Section 4.1 Rules of Exponents

Section 4.1 Rules of Exponents

THE MEANING OF THE EXPONENT

The exponent is an abbreviation for repeated multiplication.

The repeated number is called a factor. x n _{means “n factors of x.”}

The exponent tells us how many factors of x there are in the expression.

For example, 3 · 3 · 3 · 3 · 3 is abbreviated as 35_{; this means that there are five factors of 3.}

In this abbreviated form, 35 _{is called the exponentialform; 5 is called the exponent and the repeated factor, 3,}

is called the base.

Of course, 35 _{actually has a numerical value; 3}5 _{= 243. This can be found by using repeated multiplication:}

3 · 3 · 3 · 3 · 3 = 243; this value, 243, is called the 5thpower of 3. Sometimes we use the word “power” in place of the word “exponent.”

Where 35 _{is called the exponential form, 3 · 3 · 3 · 3 · 3 is called the expanded form.}

One of the most important things to know about exponents is this:

Exponents have more meaning than they have value.

As has been mentioned, the exponent is an abbreviation for repeated multiplication, and xn _{means “n factors}

of x.” To understand this idea better, it might be best to represent exponents (at least for now) in words instead of as numbers.

For example, instead of writing 23_{, we could write 2}three_{. It means, of course, three factors of 2. Writing the}

exponential form like this will help us to realize that the value of an exponent is not as important as its meaning. So, 2three means three factors of 2; and 2four means four factors of 2. Writing them this way should cause one to think more of the meaning of the exponent and not of the value of the exponent. Should we, then, ignore the value of the exponent? No, but we do need to think of its meaning first.

We might see 23_{, which is 2 · 2 · 2 = 8, but our thinking needs to be 2}three_.

This new way of writing exponents will be used throughout this section as we develop the rules of exponents. Though the meaning of the exponent will be emphasized, we will not ignore its value.

THE FIRST RULES OF EXPONENTS

Example 1: For each of the following, state its meaning, write it in expanded form and then evaluate the

expression.

a) 25 b) (- 5)3 c) 18 _{d) 0}6

Answer: The exponent indicates the number of factors that are to be multiplied. (Some of the

multiplication will need to be done off to the side as scratch work.) a) 25 _{means five factors of 2 = 2 · 2 · 2 · 2 · 2 = 32}

b) (- 5)3 _{means three factors of - 5 = (- 5) · (- 5) · (- 5) = - 125}

c) 1 · 1 · 1 · 1 · 1 · 1 · 1 · 1 = 1 d) 0 · 0 · 0 · 0 · 0 · 0 = 0

Notice, in particular, problems (c) and (d) of the Example 1: (d) 16 = 1 and (e) 04 = 0. Neither of these should be a surprise. In fact, there is a rule about these very ideas:

Exponent Rules for 0 and 1: If n is a positive number, then

a) 0n = 0: 0 raised to any positive exponent will always be 0. b) 1n _{= 1: 1 raised to any positive exponent will always be 1.}

Exercise 1: For each of the following, state its meaning, write it in expanded form and then evaluate

the expression

a) 43 means = =

b) (- 12)2 _{means = =}

c) 34 _{means = =}

d) 16 _{means = =}

In keeping with that most basic understanding of exponents, we can introduce a few rules.

The First Power Rule: For any real number x, x1 _{= x.}

in other words, xone means one factor of x

This can be seen in the prime factorization of a number such as 45: 45 = 3 · 3 · 5. Here we can say that 45 has two factors of 3 and one factor of 5. For this reason, the prime factorization could also be written as

45 = 32 · 51 : two factors of 3 and one factor of 5. Similarly, we could write the prime factorization of 250 = 2 · 5 · 5 · 5 as

250 = 21 · 53 : one factor of 2 and three factors of 5.

