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1.3: Powers and Exponents

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Introduction

Science Class

When Michaela and Lauren arrived at science class they were surprised to see that there were big changes going on in science class too. On each desk was an Astronomy book about stars and planets along with a journal.

“This is a lot of changes to get used to,” Lauren said sitting next to Michaela.

“I agree. First wood shop, then gym and now science,” Michaela agreed.

Sure enough, when Ms. Richards arrived the students realized that they were about to branch off into the universe with Astronomy. While Astronomy had been taught before in sixth grade, this was eighth grade and something brand new.

“We will be using a lot of math,” Ms. Richards explained. “One of the reasons that we decided to include Astronomy is because it has a lot of connections with mathematics. You will be using your algebra in this class as well. Let’s take a look at this.”

Ms. Richards took a pen and wrote these values on the white board.

93 \times 10^6

2.39 \times 10^5

“These are two distances in the galaxy. The first one is the distance from the Earth to the Sun and the second is from the Earth to the Moon. Your first task is to figure out how far this actually is,” Ms. Richards explained.

Michaela and Lauren took out their notebooks and got right down to work.

You need to get right down to work too. Use the information in this lesson on exponents to help you with the problem. While scientific notation isn’t specifically covered, you should be able to figure out how to work with these values once you learn about exponents.

What You Will Learn

By the end of this lesson, you will be able to demonstrate the following skills.

  • Evaluate numerical and variable expressions involving powers.
  • Extend order of operations to include evaluating powers.
  • Write variable expressions to represent and solve real-world problems involving operations and exponents.

Teaching Time

I. Evaluate Numerical and Variable Expressions Involving Powers

Let’s think back to the last lesson where we began working with numerical and variable expressions. You will be working with them again in this section, so let’s be sure that you understand each.

A numerical expression is a group of numbers and operations that represent a quantity, there isn’t an equal sign.

A variable expression is a group of numbers, operations and variables that represents a quantity, there isn’t an equal sign.

In the last lesson, you worked on evaluating numerical and variable expressions using the order of operations. You worked with expressions that had combinations of operations and you worked with expressions that contained grouping symbols such as parentheses and brackets. Now we are going to add a new think to our work. Let’s talk about powers.

A power is a number with an exponent and a base. An exponent is a little number that shows the number of times a base is multiplied by itself. The base is the regular sized number that is being worked with.

Take a minute to write these definitions in your notebook.

Now let’s look at an example.

Example

4^2 = 16

What happened here?

We can break down this problem to better understand powers and exponents. In the power 4^2 or four-squared, four is the base and two is the exponent. 4^2 means four multiplied two times or 4 \times 4. Therefore, 4^2 is sixteen.

Example

Evaluate 6^3

First, we have to think about what this means. It means that we take the base, 6 and multiply it by itself three times.

& 6 \times 6 \times 6\\& 36 \times 6 \\& 216

The answer is 216.

Example

Evaluate (-8)^2

To work on this one, we have to work on remembering integer rules. Think back remember that we multiply a negative times a negative to get a positive.

-8 \cdot -8 = 64

The answer is 64.

This is called evaluating a power.

Let’s look at evaluating powers within expressions.

Example

Simplify the expression 6^4 + 2^5 + 12.

Step 1: Simplify 6^4.

6^4 = 6 \times 6 \times 6 \times 6 = 1,296

Step 2: Simplify 2^5.

2^5 = 2 \times 2 \times 2 \times 2 \times 2= 32

Step 3: Add to solve.

1,296 + 32 + 12 = 1,340

The answer is 1,340.

We can also evaluate variable expressions by substituting given values into the expressions.

Example

Evaluate the expression 4a^2 when a = 3.

Step 1: Substitute 3 for the variable “a.”

4(3)^2

Step 2: Simplify the powers.

& 4(3)^2\\& 4(3 \cdot 3)\\& 4(9)

Step 3: Multiply to solve.

