Explainer: Simplifying Monomials: Negative Exponents

In this explainer, we will learn how to simplify monomials with negative exponents.

Let us start by recalling what we know about positive exponents. For an expression 24, this means 2ร—2ร—2ร—2.

Here, the exponent, 4, is informing us of the number of times 2 has been multiplied by itself. Equally, if we consider the expression ๐‘ฅ๐‘›, if ๐‘› is a positive integer, the ๐‘› is telling us the number of times ๐‘ฅ is multiplied by itself. What then happens if we have a negative exponent, such as 2โˆ’3?

How do we multiply a number by itself โˆ’3 times? This, of course, is not a sensible way of introducing negative exponents. Let us start by introducing a pattern of reducing powers of 2: 24=16,23=8,22=4,21=2,20=1.

Looking at this pattern, it looks like each reduction of the exponent, of 2, by one results in the number on the right-hand-side being divided by 2. If we were to continue this pattern, it would be reasonable to assume that an exponent of โˆ’1 (that is, 2โˆ’1) would be equal to 1รท2, which is 12. Similarly, an exponent of โˆ’2 would be 12รท2, which equals 14. We could continue the list as follows: 24=16,23=8,22=4,21=2,20=1,2โˆ’1=12,2โˆ’2=14,2โˆ’3=18,2โˆ’4=116.

Is our assumption correct? Let us see if we can justify this by using the quotient rule for exponents. Recall that ๐‘ฅ๐‘Ž๐‘ฅ๐‘=๐‘ฅ๐‘Žโˆ’๐‘.

If we consider the expression 2326,

using the quotient rule of exponents, we can simplify this to 23โˆ’6=2โˆ’3.

Equally, we can rewrite the expression as 2ร—2ร—22ร—2ร—2ร—2ร—2ร—2

by expanding the exponents. We can then divide the top and bottom by 2 three times, which gives us 2ร—2ร—22ร—2ร—2ร—2ร—2ร—2.

This simplifies to 123,

which can be evaluated to 18.

If we refer back to our list of exponents of 2, we can see that our assumption, at least for the value of 2โˆ’3, is indeed correct. We could take a very similar approach to show that all of our list of powers of 2 was correct. One point to highlight in the above construction is the fact that we showed 2โˆ’3=123.

This can be generalized for any nonzero ๐‘ฅ raised to a negative exponent.

Key Information: Negative Exponents

For any nonzero ๐‘ฅ, we have that ๐‘ฅโˆ’๐‘˜=1๐‘ฅ๐‘˜.

This is, in fact, a special case of the quotient rule of exponents. Recall again that ๐‘ฅ๐‘Ž๐‘ฅ๐‘=๐‘ฅ๐‘Žโˆ’๐‘.

In the above generalization, ๐‘Ž=0 and ๐‘=๐‘˜. We have, therefore, that ๐‘ฅ0โˆ’๐‘˜=๐‘ฅ0๐‘ฅ๐‘˜.

As ๐‘ฅ0=1, for any nonzero ๐‘ฅ, we can simplify this to 1๐‘ฅ๐‘˜.

Before looking at a few examples, let us state all the rules of exponents that are often used in combination with what we have just discovered about negative exponents. We have the product rule, the quotient rule, the power rule, and the zero exponent property.

Key Information: Exponent Rules

  1. Product rule of exponents: ๐‘ฅ๐‘Žร—๐‘ฅ๐‘=๐‘ฅ๐‘Ž+๐‘
  2. Quotient rule of exponents: ๐‘ฅ๐‘Žรท๐‘ฅ๐‘=๐‘ฅ๐‘Žโˆ’๐‘
  3. Power rule of exponents: (๐‘ฅ๐‘Ž)๐‘=๐‘ฅ๐‘Ž๐‘
  4. Zero exponents: ๐‘ฅ0=1

Example 1: Evaluating a Number Raised to a Negative Exponent

Evaluate 14โˆ’2.

Answer

Recall that for any nonzero ๐‘ฅ, we have that ๐‘ฅโˆ’๐‘Ž=1๐‘ฅ๐‘Ž.

Using this, we can rewrite the expression to be 14โˆ’2=1142.

We can now evaluate 142, which equals 196, to find that the original expression is equal to 1196.

Example 2: Evaluating Expressions Containing Negative Exponents

Evaluate 2โˆ’224.

Answer

Recall that for any nonzero ๐‘ฅ, we have that ๐‘ฅ๐‘Ž๐‘ฅ๐‘=๐‘ฅ๐‘Žโˆ’๐‘,

and that ๐‘ฅโˆ’๐‘Ž=1๐‘ฅ๐‘Ž.

Using the quotient rule of exponents, we can rewrite our expression as follows: 2โˆ’224=2โˆ’2โˆ’4=2โˆ’6.

Using the negative exponent rule, we can rewrite this as 2โˆ’6=126.

At this point, we can evaluate the expression. Using the fact that 26=64, we have a final solution of 126=164.

Example 3: Recognizing Equivalent Expressions Involving Negative Exponents

Which of the following is equivalent to 3โˆ’3?