Example 2: Evaluate (or simplify) the expression.

a) 21 b) (- 5)1 c)

( )

45

1

d) x1

Answer: Use the First Power Rule:

a) 21 = 2 b) (- 5)1 = - 5 c)

( )

45 1

= 4₅ d) x1 = x

Exercise 2 Evaluate (or simplify) the expression

a) 41 = b) (- 12)1 = c) y1 = d)

( )

23

1

=

THE PRODUCT RULE

If 2three means three factors of 2 and 2four means four factors of 2 , then what is the meaning of 2three · 2four_{? This is three factors of 2 times four (more) factors of 2, 2}three _{· 2}four _{can be expanded to (2·2·2)·(2·2·2·2)}

= 27_{, or 2}seven_{, seven factors of 2.}

Look at what happened: 2three _{· 2}four _{= 2}seven_{. What is being suggested here? That exponents have more}

meaning than value. “Three factors of 2 times four more factors of 2 results in a total of seven factors of 2.”

This leads to an important rule of exponents, the product rule:

The Product Rule for Exponents:

If x is a non-zero base and a and b are positive integer exponents, then

The product rule is true only when the base is the same for each factor: when multiplying, just add the exponents.

Let’s better understand the product rule by taking a look at how we can manipulate the following repeated multiplication.

x · x · x · x · x · x · x · x · x · x = x10.

Notice that there are ten factors of x (you may want to count them). This could also be written as (x · x · x · x) · (x · x · x · x · x · x) = x10.

Notice that there are still 10 factors of x, but they’ve been broken up so that we now have

( )

x4 ·

( )

x6 =

(

xfour

)

·

(

xsix

)

= xten = x10

or just x4 · x6 = x10.

This says that we have four factors of x and sixmorefactors of x for a total of ten factors of x. Let’s try it again: (x · x · x · x · x · x · x) · (x · x · x) = x10.

There are still ten factors of x, but this time they’ve been broken up so that we now have xseven · xthree = xten.

or x7 · x3 = x10.

Again, this time we have seven factors of x and threemorefactors of x, for a total of ten factors of x.

Let’s try it again: (x · x) · (x · x · x · x · x · x · x · x) = x10 xtwo_{· x}eight _{= x}ten_.

which is the same as saying x2 _{· x}8 _{= x}10_.

One more time: (x · x · x · x · x) · (x · x · x · x · x) = x10

xfive· xfive = xten. which is the same as saying x5 · x5 = x10.

Now you try it. I’ll give you nine factors of y, and you find four different ways to rewrite them as I have, above. Write each with all nine factors split into two groups; then write each in its exponential form. I’ll show you one possible situation; you find four others.

y · y · y · y · y · y · y · y · y = y9.

My example: (y · y) · (y · y · y · y · y · y · y) = y9

which is the same as: ytwo _{· y}seven _{= y}nine _.

or: y2 _{· y}7 _{= y}9 _.

1. y·y·y·y·y·y·y·y·y = y9 2. y·y·y·y·y·y·y·y·y = y9 so, = y9 so, = y9

3. y·y·y·y·y·y·y·y·y = y9 4. y·y·y·y·y·y·y·y·y = y9 so, = y9 so, = y9

Here is another way to look at the product rule. We can expand each factor and count the number of repeated factors.

Example 3: Rewrite each product and power it in its expanded form; then combine all of the factors and

abbreviate the result using an exponential form.

a) b3 · b2 b) y3 · y4 c) x5 · x

Answer: Notice that each factor in each problem has the same base. This is important for the pattern

to be recognized.

a) b3 · b2 = (b · b · b) · (b · b) = b5 ; bthree · btwo = bfive b) y3 · y4 = (y · y · y) · (y · y · y · y) = y7 ; ythree · yfour = yseven c) x5 · x = (x · x · x · x · x) · (x) = x6 ; xfive · xone = xsix

Exercise 3 Rewrite each product and power in its expanded form; then combine all of the factors and abbreviate the result using an exponential form.

a) x5 · x4 = = b) y2 · y3 = = c) c6 · c2 = = d) w3 _{· w}3 _{= =}

e) k1 _{· k}1 _{= =}

f) m · m3 _{= =}

g) x5 _{· x = =}

Now let’s apply the product rule directly.

