### Video Transcript

In this video, we will learn how to
identify write, evaluate, and analyze exponential functions.

An exponential function is a
function with a rule of the form π of π₯ equals π to the π₯, where the constant
real number π is called the base, with π greater than zero and π not equal to
one, and π₯ is called the exponent, which can be any real number. Letβs take a look at π of π₯ for
some positive integer values of π₯. For π₯ equals one, we have π to
the first power, which is just π. For π₯ equals two, we have π to
the second power or π squared, which is equal to π times π. For π₯ equals three, we have π to
the third power or π cubed equal to π times π times π. And for π₯ equals four, we have π
to the fourth power equal to π times π times π times π.

We can see that the previous value
of π of π₯ is multiplied by π every time π₯ increases by one. So π of two is equal to π of one
times π. π of three is equal to π of two
times π, and π of four is equal to π of three times π. This alludes to a more general
property of the exponential function. For the exponential function π of
π₯ equals π to the π₯, the value of π of π₯ is always the product of π of π₯
minus one and π, which means that π is always the quotient of π of π₯ and π of
π₯ minus one. That is, π of π₯ equals π of π₯
minus one times π, which in turn means that π equals π of π₯ divided by π of π₯
minus one. This holds true not only for
integer values of π₯ but also for any real number value of π₯. We can often use the relationship
between π, π of π₯, and π of π₯ minus one to find the value of π from a graph or
table.

Consider the graph which shows π¦
equals π of π₯ equals π to the π₯ for a specific value of π. Notice that the graph intersects
the π¦-axis at one. This will be the case for any real
number value of π such that π greater than zero and π not equal to one. We can find the value of π in the
function that the graph represents by selecting two points that the graph passes
through with π₯-coordinates that differ by one. It does not matter which points we
choose, but both should have π₯- and π¦-coordinates that are easy to identify. In this case, it is easiest for us
to select points with π₯-coordinates that are consecutive integers. So letβs pick π₯ equals negative
three and π₯ equals negative two. From the graph, we can see that
these have π¦-coordinates of eight and four, respectively. So we have π of negative three
equals eight and π of negative two equals four.

Recall from the previous definition
that the base π is equal to π of π₯ divided by π of π₯ minus one regardless of
which value of π₯ we choose. So we can choose our value of π₯ to
be negative two. So π₯ minus one is negative
three. So we have π equals π of negative
two over π of negative three. And substituting in these values
obtained from the graph, we get π equals four over eight. So we have π equals one-half. And therefore π of π₯ equals
one-half to the π₯.

Next letβs look at a table
representing the function π of π₯ equals π to the π₯ for another specific value of
π. For π₯ equals one, we have π of π₯
equals two. For π₯ equals two, we have π of π₯
equals four. For π₯ equals three, we have π of
π₯ equals eight. And for π₯ equals four, we have π
of π₯ equals 16. Since the values of π₯ in the table
differ by one, we could use the same procedure as before to determine the value of
π, finding either the quotient of four and two or eight and four or 16 and
eight. However, this time we will instead
substitute the first pair of π₯- and π¦-values in the table into the rule π of π₯
equals π to the π₯.

So substituting one for π₯ and two
for π of π₯, this implies that two equals π to the first power and therefore π
equals two. So far, we have only considered
exponential functions of this particular form. But exponential functions can take
a more general form than this. We can multiply the base by some
constant, π, and we can multiply the exponent by a constant, πΌ, and add a constant
to it, π½. And then finally, we can add a
constant to the end of the function, π. These are just the standard linear
transformations of a function where π stretches the graph of the function along the
π¦-axis by a factor of π, πΌ squashes the graph of the function along the π₯-axis
by a factor of πΌ, π½ translates the function along the π₯-axis by negative π½, and
π translates the function along the π¦-axis by π.

So some other examples of
exponential functions might include π of π₯ equals four times two to the π₯ or β of
π₯ equals π to the π₯ plus one or π equals nine to the four π₯ plus three. For π of π₯, the base is still
two, and the exponent is still π₯. For β of π₯, the base is π and the
exponent is π₯. And for π of π₯, the base is nine
and the exponent is four π₯ plus three. Letβs look at an example of how to
find an exponential functionβs base and exponent.

What are the base and exponent of
the function π of π₯ equals five to the power of π₯ minus five?

Recall that an exponential function
in its simplest form is of the form π of π₯ equals π to the π₯, where π is the
base and π₯ is the exponent. We know that the base π is a real
number, such that π is greater than zero and π is not equal to one and the
exponent π₯ can be any real number. In the given function, we have a
base of five and this is raised to the power of π₯ minus five. This still fulfills the criteria of
an exponential function. So the base is five and the
exponent is π₯ minus five.

In the next problem, we are given a
table of values for an exponential function and are asked to find the functionβs
equation.

