### Video Transcript

In this video, we will learn how to
simplify monomials involving single and multiple variables using the quotient
rule. We will begin by recalling what we
mean by exponents or indices and then define the quotient rule of exponents.

Letβs firstly consider the
expression four to the fifth power or four to the power of five. The four is called the base and the
five is called the exponent. These are sometimes known as
indices or powers. The exponent tells us the number of
times that the base has been multiplied by itself. In this case, four to the power of
five is equal to four multiplied by four multiplied by four multiplied by four
multiplied by four. There are five fours. If we evaluate this, we get
1,024. This can also be done using a
scientific calculator as shown.

Often in mathematics, weβre asked
to simplify expressions involving exponents but are not always required to evaluate
them. This is particularly true with
large exponents where the calculation would be long and cumbersome. Letβs look at how we can simplify
an expression that involves a quotient of two exponential expressions.

Simplify five to the sixth power
divided by five cubed.

We can begin this question by
writing the top and bottom of the expression in expanded form. Five to the sixth power is the same
as six fives multiplied together. Five cubed or five to the third
power is the same as three fives multiplied together. We recall that dividing the top and
bottom of a fraction by the same number does not change its value. We can therefore divide the top and
bottom of this expression by five three times. Three of the fives on the numerator
will cancel along with three of the fives on the denominator. This leaves us with five multiplied
by five multiplied by five, which is the same as five cubed or five to the third
power.

We have in effect reduced each of
the exponents by three. This leads us to a general rule if
we have a quotient. π₯ to the power of π divided by π₯
to the power of π is equal to π₯ to the power of π minus π. When dividing or finding the
quotient of two exponential terms, we can subtract the exponents or powers. This can also be written as π₯ to
the power of π divided by π₯ to the power of π is equal to π₯ to the power of π
minus π. This is known as the quotient rule
of exponents.

We will now look at some examples
that we can solve using this rule.

Simplify π₯ to the sixth power
divided by π₯ to the fourth power.

We recall that the quotient rule of
exponents states that π₯ to the power of π divided by π₯ to the power of π is
equal to π₯ to the power of π minus π. When finding the quotient of two
exponential terms with the same base, we can subtract the exponents or powers. We can therefore work out the
simplified version of our expression by subtracting four from six. This means that π₯ to the sixth
power divided by π₯ to the fourth power is equal to π₯ squared.

An alternative way of solving this
problem if we didnβt recall the rule would be to write both terms out in full. π₯ to the sixth power is the same
as six π₯βs multiplied together. In this case, we have used a dot to
represent βmultipliedβ so we donβt get confused with the letter x. π₯ to the fourth power can be
written in the same way. We can then divide the numerator
and denominator by π₯ four times which in effect cancels four of the π₯βs on the top
and bottom. Weβre left with π₯ multiplied by π₯
which, once again, is equal to π₯ squared.

This method is fine if our
exponents are small. However, it becomes more cumbersome
if weβre dealing with larger exponents. This will be the case in our next
example.

Simplify four to the 17th power
divided by four to the ninth power.

We could begin this question by
writing both the numerator and denominator out in full. The numerator would consist of 17
fours being multiplied together. This method would be very time
consuming. So instead, we will use the
quotient rule for exponents. This states that π₯ to the power of
π divided by π₯ to the power of π is equal to π₯ to the power of π minus π. When finding the quotient of two
exponential terms with the same base, we can subtract the exponents. This means that four to the 17th
power divided by four to the ninth power can be rewritten as four to the power of 17
minus nine. As 17 minus nine is equal to eight,
our answer becomes four to the eighth power.

Our next two examples involve more
complicated problems where there are more than two terms.

Simplify π₯ to the 23rd power
multiplied by π₯ to the 35th power divided by π₯ to the 17th power where π₯ is not
equal to zero.

In order to solve this problem, we
need to recall two of our rules or laws of exponents or indices. Firstly, we have the product
rule. This states that π₯ to the power of
π multiplied by π₯ to the power of π is equal to π₯ to the power of π plus
π. The quotient rule, on the other
hand, states that π₯ to the power of π divided by π₯ to the power of π is equal to
π₯ to the power of π minus π. When multiplying two terms with the
same base, we add the exponents, whereas when dividing, we subtract the
exponents.