The Product Rule for Exponents:

If x is a non-zero base and a and b are positive integer exponents, then

x

a

_{· x}

b

_{= x}

a

+

b

Example 4: Use the product rule for exponents to write each of these as one base with one exponent.

a) b3 _{· b}2 _{b) y}3 _{· y}4 _{c) x}5 _{· x}

Answer: Notice that each factor in each problem has the same base. This is important for the pattern

to be recognized.

a) b3 _{· b}2 _{= b}3 + 2 _{= b}5 _{(three factors of ‘b’ and two more factors of ‘b’)}

b) y3 · y4 = y3 + 4 = y7 (three factors of ‘y’ and four more factors of ‘y’) c) x5 _{· x = x}5 + 1 _{= x}6 _{(remember that x = x}1₎

Exercise 4 Use the product rule for exponents to write each of these as one base with one exponent.

a) x5 · x4 = b) x3 · x6 =

c) y2 · y3 = d) c6 · c2 =

e) w3 · w3 = f) k1 · k4 =

g) m · m3 _{= h) x}5 _{· x =}

Please note: when working with the product of two exponent expressions with the same base, there are only two ways to combine the two bases into one (but they must first be identical bases):

1) write each in expanded form and count 2) use the product rule for exponents: the number of factors of the base, as in as in

x4 _{· x}3 _{= (x · x · x · x) · (x · x · x) = x}7_{. x}4 _{· x}3 _{= x}4 + 3 _{= x}7_.

It’s also very important to note that the product rule for exponents cannot be used if the bases are not the same. For example, x3 · y5 is nothing more than x3 · y5. This could be thought of as xthree · yfive, three factors of x and five factors of y, but we don’t have eight factors of any one variable.

THE ZERO POWER RULE

You know that 1 is the identity for multiplication. This means that the product of any number, A, and 1 is always that number: A · 1 = A. We can use this idea, along with the product rule, to introduce a new rule—the

zero powerrule.

Consider the product a4 · 1 = a4. Also consider the product a4 · a0 = a4+0 = a4. So, a4 · a0 = a4 and a4 ·

1

= a4;

This could only mean that a0 _and

₁

_{are the same; in other words, a}0 ₌

₁

_.

The Zero Power Rule: For any non-zero real number x, x0 _{= 1.}

in other words, xzero _{means no factor of x}

The Exception: The rule itself is explained above, but the exception needs a brief mention.

00 is undefined because it puts two rules in conflict with one another, and that’s not allowed in mathematics:

The Exponent Rule for 0 would state 00 = 0 ,

but The Zero Power Rule would state 00 = 1 !

But we can’t have it both ways: 00 can’t equal both 0 and 1. Therefore, we say it can’t happen at all; in other words, 00 is undefined.

Let’s put The Zero Power Rule into practice:

Example 5: a) b0 _{= 1 b) 10}0 _{= 1 c) 9}0 _{= 1}

d) (- 2)0 = 1 e) (- 3)0 = 1 f) x0 = 1

Exercise 5 Use the Zero Power Rule to evaluate each.

a) w0 _{= b) 7}0 _{= c) (- 4)}0 _{= d) 14}0 ₌

e) y0 _{= f) 1}0 _{= g) 3}0 _{= h) 0}0 ₌

To further understand the zero power rule, that a0 _{= 1, we refer back to a}4 _{· a}0 _{= a}4 _{and recognize that a}0

doesn’t “add” any more factors to the product. Note that it doesn’t make the entire product zero.

Example 6: Use the product rule for exponents to write each of these as one base with one exponent.

a) b3 · b0 b) y0 · y4 c) x0 · x

Answer: Notice that each factor in each problem has the same base. This is important for the pattern

to be recognized.

a) b3 · b0 = b3 + 0 = b3 (three factors of ‘b’ and no more factors of ‘b’)

b) y0 _{· y}4 _{= y}0 + 4 _{= y}4 _{(no factors of ‘y’ and four factors of ‘y’)}

Exercise 6 Use the product rule for exponents to write each of these as one base with one exponent.

a) x5 · x0 = b) x0 · x6 = c) y0 · y2 = d) c6 · c0 = e) w9 _{· w}1 _{= f) k}1 _{· k}4 ₌

g) m · m0 _{= h) x}0 _{· x}0 ₌

THE QUOTIENT RULE

A product is an expression of multiplication; a quotient is an expression of division. Usually, a quotient is expressed as a fraction, and we know that we can simplify fractions by canceling common factors in the

numerator and the denominator.

