Exponents of Negative Numbers

Powers of Negative Numbers - Examples, Exercises and Solutions
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Exponentiation of Negative Numbers

Negative number raised to an even power

Raising any negative number to an even power will result in a positive outcome.
When nn is even:
(βˆ’x)n=xn(-x)^n=x^n

Negative number raised to an odd power

Raising any negative number to an odd power will result in a negative outcome.
When nn is odd:
(βˆ’x)n=βˆ’(x)n(-x)^n=-(x)^n

What is the difference between a power that is inside parentheses and one that is outside of them?

When the exponent is outside the parentheses - it applies to everything inside them.
When the exponent is inside the parentheses - it applies only to its base and not to the minus sign that precedes it.

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Test yourself on powers of negative numbers!

einstein

Solve the following expression:

\( \)\( (-8)^2= \)

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Exponentiation of Negative Numbers

In this article, you will learn everything you need to know about the exponentiation of negative numbers and understand the difference between a power that is inside the parentheses and another that appears outside of them.
Shall we start?

Laws of Exponents in Negative Numbers

So far, we have learned to solve powers of positive numbers always obtaining positive results.
When a negative number is raised to a certain power, the result can be either positive or negative.

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Test your knowledge
Question 1

\( (-2)^7= \)

Question 2

\( 9= \)

Question 3

\( \)\( -(2)^2= \)

Negative number raised to an even power

Raising any negative number to an exponent that is an even number, even power, will result in a positive outcome.

For example

(βˆ’3)2=(-3)^2=
If we want to simplify the exercise we will get: (βˆ’3)Γ—(βˆ’3)=(-3) \times (-3)=
Negative times negative = Positive
Therefore, the result will be 99.
When the base is a negative number and the exponent is even, we can ignore the minus sign. Let's formulate it like this:
When nnΒ is even:
(βˆ’x)n=xn(-x)^n=x^n

Negative number raised to an odd power

When raising any negative number to an exponent that is an odd number, odd power, the result will be negative.

For example

(βˆ’3)3=(-3)^3=
If we want to simplify the exercise we will get: (βˆ’3)Γ—(βˆ’3)Γ—(βˆ’3)=(-3) \times (-3) \times (-3)=
Negative times negative = Positive
Positive times negative = Negative
Therefore, the result will be βˆ’27-27.
When the base is a negative number and the exponent is odd, we cannot ignore the minus sign, the result will always be negative.
Let's formulate it as a rule:
When nn is odd:

(βˆ’x)n=βˆ’(x)n(-x)^n=-(x)^n

Do you know what the answer is?
Question 1

\( -(-1)^{80}= \)

Question 2

\( 36= \)

Question 3

\( 49= \)

Let's Practice

Solve the exercise

(βˆ’4)3=(-4)^3=

Solution:
In this exercise, the exponent is odd.
Therefore, the result must necessarily be negative.
We will obtain:
(βˆ’4)Γ—(βˆ’4)Γ—(βˆ’4)=βˆ’64(-4) \times (-4) \times (-4)=-64

Solve the exercise

(βˆ’2)4=(-2)^4=

Solution:
In this exercise, the exponent is even. Consequently, we can ignore the minus sign and the result will be positive.
We will obtain:
(βˆ’2)Γ—(βˆ’2)Γ—(βˆ’2)Γ—(βˆ’2)=16(-2) \times (-2) \times (-2) \times (-2)=16

Check your understanding
Question 1

\( -6^2= \)

Question 2

\( 64= \)

Question 3

\( 8= \)

Solve the exercise

(βˆ’5)5=(-5)^5=

Solution:
In this exercise, the exponent is odd. Consequently, the result will be negative.
We will obtain:
(βˆ’5)Γ—(βˆ’5)Γ—(βˆ’5)Γ—(βˆ’5)Γ—(βˆ’5)=βˆ’3125(-5) \times (-5) \times (-5) \times (-5) \times (-5)=-3125

What is the difference between a power that is inside parentheses and one that is outside of them?

It is important that you know that the difference is very large.

