man’s
body and clothes only 3,000 radioactive
C14 atoms were present instead of 10,000
atoms expected for a recently deceased
person. This has greatly confused the
police who had assumed that the murder was
a recent event.

Given this information the age of the man can be
approximated.
 

It has been 9952.81285457 years since the man
was murdered. This mystery will haunt mankind for
eternity. Carbon 14 was indispensable useful in
dating the age of this man.