For example, you can probably see that 24_{30 can reduce by a factor of 6:} 24 ÷ 6_{30 ÷ 6 =} 4_{5 . Recognizing} common factors, especially using prime factorization, is another method that works quite well. Writing the numerator and denominator in its prime factored form allows for easy cancellation of common factors:

We can do the same for a quotient such as x

5

x3 . If we write it in its expanded form, then we’ll be able to see that many of the factors will cancel:

Of course, we can cancel only common factors, and because the bases are the same, both x, the numerator and denominator have plenty of common factors to cancel.

Another way to show the canceling process is to group the same number of common factors in both the numerator and denominator. For example, with x

5

x3 we can expand both the numerator and denominator and then group three factors of x in each:

x5 x3 =

x·x·x·x·x

Example 7: Write both the numerator and the denominator in expanded form; then simplify the fraction by canceling any common factors.

a) x

8

x2 =

x·x·x·x·x·x·x·x

x · x = (x·x)·x(x · x)·x·x·x·x·x = x · x · x · x · x · x = x6

b) y

7

y6 =

y·y·y·y·y·y·y

y · y · y · y · y · y = (y(y · y · y · y · y · y) = y or y·y·y·y·y·y)·y 1

c) w

4

w4 =

w · w · w · w w·w·w·w =

(w · w · w · w) (w·w·w·w) = 1

d) p

5

p0 =

p·p·p·p·p

1 = p · p · p · p · p = p5

Exercise 7 Write both the numerator and the denominator in expanded form; then simplify the

fraction by canceling any common factors.

a) x

5

x4 =

=

b) w

8

w5 =

=

c) c

6

c1 =

=

d) p

4

p0 =

=

e) y

9

y 7 =

=

f) a

6

a6 =

=

g) m

8

m4 =

=

h) x

7

x =

Let’s return to the example of simplifying x

5

x3  

  xfive xthree :

In this example, the numerator has five factors of x and the denominator has three factors of x. In the canceling process, all three of the denominator factors cancel with three of the numerator factors leaving two factors of x in the numerator. It’s as if we are eliminating three factors of x. This process suggests a subtraction:

five “take away” three is two 5 – 3 = 2

Of course, in this process we are subtracting the exponents, just as we are dividing out the number of common factors. The rule that supports this type of dividing out is called the quotient rule.

The Quotient Rule (for Exponents):

If x is a non-zero base and a and b are positive integer exponents, then

x

a

x

b

= x

a

–

b

This works only when the base is the same for each factor: when dividing, just subtract the exponents.

Example 8: Use the Quotient Rule to simplify each of these.

a) x

8

x2 b)

y7

y6 c)

w4

w4 d)

p5

p0

Answer: Make sure the bases are the same. Then, subtract the exponents to get the result.

a) x

8

x2 = x8–2 = x 6 b) y7

y6 = y7–6 = y 1 or just y

c) w

4

w4 = w4–4 = w0 = 1 d) p5

Exercise 8 Use the Quotient Rule to simplify each of these.

a) x

5

x4 = b)

w8

w5 = c) c

6

c1 = d)

p4

p0 = e) y

9

y 7 = f)

a 6

a 6 =

g) v

9

v3 = h)

m8

m4 = i) x

4

x = j) k

8

k7 =

THE ZERO POWER RULE, REVISITED

We can use the Quotient Rule to further explain why x0 =

1

. We know, of course, that any number divided by itself is 1: A_{A =}

1

. This is also true when the number is raised to a power. For example, consider 2

3

23 ,

which is just

1

. We can consider this fraction in a variety of ways, but it always reduces to

1

: (1) Expanding the numerator and denominator: 2

3

23 =

2 · 2 · 2 2·2·2 =

(2 · 2 · 2) (2·2·2) = 1

(2) Evaluating the numerator and denominator: 2

3

23 =

8 8 = 1

(3) Using the Quotient Rule: 2

3

23 = 23–3 = 2 0 = 1 Of course, (3) shows us again that x0 = 1.

The zero power rule can be confusing. When seeing it or using it, you must first think and remember that

exponents have more meaning then they have value. In other words, think about what the zero power means

THE DISTRIBUTIVE PROPERTIES FOR EXPONENTS

To this point we have used the rules of exponents with bases that are integers. However, it’s quite possible that the base can be a fraction.