& 4(9)\\& 36

The answer is 36.

Example

Evaluate the expression 5b^4 + 17. Let b=5.

Step 1: Substitute 5 for “b.”

5(5)^4 + 17

Step 2: Simplify the powers.

& 5(5 \cdot 5 \cdot 5 \cdot 5) + 17\\& 5(625) + 17

Step 3: Multiply then add to solve.

& 5(625) + 17\\& 3,125 + 17 = 3,142

The answer is 3,142.

II. Extend Order of Operations to Include Evaluating Powers

Now that you have learned how to evaluate powers and to apply evaluating powers to numerical and variable expressions, you are ready to apply this information to evaluating expressions with the order of operations.

To do this, we are going to need to refer back to the order of operations.

Order of Operations

P parentheses or grouping symbols

E exponents

MD multiplication and division in order from left to right

AS addition and subtraction in order from left to right

Here is the order of operations once again. You will be working with the order of operations from now on so it is a good idea to commit it to memory.

Now look at the E. That E refers to exponents and powers and evaluating exponents in the order of operations. You can see that you evaluate the powers right after the grouping symbols.

Example

Evaluate the expression -12^3 + 7y^2 + 12 for y=6.

Step 1: Before performing the order of operations, substitute 6 for “y.”

-12^3 + 7(6)^2 + 12

Step 2: Perform the calculations inside the parentheses.

-12^3 + 7(36) + 12

Step 3: Perform the calculations with exponents.

& -12^3 + 7(36) + 12\\& -1,728 + 7(36) + 12

Step 4: Multiply

& -1,728 + 7(36) + 12\\& -1,728 + 252 + 12

Step 5: Add

-1,728 + 252 + 12 = - 1,464

The answer is -1,464.

Now let’s look at another example.

Example

Evaluate the expression 8h^2 + [51 \div (4 \cdot 4.25)] - 52 \div 5. Let h=4.

It does look complicated, but if it helps, think of this as a series of steps. The order of operations is your guide. If you follow the order of operations then working through a problem such as this one becomes much easier.

& P && Step \ 1: \ \text{Substitute} \ 4 \ \text{for} \ ``h.''\\& && 8h^2 + [51 \div (4 \cdot 4.25)] - 52 \div 5\\& && 8(4)^2 + [51 \div (4 \cdot 4.25)] - 52 \div 5\\& && Step \ 2: \ \text{Remember PEMDAS. Therefore, perform the operation inside the}\\& && \text{grouping symbols first. Recall that order of operations must be followed}\\& && \text{inside grouping symbols also. In this case, multiply} \ 4 \times 4.25 \ \text{before dividing}\\& && \text{by} \ 51.\\& && 8(4)^2 + [51 \div (4 \cdot 4.25)] - 52 \div 5\\& && 8(4)^2 + [51 \div 17] - 52 \div 5\\& && 8(4)^2 + 3 - 52 \div 5\\& E && Step \ 3: \ \text{The next step in order of operations is to simplify the numbers with}\\& && \text{exponents.}\\& && 8(4)^2+ 3 - 52 \div 5\\& && 8(4 \cdot 4) + 3 - 5 \cdot 5 \div 5\\& && 8(16) + 3 - 25 \div 5\\& M && Step \ 4: \ \text{Multiply}\\& && 8(16) + 3 - 25 \div 5\\& && 128 + 3 - 25 \div 5\\& D && Step \ 5: \ \text{Divide}\\& && 128 + 3 - 25 \div 5\\& && 128 + 3 - 5\\& A && Step \ 6: \ \text{Add}\\& && 128 + 3 - 5\\& && 131 - 5\\& S && Step \ 6: \ \text{Subtract}\\& && 131 - 5 = 126

The answer is 126.

We can also evaluate variable expressions that have more than one variable. Notice that a different value has been given for x and y. You simply substitute the given values into each expression and evaluate it for the quantity of the expression.