  1. 27
  2. โˆ’9
  3. โˆ’19
  4. 127
  5. 19

Answer

Here, we have a number raised to the power of โˆ’3. A common mistake would be to evaluate this to โˆ’9, which is the product of the two numbers. This, however, is not correct. We need to recall the negative exponent rule: for any nonzero ๐‘ฅ we have that ๐‘ฅโˆ’๐‘Ž=1๐‘ฅ๐‘Ž.

We can use this to rewrite our expression: 3โˆ’3=133.

This, however, is not the same as any of the options. We need to evaluate the denominator of the expression: 133=127.

Our answer is, therefore, D.

All of the rules that we have been using for the previous examples can be applied to both numerical and algebraic expressions. Let us now move on to look at some algebraic examples.

Example 4: Evaluating Algebraic Expressions Containing Negative Exponents

Simplify ๏€น5๐‘ฅโˆ’8๏…2๏€น6๐‘ฅ2๏…2.

Answer

To answer this question, let us first recall the power rule of exponents: (๐‘ฅ๐‘Ž)๐‘=๐‘ฅ๐‘Ž๐‘.

If we distribute the power across each of the terms and apply the power rule of exponents, we get ๏€น52๐‘ฅโˆ’8ร—2๏…๏€น62๐‘ฅ2ร—2๏….

Simplifying each term gives us ๏€น25๐‘ฅโˆ’16๏…๏€น36๐‘ฅ4๏….

We can then reorder the two terms: 25ร—36ร—๐‘ฅโˆ’16ร—๐‘ฅ4.

At this point, you can see that we can multiply together the two numbers and use the product rule of exponents to simplify the ๐‘ฅ terms. This gives us 900๐‘ฅโˆ’16+4.

Simplifying, we find that our answer is 900๐‘ฅโˆ’12.

This is our answer. However, it is worth noting that we can present this in a different form that would be equally valid. If we recall the rule for negative exponents, which states that for any nonzero ๐‘ฅ, ๐‘ฅโˆ’๐‘Ž=1๐‘ฅ๐‘Ž,

we can rewrite the ๐‘ฅโˆ’12 as 1๐‘ฅ12. This means that our solution can be written as 900ร—1๐‘ฅ12,

which is the same as 900๐‘ฅ12.

We will finish by looking at one final example that combines multiple exponent rules.

Example 5: Combining Exponent Rules to Simplify Expressions Containing Negative Exponents

Simplify ๏€น๐‘ฅ8๏…โˆ’2ร—๏€น๐‘ฅโˆ’6๏…4๐‘ฅโˆ’8ร—๐‘ฅโˆ’3, given that ๐‘ฅโ‰ 0.

Answer

Let us start by recalling the product and power rules of exponents:

  • Product rule of exponents: ๐‘ฅ๐‘Žร—๐‘ฅ๐‘=๐‘ฅ๐‘Ž+๐‘
  • Power rule of exponents: (๐‘ฅ๐‘Ž)๐‘=๐‘ฅ๐‘Ž๐‘

Using the power rule first, we can rewrite the top of the rational expression as follows: ๐‘ฅ8ร—โˆ’2ร—๐‘ฅโˆ’6ร—4๐‘ฅโˆ’8ร—๐‘ฅโˆ’3.

Simplifying, we get ๐‘ฅโˆ’16ร—๐‘ฅโˆ’24๐‘ฅโˆ’8ร—๐‘ฅโˆ’3.

If we now use the product rule of exponents, we rewrite the top and the bottom of our expression to give us ๐‘ฅโˆ’16+(โˆ’24)๐‘ฅโˆ’8+(โˆ’3).

Simplifying, we get ๐‘ฅโˆ’40๐‘ฅโˆ’11.

At this stage, let us recall the quotient rule of exponents: ๐‘ฅ๐‘Žรท๐‘ฅ๐‘=๐‘ฅ๐‘Žโˆ’๐‘.

Using this, we can rewrite the expression as follows: ๐‘ฅโˆ’40โˆ’(โˆ’11).

Being very careful with negatives and remembering that subtracting a negative is the same as adding, we can simplify the expression to give us ๐‘ฅโˆ’29.

This is our answer, but it is worth noting that this can be rewritten using the rule for negative exponents. Remember that for any nonzero ๐‘ฅ, we have that ๐‘ฅโˆ’๐‘Ž=1๐‘ฅ๐‘Ž.

This means that we can rewrite our answer as 1๐‘ฅ29.

Key Points

  1. Negative exponents are defined as follows: for any nonzero ๐‘ฅ, we have that ๐‘ฅโˆ’๐‘˜=1๐‘ฅ๐‘˜.
  2. When working with negative exponents, we may need to use the following rules:
    1. Product rule of exponents: ๐‘ฅ๐‘Žร—๐‘ฅ๐‘=๐‘ฅ๐‘Ž+๐‘.
    2. Quotient rule of exponents: ๐‘ฅ๐‘Ž๐‘ฅ๐‘=๐‘ฅ๐‘Žโˆ’๐‘.
    3. Power rule of exponents: (๐‘ฅ๐‘Ž)๐‘=๐‘ฅ๐‘Ž๐‘.
    4. Zero exponents: for any nonzero ๐‘ฅ, ๐‘ฅ0=1.

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