For example, consider

( )

ab 3

: when expanded it becomes _{b ·} a a_{b ·} a_{b = b}a·_·ba·_·a_{b =} a

3

b3 .

Similarly, consider (a·b)3: when expanded it becomes... (a·b) · (a·b) · (a·b)

The associative property allows us to write these without parentheses = a · b · a · b · a · b

the commutative property allows us to switch these around = a · a · a · b · b · b

the associative property allows us to regroup them = (a · a · a) · (b · b · b)

and the definition of the exponent allows to abbreviate it as = a3 _{· b}3

In general, if a quantity is a product (multiplication) or a quotient (such as a fraction), then we can distribute

an exponent to each part.

The Distributive Properties for Exponents:

a) If x and y are any bases, then

(x·y)

a

= x

a

· y

a

b) If x and y are any bases, and y ≠ 0, then _

x

_y

_ a

=

x

a

y

a

The reason the Distributive Properties for Exponents work so well for multiplication is that exponents are

compatible with multiplication and division. Consider this:

Multiplication is the abbreviation for repeated addition, so multiplication is

compatible with the Distributive Property for Addition:

a(b + c) = a·b + a·c

Likewise, an exponent is the abbreviation for repeated multiplication, so the exponent is compatible with the Distributive Property for Exponents.

(x·y)a _{= x}a_·ya

However, the Distributive Property for Exponents is not compatible with addition; in other words

Example 9: Use the Distributive Properties for Exponents to rewrite each of these. Simplify if possible.

a) (a·b)5 _{b) (3·w)}4 _{c) }2_

p

5

d) 7y_w  2

e) (- 2·m)3 _{f) (- 5·r)}2

Answer: a) (a·b)5 _{= a}5_·b5 _{b) (3·w)}4 _{= 3}4_·w4 _{= 81w}4

c) 2_p  5

= 2

5

p5 = 32

p5 d)  

7y w

2

= (7y)

2

w2 = 72·y2

w2 = 49y2

w2 e) (- 2·m)3 = (- 2)3m3 = - 8m3 f) (- 5·r)2 = (- 5)2r2 = 25r2

Exercise 9 Use the Distributive Properties for Exponents to rewrite each of these. Simplify if

possible.

a) (w·y)6 _{= b) (4p)}3 ₌

c) (- 2·x)4 _{= d) (- 10c)}3 ₌

e) a_b 

9

= f) 5_y 

3

=

g) x₆ 

2

= h) 3y₂ 

3

=

i) -11x9 

2

= j) -2d 1k

5

=

k) (2x)1 = l) (10c)0 =

THE POWER RULE OF EXPONENTS

The last rule to be presented in this section is the Power Rule (of Exponents):

The Power Rule (of Exponents)

also known as “The Power of a Power”:

(x

a

₎

b

_{= x}

_a·b

The explanation to this rule is best seen in an example. Consider (x2)3; to best understand this, we should refer back to the definition of the exponent, an abbreviation for repeated multiplication.

(x2)3, or (x2)three, means three factors of x2: x2 · x2 · x2 = x2 + 2 + 2 = x6.

Notice that in x2 + 2 + 2 _{we actually have repeated addition in the exponent,}

this repeated addition could be abbreviated as xthree 2’s _{= x}3·2 _{(or x}2·3_{) = x}6_.

Notice also that (x3)2 = x3·2 = x2·3 = x6; this suggests that the commutative property can also be applied.

Let’s look at two numerical examples to see it just one more way; consider both (22₎3 _{and (2}3₎2

First, recall that 22 _{= 4 8}2 _{= 64}

23 _{= 8 4}3 _{= 64}

26 _{= 64}

So, in applying the order of operations to (22₎3_{, we get (4)}3 _{which is 64;}

and, in applying the order of operations to (23₎2_{, we get (8)}2 _{which is 64.}

Lastly, in applying the power rule to (22₎3_{, we get 2}2·3 _{= 2}6 _{which is 64.}

Example 10: Use the Power Rule to rewrite each of these.

a) (w4)3 b) (x3·y)5 c) _a _

4

p3

2

Answer: Notice that, in (b) and (c), the power rule can be combined with the distributive rules.