Example

Evaluate the expression 4x^3 - (3y \div 9) + 12. Let x=3 and y=9.

& 4x^3 - (3y \div 9) + 12 \ (\text{Substitute the variables})\\& 4(3)^3 - [(3 \times 9) \div 9] + 12 \ (\text{Parentheses})\\& 4(3)^3 - [27 \div 9] + 12\\& 4(3)^3 - 3 + 12 \ (\text{Exponents})\\& 4(3 \times 3 \times 3) - 3 + 12\\& 4(27) - 3 + 12 \ (\text{Multiply})\\& 108 - 3 + 12 \ (\text{Add and then Subtract from left to right})\\& 105 + 12\\& 117

The answer is 117.

When you have variable and numerical expressions with powers in them, you can use the order of operations to evaluate the expressions. Remember not to get stuck if the problem seems complicated. Stick to the order of operations and you will be able to evaluate the expression.

III. Write Variable Expressions to Represent and Solve Real-World Problems Involving Operations and Exponents

We can use exponents to express real-life situations too. An example of a situation like this can be found when we look at geometric progressions.

A geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number. For example, the sequence 2, 4, 8, 16 is found by multiplying the previous number by two. Exponents can be used to express geometric progression.

Example

Ben’s pet mice, Spike and Betty, double in number each month. Write a variable expression to determine the number of mice after “n” months. Then evaluate the expression for twelve months.

To determine the number of mice for twelve months, double the original amount (2) for twelve months.

Month 1: 2 \times 2

Month 2: 4 \times 2

Month 3: 8 \times 2

Month 4: 16 \times 2

Month 5: 32 \times 2

Month 6: 64 \times 2

Month 7: 128 \times 2

Month 8: 256 \times 2

Month 9: 512 \times 2

Month 10: 1,024 \times 2

Month 11: 2,048 \times 2

Month 12: 4,096 \times 2

You can write an expression to represent the work you did above. By writing the expression, you are creating a way to determine the number of mice at any time in a much quicker way.

y = 2 \cdot 2^x

To evaluate the expression for twelve months, substitute “12” for “x.”

y & = 2 \times 2^x\\y & = 2 \times 2^{12}\\y & = 2 \times (2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2)\\y & = 2 \times 4,096\\y & = 8,192

There will be 8,192 mice after 12 months.

Example

Would you rather have one million dollars or a penny doubled each day for thirty days?

Number of pennies on Day One: 1

Number of pennies on Day Two: 1 + 2 = 3

Number of pennies on Day Three: 1 + 2 + 4 = 7

Number of pennies on Day Four: 1 + 2 + 4 + 8 = 15

Number of pennies on Day Five: 1 + 2 + 4 + 8 + 16 = 31

Number of pennies on Day Six: 1 + 2 + 4 + 8 + 16 + 32 = 63

Number of pennies on Day Seven: 1 + 2 + 4 + 8 + 16 + 32 + 64 = 127

Number of pennies on Day Eight: 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255

Number of pennies on Day Nine: 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 = 511

Number of pennies on Day Ten: 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 = 1,023

An expression can be written to represent the work completed above: 2^n - 1.

Use the expression to determine the number of pennies on Day Ten.

& 2^{10} - 1\\& (2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2) - 1\\& 1,024 - 1\\& 1,023

Use the expression to determine the number of pennies on Day Thirty.

& 2^{30} - 1\\& 1,073,741,824 - 1\\& 1,073,741,823

Since there are one hundred pennies in a dollar, divide the amount of pennies by one hundred to determine the pennies worth.

1,073,741,823 \div 100 = \$ 10,737,418.23

Therefore, a penny doubled each day for thirty days is worth more than one million dollars.

In these examples you can see that we used an exponent and a variable for the number of months. This was a way to use a variable expression to write an equation.

