### Video Transcript

In this video, we will learn how to
identify the base and exponent in power formulas, write them in exponential and
expanded forms, and evaluate simple powers.

We begin by recalling that we can
represent repeated multiplication as a power. For example, two to the fifth power
is defined as the product of five twos as shown. We call two the base and five the
exponent. We can extend this definition to
general rational bases. In this case, if π is a positive
integer and π over π is a rational number, then π over π to the πth power will
be the product of π lots of π over π. For example, we can evaluate
one-half to the third power, or one-half cubed, by multiplying one-half by one-half
and one-half again. Recalling our rules for multiplying
fractions, we simply multiply the numerators and denominators separately. This gives us one over eight, or
one-eighth.

We can also follow the same process
in reverse. Letβs imagine we want to write the
fraction 27 over eight in exponential form. We begin by factoring the numerator
and denominator into primes as follows. 27 is equal to three multiplied by
three multiplied by three, and eight is equal to two multiplied by two multiplied by
two. We can then split the
multiplication, giving us three over two multiplied by three over two multiplied by
three over two. As such, 27 over eight is the
product of three lots of three over two and can therefore be written as three over
two cubed.

Letβs now consider some examples
involving the powers of rational numbers.

What terminology do we use to
describe the half in the expression one-half to the fifth power and the five in the
expression one-half to the fifth power?

We begin by recalling that an
expression of the form π to the πth power is called an exponential expression or
the πth power of π. We call π the base of the
expression and π the exponent or power. In this question, we have the
expression one-half to the fifth power. The one-half is the number that is
being taken to a power, and the five is the power itself. We can therefore conclude that in
the expression one-half to the fifth power, one-half is called the base of the
expression and five is called the exponent of the expression.

In our next example, we will
simplify an expression by rewriting it in exponential form.

What is four over 11 multiplied by
four over 11 multiplied by four over 11 multiplied by four over 11 multiplied by
four over 11 multiplied by four over 11 multiplied by four over 11? Is it option (A) four elevenths to
the power of negative seven? (B) Four elevenths to the power of
seven. (C) Four elevenths to the power of
nine. (D) Seven elevenths to the power of
four. Or (E) twenty-eight elevenths to
the power of seven.

We could evaluate this expression
by multiplying all of the numerators and all of the denominators. This would give us the following
expression which we could calculate with or without a calculator. However, the five options in this
question are given as powers. This means that instead of
evaluating the expression, we can simplify by recalling that repeated multiplication
can be written in exponential form. In particular, in this question, we
are multiplying seven lots of four elevenths. We know that the product of seven
lots of four elevenths can be written by raising four elevenths to an exponent of
seven. And as such, we can conclude that
the correct answer is option (B). The expression in the question is
equivalent to four elevenths to the power of seven.

We will now use the information we
have seen so far to define a key property of the powers of rational numbers. Since a positive integer power of a
rational base is defined by repeated multiplication, we can show that if π is a
positive integer and π over π is a rational number, then π over π to the πth
power is equal to π to the πth power divided by π to the πth power. In other words, we can raise the
numerator and denominator to the power separately.

Letβs now consider an example where
we can apply this.

Find the value of negative
six-fifths cubed, giving your answer in its simplest form.

We can evaluate the expression in
this question in two ways. Firstly, we recall that negative
six over five cubed can be written as repeated multiplication. It is the product of three lots of
negative six over five as shown. We can then multiply the numerators
and denominators separately, giving us negative six multiplied by negative six
multiplied by negative six over five multiplied by five multiplied by five. Multiplying three negative numbers
gives a negative answer, so the numerator is negative 216. The denominator simplifies to
125. The value of negative six over five
cubed is negative 216 over 125.

We can also evaluate the expression
by recalling the general result for powers of rational numbers. If π is a positive integer and π
over π is a rational number, then π over π to the πth power is equal to π to
the πth power divided by π to the πth power. This means that we can rewrite
negative six over five cubed as negative six cubed over five cubed, which once again
simplifies to negative 216 over 125.

In our next example, we will
consider a problem in context.

Find an expression for the volume
of the given cube whose side lengths are two π₯ over five.

We begin by recalling that the
volume of a cube is given by the cube of its side length. This means that if a cube has side
length π, then its volume is π multiplied by π multiplied by π, which is π
cubed. In this question, we are told that
the side length is two π₯ over five. This means that its volume is given
by the expression two π₯ over five cubed. Recalling that for any rational
number π over π and any integer π π over π to the πth power is equal to π to
the πth power divided by π to the πth power, our expression simplifies to two π₯
cubed over five cubed.

Next, we recall that to multiply
monomials, we multiply the coefficients and add the powers of the shared
variables. This means that two π₯ multiplied
by two π₯ multiplied by two π₯ is equal to eight π₯ cubed. And since five cubed is 125, our
expression simplifies to eight π₯ cubed over 125. This is an expression for the
volume of a cube whose side lengths are two π₯ over five.

We will now consider one final
example where we need to evaluate an algebraic expression using the results for the
powers of rational numbers.

If π₯ is equal to three over two
and π¦ is equal to negative four over five, find the value of π₯ squared π¦ minus
π₯π¦ cubed, giving your answer as a fraction in simplest form.

We will begin this question by
substituting the values of π₯ and π¦ into our expression. π₯ squared π¦ minus π₯π¦ cubed is
therefore equal to three over two squared multiplied by negative four over five
minus three over two multiplied by negative four over five cubed. Our next step is to evaluate the
powers by recalling that if π is a positive integer and π over π is a rational
number, then π over π to the πth power is equal to π to the πth power over π
to the πth power. This means that three over two
squared is equal to three squared over two squared, which is equal to nine over
four. In the same way, negative four over
five cubed is equal to negative four cubed over five cubed. And this is equal to negative 64
over 125. Our expression therefore simplifies
to nine over four multiplied by negative four over five minus three over two
multiplied by negative 64 over 125.

We can simplify the first part of
our expression by dividing the numerator and denominator by four. In the same way, we can divide the
numerator and denominator of the second part of our expression by two. And our expression simplifies to
negative nine-fifths minus negative 96 over 125. This in turn simplifies to negative
nine over five plus 96 over 125. Finally, in order to add fractions,
we must have the same denominators, so we multiply the numerator and denominator of
our first fraction by 25. The expression becomes negative 225
over 125 plus 96 over 125. Finally, we add the numerators,
giving us negative 129 over 125. Since there are no shared factors
apart from one, this is the value of π₯ squared π¦ minus π₯π¦ cubed as a fraction in
its simplest form.

We will now finish this video by
recapping the key points. We saw that in an expression of the
form π to the πth power, we call π the base and π the power or exponent. We define positive integer powers
by repeated multiplication, known as expanded form. In general, if π over π is a
rational number and π is a positive integer, then π over π to the πth power is a
product of π lots of π over π as shown. Finally, we saw that we can
evaluate the power of a rational number by evaluating the power of the numerator and
denominator separately. If π is a positive integer and π
over π is a rational number, then π over π to the πth power is equal to π to
the πth power divided by π to the πth power.