### 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.