A wooden artifact from an ancient tomb
contains 60% of the carbon-14
that is present in living trees.
How long ago, to the nearest year, was the artifact made?
And we're given the fact that half-life
of carbon-14 is 5,730 years.
We'll solve this problem using
the exponential function given here below
where P of t is the amount after time t,
P sub zero is the starting amount,
and k would be the exponential decay rate.
To determine how long ago this artifact was made
we have to first find k, the exponential decay rate.
We can find k, since we're given
that the half-life of carbon-14 is 5,730 years.
This means for example,
if we start with one gram of carbon-14,
we could say that P sub zero equals one gram.
Then after 5,730 years, half of this would remain,
meaning P of 5,730
would be equal to one-half of a gram.
We can use this information to find k
and then answer the question.
So with this information,
this would give us the equation
one-half equals one times e,
raised to the power of k times t
or k times 5,730.
So we have our exponent as 5,730k.
So again, we're starting with one gram
and after 5,730 years, we have half a gram
because the half-life is 5,730 years.
Now we'll solve this for k.
We can just drop this one if we want.
So we'd have one-half equals e,
raised to the power of 5,730k.
Now we'll take the natural log
of both sides of the equation.
So we have natural log one-half equals,
here we'll apply the power property of logs,
so we can take this exponent and write this as a product
of the exponent and natural log e.
So we'd have 5,730k
times natural log e,
but natural log e equals one,
so to solve for k, we'll divide both sides by 5,730
to determine the exponential decay rate.
Now we'll go to the calculator.
So we have natural log one-half
or natural log point five,
divided by 5,730.
Notice e negative four on the right,
which means this is scientific notation,
should have it convert to a decimal,
with a decimal point to the left four times.
So we'd have negative point zero, zero, zero,
one, two, zero, nine, six, eight.
Sometimes you'll see a zero
to the left of the decimal point as well here,
which means for a carbon-14, the exponential decay model
would be P of t equals
P sub zero times e,
raised to the power of negative zero point zero, zero, zero,
one, two, zero, nine, six, eight times t.
Now we can finally answer the question,
if 60% of carbon-14 is present,
then how long ago was the artifact made?
So in this case, we could say that
if P sub zero equals one gram,
then P of t would be equal
to 60% as a decimal, zero point six grams.
Now that we have k, we can perform substitutions
for P sub zero, P of t,
and solve for t to the nearest year.
Our equation would be zero point six
equals one times e, raised to the power
of negative zero, point zero, zero, zero
one, two, zero, nine, six, eight times t.
Of course we can drop this one.
And now to solve for t, we'll take the natural log
of both sides of the equation.
On the right side here,
we'll apply the power property of logarithms
and write this exponent as a product
with natural log e.
So we'll have natural log zero point six
equals negative zero point zero, zero, zero,
one, two, zero, nine, six, eight t
times natural log e.
But again natural log e is equal to one,
so to solve for t, we'll divide
by the coefficient of t on both sides.
Notice on the right, this simplifies nicely to t,
and therefore this quotient here
will tell us how old the artifact is.
So we have natural log point six
divided by negative zero point zero, zero, zero,
one, two, zero, nine, six, eight.
To the nearest year, we can say that the artifact
is approximately 4,223 years old.
I hope you found this helpful.