When the exponent appears outside of the parentheses
We multiply the number inside the parentheses by itself, as many times as indicated by the number representing the exponent.
For example:
(βˆ’4)2=(-4)^2=
(βˆ’4)Γ—(βˆ’4)=16(-4) \times (-4)=16

On the other hand, when the exponent is inside the parentheses (sometimes, without any parentheses)
Thus:
(βˆ’42)=(-4^2 )=
or
βˆ’(42)=-(4^2 )=
or
βˆ’42=-4^2=
The exponent applies only and exclusively to the base number and not to the minus sign that precedes it.
Therefore, we will calculate the power and add the minus as an annex.
We will obtain:
βˆ’42=βˆ’16-4^2=-16

We have obtained 22 different answers! That's why it is necessary to pay close attention to understand well to which part of the exercise the power applies.
If the exponent is outside the parentheses - it applies to everything inside them.
If the exponent is inside the parentheses - it applies only to its base and not to the minus sign that precedes it.

Let's see how this is done in slightly more complicated exercises:

Do you think you will be able to solve it?
Question 1

\( \)\( -(-1)^{100}= \)

Question 2

\( \)\( (-1)^{99}= \)

Question 3

\( \)\( -(-2)^3= \)

Solve the exercise

(βˆ’2)4βˆ’32=(-2)^4-3^2=

Solution:
Let's start with (βˆ’2)4(-2)^4
The positive exponent is outside the parentheses, therefore, it applies to the entire βˆ’2-2.
We will obtain: (βˆ’2)4=16(-2)^4 = 16
Let's rewrite the exercise, we will get:
16βˆ’32=16-3^2=
Now let's continue with the other part of the exercise.
In the second monomial, there are no parentheses, meaning the exponent applies only to the 33 without taking into account the minus sign that precedes it.
We know that
32=93^2= 9
Therefore, let's rewrite the exercise in the following way:
16βˆ’9=716-9=7
Note –> Although it's true that the power is positive, it does not apply to the entire βˆ’3-3 therefore, we will not write 99 but,βˆ’9-9.

Examples and exercises with solutions on Exponentiation of negative numbers

Exercise #1

Solve the following expression:

(βˆ’8)2= (-8)^2=

Video Solution

Step-by-Step Solution

When we have a negative number raised to a power, the location of the minus sign is very important.

If the minus sign is inside or outside the parentheses, the result of the exercise can be completely different.

 

When the minus sign is inside the parentheses, our exercise will look like this:

(-8)*(-8)=

Since we know that minus times minus is actually plus, the result will be positive:

(-8)*(-8)=64

 

Answer

64 64

Exercise #2

(βˆ’5)βˆ’3=? (-5)^{-3}=\text{?}

Video Solution

Step-by-Step Solution

First let's recall the negative exponent rule:

bβˆ’n=1bn b^{-n}=\frac{1}{b^n} We'll apply it to the expression we received:

(βˆ’5)βˆ’3=1(βˆ’5)3 (-5)^{-3}=\frac{1}{(-5)^3} Next let's recall the power rule for expressions in parentheses:

(xβ‹…y)n=xnβ‹…yn (x\cdot y)^n=x^n\cdot y^n And we'll apply it to the denominator of the expression we received:

1(βˆ’5)3=1(βˆ’1β‹…5)3=1(βˆ’1)3β‹…53=1βˆ’1β‹…53=βˆ’153=βˆ’1125 \frac{1}{(-5)^3}=\frac{1}{(-1\cdot5)^3}=\frac{1}{(-1)^3\cdot5^3}=\frac{1}{-1\cdot5^3}=-\frac{1}{5^3}=-\frac{1}{125} In the first step, we expressed the negative number inside the parentheses in the denominator as a multiplication between a positive number and negative one, and then we used the power rule for expressions in parentheses to expand the parentheses, and then we simplified the expression.

Let's summarize the solution to the problem:

(βˆ’5)βˆ’3=1(βˆ’5)3=1βˆ’53=βˆ’1125 (-5)^{-3}=\frac{1}{(-5)^3} =\frac{1}{-5^3}=-\frac{1}{125}

Therefore, the correct answer is answer B.

Answer

βˆ’1125 -\frac{1}{125}

Exercise #3

(βˆ’2)7= (-2)^7=

Video Solution

Answer

βˆ’128 -128

Exercise #4

9= 9=

Video Solution

Answer

(βˆ’3)2 (-3)^2

Exercise #5

βˆ’(2)2= -(2)^2=

Video Solution

Answer

βˆ’4 -4

Test your knowledge
Question 1

\( \)\( -(-6)^2= \)

Question 2

\( \)\( -(7)^2= \)

Question 3

Solve the following expression:

\( \)\( (-8)^2= \)

Start practice

More Questions

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