Welcome back to part two of the lesson
on exponential functions with base e.
For a quick review, an exponential function with base e
is in the form of f[x] equals a
times e raised to the power of kx,
where a is the initial value or starting amount.
If k is positive, or greater than zero,
then the function represents exponential growth
with a continuous growth rate of k,
and if k is negative, or less than zero,
the function represents exponential decay
with a continuous decay rate of absolute value of k.
Let's go and take a look at our second example.
We're told carbon-14 has a continuous decay rate
of 0.012097 percent per year.
Suppose 50 grams of carbon-14 is present.
Write the exponential function to model N of t,
the amount of carbon-14 present after t years.
How much carbon-14 will remain after 120 years?
And finally, if 20 grams of carbon-14 remain,
how old is the artifact?
So because we know we have a continuous decay rate here,
we can model the amount of carbon
using an exponential function with base e.
For part a, we'll have an exponential function
to form N of t...
equals a times e raised to the power of kt.
Remember, a is the initial or starting amount,
so in this case, because we're beginning
with 50 grams of carbon-14, a is equal to 50.
And now for k...
because we have a continuous decay rate or 0.012097 percent,
k is going to be negative 0.012097 percent.
But k must be expressed as a decimal,
so as a decimal, we would have negative 0.00012097.
When you convert a percent to a decimal,
remember we drop the percent sign
and move the decimal point to the left two places,
or divide by 100.
Which means the exponential function
to model the amount of carbon-14 present after t years
is going to be N of t equals 50
times e raised to the power of negative 0.00012097 times t.
And now I'll be using this exponential function,
our exponential decay function, to answer parts b and c.
Part b asks how much carbon-14 will remain after 120 years.
But since t is equal to the number of years,
we know that t equals 120.
So to determine the amount of carbon
present after 120 years,
we need to find the function value N of 120,
which would be 50 times e raised to the power
of negative 0.00012097 times 120.
So now we'll go to the calculator
and approximate this function value.
So we have 50, and then for e,
we can press Second Natural Log,
which brings up e raised to the power of,
let's go ahead and put this exponent in parentheses
so open parenthesis, negative 0.00012097, times 120.
Now in the newer version of 84, like I have here,
the parentheses around the exponent is not required,
but it is always nice to clarify
that product is the exponent.
We press Enter.
If we round to three decimal places,
we'd have approximately 49.279 grams
of carbon-14 present after 120 years.
Let's go and write this as a sentence.
Now for part c, we're asked
if 14 grams of carbon-14 remain,
how old is the artifact.
Now we're given N of t would determine the value of t.
Here we're given N of t equals 20 grams...
and we want to solve the equation,
20 equals 50 times e raised to the power
of negative 0.00012097 times t.
And we'll actually solve this two ways.
Just in case you don't know about logarithms yet,
we'll solve this graphically,
then we'll come back and solve this using logarithms.
So to solve this graphically,
we would enter the right side
in the graphing calculator in y1,
and we enter the left side in y2,
then we graph the two functions
to determine the point of intersection,
which will give us the value of t
that satisfies this equation.
So going back to the calculator, we'll press Y Equals,
we'll clear out any old functions here,
and in y1, we'll enter the right side of 50,
second natural log for e
raised to the power of negative 0.00012097...
and then times t, which we'll use x instead,
so here's the x key, Enter.
Then in y2, we enter 20, Enter.
Before we graph, we do want to adjust the window,
so let's press the Window key.
Now for the x values, let's go
from negative 500 to let's say 10,000...
and we'll scale it by 500s.
Now for the y axis, we know the maximum function value
would be 50 when t equals zero.
Let's scale the y axis from, let's say,
negative 10 to 60, and scale this by 10s.
Now let's press Graph.
If it doesn't look good, we can always come back
and change the scaling on the axes.
We'll press Graph.
So there's the exponential in the k function,
and there's y equals 20.
And notice how we do have
the point of intersection on the screen,
so our window is fine.
Then we'll press Second Trace for the calculation menu,
option five for intersection.
Now I'll press Enter three times.
Enter, Enter, Enter.
Rounding to two decimal places,
x is approximately 7,574.53 for in this case t.
So graphically, we now know that t
is going to be approximately 7,574.53,
and this would be years.
Again, let's also solve this
algebraically using logarithms.
So to solve this algebraically,
the first step would be to isolate the exponential part,
which means we would divide both sides by 50.
Simplifying, 20 divided by 50 is equal to 0.4.
We have 0.4 equals e raised to the power
of negative 0.00012097 t.
And here's where we're going to use logarithms,
and because we have a t here,
it'll be more convenient to use natural log.
So we'll take natural log on both sides of the equation,
and now on the right side, we're going to apply
the power property of logarithms,
which allows us to take this exponent here
and write this as a product, where we have
the exponent times the natural log.
So we have natural log 0.4...
equals, here we'd have negative 0.00012097 t
times natural log e.
But because e to the one equals e,
natural log e is equal to one,
therefore in order to solve for t,
we just need to divide both sides by this coefficient.
Now simplifying on the right, we just have t,
and then we'll approximate this quotient
on the graphing calculator.
Go back to the home screen,
and we'll press Natural Log 0.4
divided by negative 0.00012097.
Enter, and of course we get
the same answer that we did graphically,
approximately 7,574.53.
Again, this would be years.
So let's enter the question as a complete sentence.
If 20 grams of carbon-14 remain...
the artifact is approximately 7,574.53 years old.
Okay, I hope you found this helpful.