Real-Life Example Completed

Science Class

Here is the problem from the introduction once again. Now you will need to apply what you have learned about exponents. Reread the problem first. Then, you need to figure out the distance from the Earth to the Moon and from the Earth to the Sun. Your answer will have two parts to it.

When Michaela and Lauren arrived at science class they were surprised to see that there were big changes going on in science class too. On each desk was an Astronomy book about stars and planets along with a journal.

“This is a lot of changes to get used to,” Lauren said sitting next to Michaela.

“I agree. First wood shop, then gym and now science,” Michaela agreed.

Sure enough, when Ms. Richards arrived the students realized that they were about to branch off into the universe with Astronomy. While Astronomy had been taught before in sixth grade, this was eighth grade and something brand new.

“We will be using a lot of math,” Ms. Richards explained. “One of the reasons that we decided to include Astronomy is because it has a lot of connections with mathematics. You will be using your algebra in this class as well. Let’s take a look at this.”

Ms. Richards took a pen and wrote these values on the white board.

93 \times 10^6

2.39 \times 10^5

“These are two distances in the galaxy. The first one is the distance from the Earth to the Sun and the second is from the Earth to the Moon. Your first task is to figure out how far this actually is,” Ms. Richards explained.

Michaela and Lauren took out their notebooks and got right down to work.

Now solve the problem. Remember that your answer will have two parts to it.

Solution to Real – Life Example

Now begin by looking at the information that you have been given in the problem.

Earth to Sun 93 \times 10^6

Earth to Moon 2.39 \times 10^5

Next, we need to figure out those actually distances. First, let’s figure out the first distance. To do this, we need to understand the value of 10^6.

10^6 = 10 \times 10 \times 10 \times 10 \times 10 \times 10 = 1,000,000

Now we multiply the value times 93 which means we move the decimal point six places.

93,000,000

The distance from the Earth to the Sun is 93 million miles.

Next, we figure out the distance from the Earth to the Moon. To do this, we need to understand the value of 10^5.

10^5 = 10 \times 10 \times 10 \times 10 \times 10 = 100,000

Now we multiply.

2.39 \times 100,000 = 239,000 \ miles

There are our answers.

Vocabulary

Here are the vocabulary words that are found in this lesson.

Numerical Expression
a group of numbers and operations used to represent a quantity without an equals sign.
Variable Expression
a group of numbers, operations and variables used to represent a quantity without an equals sign.
Powers
the value of a base and an exponent.
Base
the regular sized number that the exponent works upon.
Exponent
the little number that tells you how many times to multiply the base by itself.
Geometric Progression
each term after the first is found by multiplying the previous number by a fixed non-zero number.

Time to Practice

Directions: Evaluate each power.

  1. 3^3
  2. 4^2
  3. (-2)^4
  4. (-8)^2
  5. 5^3
  6. 2^6
  7. (-9)^2
  8. (-2)^6

Directions: Evaluate each numerical expression.

  1. 6^2 + 22
  2. -3^3 + 18
  3. 2^3 + 16 - 4
  4. (-5)^2 - 19
  5. (-7)^2 + 52 - 2
  6. 18 + 9^2 - 3
  7. 22 - 3^3 + 7

Directions: Evaluate each variable expression using the given values.

  1. 6a + 4^2 - 2, when a = 3
  2. a^3 + 14, when a = 6
  3. 2a^2 - 16, when a = 4
  4. 5b^3 + 12, when b = -2
  5. 2x^2 + 52, when x = 4

Directions: Evaluate each expression. Remember to follow the order of operations.

  1. 3^2 + [(5 \cdot 2) - 3] - 8 \cdot 2
  2. 5^2 + (3 + 5) - 6^2 + 2
  3. 6^3 + 5^2 + 25
  4. 16(12^3)
  5. 8^2 - (2(3^3) \div 2) + (16 \cdot 5)

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Date Created:

Jan 11, 2013

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Sep 08, 2014
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