Welcome to a lesson
on exponential functions.
An exponential function is function in the form:
f of x equals a times b, raise the power of x
where a can't equal zero, b must be greater than zero
and b can't equal one.
If a was equal to zero or b was equal to one,
we would have a constant function,
not an exponential function.
The value of a is called the initial value or starting value
which is the function value when the input x
is equal to zero
and therefore a also gives us the vertical intercept
which would be the ordered pair, zero comma a.
The base b is called the growth factor
or decay factor, where the base b equals one plus r,
where r is equal to the growth rate per unit of time
as a decimal.
So if the base b is greater than one,
then r, the growth rate, would have to be greater than zero
or positive,
and therefore the function represents exponential growth
which is the function increasing
at a constant percentage with a growth rate of r.
Now if the base b is between zero and one,
that means r the growth rate would be negative,
or less than zero,
and therefore the function represents exponential decay
which is the function decreasing at a constant percentage
and the decay rate is equal to the opposite value of r.
Now take a look at expontential growth
and decay graphically.
Here we have the graph of f of x equals three times two,
raise the power of x.
Notice how the base b is greater than one
and therefore we have expontential growth,
meaning as x increases the function values also increase.
Now because we know the base b is equal to one plus r
it's also true that r, the growth rate,
is equal to b minus one.
So in this case notice that b is equal to two
and therefore r would be equal to two minus one, or one,
so if r is equal to one,
that means we have a growth rate of 100%
which means we have a doubling function
because notice how the base is two.
Looking at the table of values
we can also see that as x increases
f of x also increases
and it actually doubles as x increases by one.
Now looking at the function here on the right
notice how the base b is between zero and one
and therefore we have expontential decay
and therefore we have a function that decreases
at a constant percentage
so as x increases the function values decrease.
And in this case, because the base b is equal to one-half,
we can say that r equals b minus one
or in this case r equals one-half minus one,
which equals negative one-half.
Notice here the growth rate is negative 0.5,
or negative 50%,
which again means we have expontential decay,
we have a decay rate of 50%.
And because the base is one-half,
we can say this is the halving function.
Notice as x increases, f of x decreases,
in fact as x increases by one
the function values decrease by 50%
or by half.
One more thing to notice about both of these functions,
is notice that a, the initial value is equal to three,
which also gives us the vertical intercept
for both functions.
Now let's look at some applications
of expontential functions.
Here were told the population of Flagstaff, Arizona
was approximately 65,000 in 2010.
It was growing at a rate of 3.9% per year.
We're asked to write the expontential function
to model the population.
Let t be the number of years since 2010
and p of t be the population in thousands.
Then we'll predict the population in the year 2017
and then predict when the population will reach 100,000.
For part a, our function p of t
is going to be in the form of a times b
raise the power of t,
again because we have a function of t,
where a is equal to the initial or starting amount.
The base b equals one plus r,
where r is equal to the growth rate.
So the first thing we might notice here
is that r is going to be 3.9%
which as a decimal
would be 0.039.
Now let's fine the value of a, the initial amount.
Because t is the number of years since 2010
the population of 2010
will help us determine the initial amount, a.
Notice how the population in 2010 is 65,000,
but because the function value, p of t,
is the population in thousands,
a is not 65,000,
a is 65.
This is all we need
in order to write our expontential function.
We would have p of t,
equals 65 times,
again the base is one plus r,
so we have one plus 0.039,
raise the power of t,
for simplifying, population function p of t
is equal to 65 times 1.039,
raise the power of t.
And now we can use this function
to answer parts b and c.
Part b, we're asked to predict the population
in the year 2017.
Remember t is the number of years after 2010,
so t is going to be equal to seven.
And if you have a hard time determining t
we can always determine t
by starting with the desired year
and subtracting the base year
which will give us the correct value of t
which in this case is seven.
So we want to find the function value p of seven
to predict the population in the year 2017.
So we have 65 times 1.039,
to the power of seven,
and now we'll go to the calculator
to approximate this function value,
remember p of seven is going to be in thousands.