Write an exponential equation in
the form π¦ equals π to the π₯ for the numbers in the table. π₯ equals two, four, and five with
corresponding π¦-values nine over 16, 81 over 256, and 243 over 1024.

Recall that an exponential function
in its simplest form takes the form of π¦ equals π to the power of π₯, where π is
the base, a real number greater than zero and not equal to one, and π₯ is the
exponent, which can be any real number. We therefore need to find the value
of π from the values in the table to solve the problem. Letβs begin by substituting one of
the pairs of π₯- and π¦-values from the table into the equation π¦ equals π to the
π₯.

Substituting in two for π₯ and its
corresponding π¦-value nine over 16, we get nine over 16 equals π squared. Solving for π, we get π equals
plus or minus the square root of nine over 16. And using surd rules, we get plus
or minus the square root of nine over the square root of 16. This simplifies to plus or minus
three-quarters. And since π must be greater than
zero, it must be the positive square root. Therefore, π equals
three-quarters. Substituting this value of π into
the equation for the function, we get our final answer π¦ equals three-quarters to
the power of π₯.

One way to verify our answer is to
check the other values in the table. Substituting in the other values of
π₯ and π¦, we get 81 over 256 equals π to the fourth and 243 over 1024 equals π to
the fifth. If we divide the second equation by
the first, we get 243 over 1024 divided by 81 over 256 equal to π to the fifth over
π to the fourth. The right-hand side here simplifies
to π to the five minus four, and π to the five minus four is just π to the one,
which is just π. And taking the quotient on the
left-hand side, we also get three-quarters.

In the process of checking our
answer, we used the quotient rule, which is one of the properties of exponents. And it states that when dividing
exponential expressions with the same base, we keep the base and find the difference
of the exponents. That is, π to the π over π to
the π equals π to the π minus π, where π is the base and π and π are the
exponents. For example, we would use the
quotient rule to show that five to the eight over five to the four equals five to
the eight minus four, which is equal to five to the four.

In the problem that follows, we are
again given a table of values for an exponential function and are asked to find the
functionβs equation. This time, however, the equation is
in a form other than π¦ equals π to the π₯.

Write an exponential equation in
the form π¦ equals π times π to the π₯ for the numbers in the table. For π₯ equal to zero, one, two, and
three, the corresponding π¦-values are 18, six, two, and two-thirds.

In this problem, we are asked to
write an exponential equation in the form π¦ equals π times π to the π₯ for the
numbers in the table. So we must find the values of both
π and π to solve the problem. Letβs begin by substituting one of
the pairs of π₯- and π¦-values from the table into the equation π¦ equals π times
π to the π₯. Substituting in the first pair of
values, so π₯ equals zero and π¦ equals 18, we get 18 equals π times π to the
zero. Since any base raised to the power
of zero is equal to one, the equation simplifies to π equals 18.

We still need to find the value of
π, however, and we can do this by substituting in another pair of values of π₯ and
π¦. So substituting in the next pair of
values, π₯ equals one and π¦ equals six, we get six equals π times π to the
one. Any base π raised to the power of
one is just itself, so this simplifies to six equals ππ. We already found the value of π to
be equal to 18. So we can substitute this into the
equation to give six equals 18π. Solving for π gives π equals
one-third. Substituting these values of π and
π into the exponential equation gives us our final answer π¦ equals 18 times
one-third to the π₯.

We can verify our answer by
substituting the last two π₯-values from the table into the equation to make sure
that they give the correct π¦-values. Substituting in π₯ equals two, we
get π¦ equals 18 times one-third squared, which is equal to 18 times one-ninth which
is equal to two, which is correct. And substituting in π₯ equals three
gives π¦ equals 18 times one-third cubed, which is equal to 18 times one over 27,
which is equal to two-thirds, which is again correct.

In the process of solving the last
problem, we used one of the properties of exponents called the zero-exponent rule,
which states that any base raised to the power of zero is equal to one. That is, π to the zero equals one,
where π is the base. For example, three to the zero is
equal to one.

In the next problem, we will see
how to find the equation of an exponential function from its graph.

Observe the given graph and then
answer the following questions. Find the π¦-intercept in the shown
graph. As this graph represents an
exponential function, every π¦-value is multiplied by π when π₯ increases by
Ξπ₯. Find π for Ξπ₯ equals one. Find the equation that describes
the graph in the form π¦ equals ππ to the power of π₯ over Ξπ₯.

Letβs begin by finding the graphβs
π¦-intercept. We can see that the line of the
graph passes through the point zero, 10. Therefore, the π¦-intercept is
10. Now, letβs find the value of π in
the equation of the graph. In the problem, we are told that
the graph represents an exponential function for which every π¦-value is multiplied
by π when π₯ increases by Ξπ₯. And weβre asked to find π when Ξπ₯
equals one. Weβve already seen that the graph
passes through the point zero, 10. And we can also see that it passes
through the point one, 20. The change in π₯ between these two
points is one, so it satisfies Ξπ₯ equals one.