We usually begin a question of this
type by simplifying the numerator and denominator first. In this question, we need to
simplify π₯ to the 23rd power multiplied by π₯ to the 35th power. As the two terms are being
multiplied, we need to add the exponents. 23 plus 35 is equal to 58. Therefore, the numerator simplifies
to π₯ to the 58th power. Our expression is therefore
simplified to π₯ to the 58th power over or divided by π₯ to the 17th power. As weβre dividing here, we need to
subtract the exponents. 58 minus 17 is equal to 41. π₯ to the 23rd power multiplied by
π₯ to the 35th power divided by π₯ to the 17th power is equal to π₯ to the 41st
power.

An alternative method here to keep
the arithmetic as simple as possible would be to divide π₯ to the 23rd power by π₯
to the 17th power first. This wouldβve given us π₯ to the
sixth power multiplied by π₯ to the 35th power. As six plus 35 is equal to 41, this
wouldβve given us the same answer.

Our next question is a problem
involving multiple variables.

Simplify π₯ to the fourth power π¦
to the fourth power multiplied by π₯ squared π¦ to the fourth power divided by π₯ to
the fourth power π¦ cubed.

This expression involves two
variables, π₯ and π¦, which we will treat separately. We also need to recall two of our
rules of exponents. These are the products in quotient
rule. The product rule states that π₯ to
the power of π multiplied by π₯ to the power of π is equal to π₯ to the power of
π plus π. The quotient rule states that π₯ to
the power of π divided by π₯ to the power of π is equal to π₯ to the power of π
minus π.

When multiplying two terms with the
same base, we add the powers. And when dividing, we subtract
them. Letβs consider our π₯-variable
first. On the top or numerator, we have π₯
to the fourth power multiplied by π₯ squared. And on the denominator, we have π₯
to the fourth power. We might notice here that we have
π₯ to the fourth power on the numerator and denominator. So we can divide both of these by
this term. This leaves us with π₯ squared. Alternatively, we couldβve added
the exponents on the numerator. Four plus two is equal to six. We couldβve then subtracted the
exponents on the denominator. And six minus four is equal to
two.

When considering the π¦-variable,
we have π¦ to the fourth power multiplied by π¦ to the fourth power divided by π¦
cubed or π¦ to the third power. Simplifying the numerator using the
product rule gives us π¦ to the eighth power as four plus four is equal to
eight. We can then use the quotient rule
to simplify π¦ to the eighth power divided by π¦ to the third power. Eight minus three is equal to
five. So our π¦-variable simplifies to π¦
to the fifth power. This means that the overall
expression simplifies to π₯ squared π¦ to the fifth power. Whilst the two variables could be
written in either order, we tend to follow the same format as the question which is
usually alphabetical order.

Our final question is a problem
involving negative exponents.

Simplify π₯ cubed or π₯ to the
third power divided by π₯ to the sixth power.

We recall that the quotient rule of
exponents states that π₯ to the power of π divided by π₯ to the power of π is
equal to π₯ to the power of π minus π. If the base is the same, we can
subtract the exponents. This means that, in our question,
π₯ cubed divided by π₯ to the sixth power is equal to π₯ to the power of three minus
six. As three minus six is equal to
negative three, our expression simplifies to π₯ to the power of negative three.

Letβs use another method to solve
this problem to understand what a negative power or exponent means. π₯ cubed is the same as π₯
multiplied by π₯ multiplied by π₯. π₯ to the sixth power involves six
π₯βs being multiplied together. This expression can be simplified
by dividing the numerator and denominator by π₯. We can repeat this three times.

The numerator has therefore
simplified to one and the denominator to π₯ multiplied by π₯ multiplied by π₯, which
is π₯ cubed. π₯ cubed divided by π₯ to the sixth
power can therefore be written as one divided by or over π₯ cubed. These two results lead us to the
general rule for negative exponents. This states that π₯ to the power of
negative π is equal to one over π₯ to the power of π or one over π₯ to the πth
power.

We will now summarize the key
points from this video. If you have a quotient of two
exponential expressions that have the same base, then we can use the quotient rule
of exponents to simplify the expression. π₯ to the power of π divided by π₯
to the power of π is equal to π₯ to the power of π minus π. This can also be written with a
division sign instead of in fractional form. For example, two to the eighth
power divided by two to the fifth power is equal to two to the power of eight minus
five, which equals two cubed.

In this video, we also used the
product rule of exponents. This states that π₯ to the power of
π multiplied by π₯ to the power of π is equal to π₯ to the power of π plus
π. When multiplying, we add our
exponents, whereas when we are dividing, we subtract them. The final question that we looked
at also led us to the rule of negative exponents, which states that π₯ to the power
of negative π is equal to one over π₯ to the power of π.