So we have 65
and then in parenthesis we have 1.039
which is raised the power of seven.
Enter.
So because this function value is in thousands,
we must round this to three decimal places,
which will be approximately 84.962.
Again this is in thousands,
so if we want to convert this to the number of people
we would multiply it by 1,000
which would give us 84,962 people.
So as a complete sentence
we can say in 2017,
the population is predicted to be 84,962.
Now for part c,
we're asked to predict
when the population will reach 100,000?
We're trying to find the year in which the population
is going to reach 100,000
so we're given a function value
and we want to find the value of t.
But because p of t is in thousands
we need to recognize here
that we're trying to solve this equation
when p of t equals 100,
because p of t equal to 100
does represent a population of 100,000.
We want to solve the equation 100 equals 65
times 1.039, raise the power of t.
And we'll solve this two ways,
we'll first solve this graphically,
just in case you haven't learned about logarithms yet
and then we'll solve it a second way using logarithms.
To solve this graphically
we would use the graphing calculator
and enter the right-side in y one,
then the left-side in y two,
then determine the point of intersection
of y one and y two
to determine the value of t
that satisfies this equation.
So with the calculator,
I've already done some of this to save some time,
we press y equals,
notice how I've entered the right-side into y one,
though I used x instead of t, for the variable,
and I entered the left-side in y two.
Now before we graph this
we want to make sure we can determine
the point of intersection
so we do have to adjust the window,
which again I've already done to save time.
Press the window key,
notice how I changed the x axis,
or the horizontal axis,
from negative five to 40, by fives,
and the y axis from negative 20 to 140, by 20s.
Now we'll go ahead and press Graph,
so there's the expontential function
and here's the constant function,
y equals 100.
Our goal here is to find the x coordinate,
or in this case the t coordinate
of this point of intersection.
So we'll press Second Trace for the Calculation menu,
Option five for Intersection
and now we can just press Enter three times,
so Enter, Enter, Enter.
Notice how we get t or in this case x,
is approximately 11.26.
So graphically we know the solution
is t is approximately 11.26.
But now let's also solve this using logarithms.
To solve this algebraically using logarithms
the first step is to isolate the expontential part
by dividing both sides by 65.
100 divided by 65 simplifies to 20/13ths.
So we have 20/13ths equals 1.039,
raise the power of t
and here's were we'll have to logarithms
in order to solve for t.
Because we'll be using the calculator
we'll using either a common log or natural log.
In this example let's use the common log,
so we'll take the common log
or log base 10 on both sides of the equation.
So now we have log of 20/13ths
equals, now on the right side we can apply
the power property of logarithms shown here below
where we can take this exponent here of t
and write this as a product.
We'd would have t
times the common log of 1.039
and therefore in order to solve for t here
we would divide both sides by common log
of 1.039.
Simplifying on the right,
this simplifies nicely to t.
And now we'll approximate this quotient on the calculator
which again we know should be 11.26.
So going back to the home screen,
we have common log of 20/13ths,
divided by the common log of 1.039,
Enter.
And notice how we do get the same value for t,
approximately 11.26.
But this tells as that we predict the population
will reach 100,000 approximately 11.26 years
after 2010.
Now determining what year this would actually be in
can be a little bit tricky.
If t was equal to 11,
then that would be the year 2010 plus 11,
which equals 2021.
But this would be the end of the year 2021,
so for any value greater than 11 and less than 12,
it would actually be sometime in the following year of 2022.
So to put this in a sentence,
we'll say we predict the population
to reach 100,000
in 2022.
Again this idea of what year it is
can be a little bit tricky,
because sometimes some textbooks
actually would take the value of t
which is approximately 11.26
and round to the nearest year of 11
and say the year is 2021.
But any value more than 11 and less than 12
would actually be in the following year
which would be the year 2022.
I think we'll stop here for part one.
In part two we'll take a look at an example
of expontential decay.
I hope you found this helpful.