We are told that the π¦-value is
multiplied by π when π₯ increases by Ξπ₯. If the π¦-value of our first point
is π¦ one and the π¦-value of our second point is π¦ two, then π equals π¦ two over
π¦ one which is equal to 20 divided by 10, which is equal to two. And finally, we need to find the
equation that describes the graph in the form π¦ equals ππ to the power of π₯ over
Ξπ₯. We are already given Ξπ₯ equals
one, and we have found the value of π equal to two. Therefore, π¦ equals π times two
to the π₯ over one, which is just equal to π times two to the π₯. To find the value of π, we can
substitute the π₯- and π¦-values of points we know to be on the graph.

Weβve already found the
π¦-intercept to be 10, so we know that the point zero, 10 lies on the graph. This means we can substitute the
values of π₯ equals zero and π¦ equals 10 into the equation which gives us 10 equals
π times two to the zero. By the zero-exponent rule, two to
the zero is just equal to one. Therefore, this gives π equals
10. This gives us our equation and the
final answer π¦ equals 10 times two to the π₯.

In the next problem, we determine
the value of an expression by evaluating an exponential function.

Given that π of π₯ equals four to
the π₯, determine the value of π of π₯ over π of π₯ minus one minus π of π₯ minus
one over π of π₯.

In this problem, we are asked to
find the value of an expression for π₯ without being given a value for π₯, hinting
that this expression has the same value regardless of the value of π₯. We are given that π of π₯ equals
four to the π₯. Therefore, π of π₯ minus one
equals four to the π₯ minus one. Therefore, π of π₯ over π of π₯
minus one equals four to the π₯ over four to the π₯ minus one. Using the quotient rule, this is
equal to four to the π₯ minus π₯ minus one, which is equal to four to the one which
is equal to four. This is also the value of the base
of the exponential function π of π₯, which is defined by π of π₯ over π of π₯
minus one.

The second part of the expression
is π of π₯ minus one over π of π₯, which is just the reciprocal of what we just
found, one over π of π₯ over π of π₯ minus one. Therefore, this is equal to one
over four or one-quarter. Therefore, π of π₯ over π of π₯
minus one minus π of π₯ minus one over π of π₯ equals four minus one-quarter,
which equals 15 over four.

The previous problem made use of
the negative-exponent rule, which states that any base raised to a negative power is
equal to one over the base raised to the exponentβs additive inverse. That is, π to the negative π
equals one over π to the π, where π is the base and π and negative π are the
exponents. For example, six to the negative
two is equal to one over six to the positive two, or six squared, which is equal to
one over 36.

In the final problem, we will
determine the values of an exponential functionβs base that result in the function
decreasing without being told the form of the function.

Consider an exponential function
with base π. For which values of π is the
function decreasing?

In this problem, we are given an
exponential function π of π₯ equals π to the π₯. Recall that the base π may be
given by π of π₯ over π of π₯ minus one. Recall that π is a constant real
number with π greater than zero and not equal to one. This means that regardless of the
value of π₯, the value of π of π₯ over π of π₯ minus one is always a constant. Therefore, for any value of π₯, the
increase in π of π₯ between π₯ minus one and π₯ is always π.

Since an exponential function is
always positive, π of π₯ over π of π₯ minus one being greater than one implies
that π of π₯ is greater than π of π₯ minus one and, therefore, that the function
is increasing. And likewise, π of π₯ over π of
π₯ minus one being less than one implies that π of π₯ is less than π of π₯ minus
one and, therefore, that the function is always decreasing.

Since the left-hand side of both of
these inequalities is equal to π, this means that π greater than one implies the
function increases and π less than one implies it decreases. Remember also that π must be
greater than zero, so this gives us our final answer. π greater than one implies the
function increases and π between zero and one implies the function decreases.

The last problem alluded to another
property of exponential functions, their monotonicity. An exponential function is a
monotonic function or a function that always increases or always decreases. If the base π is greater than one,
it is called the growth factor, and the function will always increase. If the base is greater than zero
and less than one, it is called the decay factor, and the function will always
decrease.

Letβs finish by recapping some key
points from this video. An exponential function has the
form π of π₯ equals π to the π₯, where π is the base, which is a positive real
number not equal to one, and π₯ is the exponent, which may be any real number. The quotient rule states that when
dividing exponential expressions with the same base, we keep the base and find the
difference of the exponents. That is, π to the π over π to
the π equals π to the π minus π. The zero-exponent rule states that
any base raised to the power of zero is equal to one. That is, π to the zero equals
one.

The negative-exponent rule states
that any base raised to a negative power is equal to one over the base raised to the
exponentβs additive inverse. That is, π to the negative π
equals one over π to the π. All exponential functions are
monotonic functions; that is, they always increase or always decrease. Specifically, if π is greater than
one, it is called the growth factor, and the function always increases. And if π is between zero and one,
it is called the decay factor, and the function always decreases.