(a) (w4)3 = w4·3 = w12 (b) (x3·y)5 = (x3)5·y5 = x15y5 (c) _a _

4

p3

2

= (a

4₎2

(p3)2 = a8 p6

Exercise 10 Use the Power Rule to rewrite each of these. Simplify if possible.

a) (c5₎2 _{= b) (m}4₎6 ₌

c) (3w2₎4 _{= d) (a}6_b)3 ₌

e) (2y5₎6 _{= f) (c}2_d5₎3 ₌

g) n  5

9

2

= h) __c33_

3

=

i) __wy4_ 3

= j) _p _

5

k6 4

=

k) (c2_d5₎1 _{= l) (c}2_d5₎0 ₌

m) __wy4_ 1

= n) _p _

5

k6 0

Answers to each Exercise

Section 4.1

Exercise 1: a) 43 means three factors of 4 = 4 · 4 · 4 = 64.

b) (- 12)2 means two factors of (- 12) = (- 12) · (- 12) = + 144 c) 34 means four factors of 3 = 3 · 3 · 3 · 3 = 81.

d) 16 means six factors of 1 = 1 · 1 · 1 · 1 · 1 · 1 = 1. e) 04 means four factors of 0 = 0 · 0 · 0 · 0 = 0.

Exercise 2: a) 41 _{= 4 b) (- 12)}1 _{= - 12 c) y}1 _{= y d)}

_{( )}

2

3 1

= 2₃

Exercise 3: a) x5 · x4 = (x · x · x · x · x) · (x · x · x · x) = x9

b) y2 · y3 = (y · y) · (y · y · y) = y5

c) c6 · c2 = (c · c · c · c · c · c) · (c · c) = c8 d) w3 · w3 = (w · w · w) · (w · w · w) = w6 e) k1 · k1 = (k) · (k) = k2

f) m· m3 = (m) · (m · m · m) = m4 g) x5 · x = (x · x · x · x · x) · (x) = x6

Exercise 4: a) x9 b) x9 c) y5 d) c8

e) w6 f) k5 g) m4 h) x6

Exercise 5: a) 1 b) 1 c) 1 d) 1

e) 1 f) 1 g) 1 h) undefined

Exercise 6: a) x5 b) x6 c) y2 d) c6

Exercise 7: a) x

5

x4 =

x·x·x·x·x

x·x·x·x = (x(x·x·x·x·x·x)·x) = x ·x

b) w

8

w5 =

w·w·w·w·w·w·w·w

w · w · w · w · w = (w·w(w · w · w · w · w)·w·w·w)·w·w·w = w3 c) c

6

c1 =

c·c·c·c·c·c c =

(c)·c·c·c·c·c

(c) = c5 d) p

4

p0 =

p·p·p·p

1 = p4 e) y

9

y7 =

y · y · y · y · y · y · y · y · y y·y·y·y·y·y·y =

(y · y · y · y · y · y · y) · y · y

(y·y·y·y·y·y·y) = y2 f) a

6

a6 =

a·a·a·a·a·a

a·a·a·a·a·a = (a(a··aa··aa··aa··aa··a)a) = 1

g) m

8

m4 =

m·m·m·m·m·m·m·m

m · m · m · m = (m·m·(m · m · m · m)m·m)·m·m·m·m = m4 h) x

7

x1 =

x·x·x·x·x·x·x

x =

(x)·x·x·x·x·x·x

(x) = x6

Exercise 8: a) x 1 _{or just x b) w} 3 _{c) c} 5 _{d) p} 4

e) y 2 _{f) a} 0 _{= 1 g) v} 6 _{h) m} 4

i) x 3 _{j) k} 1 _{or just k}

Exercise 9: a) w 6_y 6 _{b) 64p}3 _{c) 16x}4 _{d) - 1,000 c}3

e) a

9

b9 f)

125

y3 g)

x2

36 h)

27y3

8 i) 121x

2

81 j)

-1k5

32d5 k) 2x l) 1

m) x₉ n) 1

Exercise 10: a) c10 _{b) m}24 _{c) 81w}8 _{d) a}18_b3

e) 64y30 f) c6d15 g) n

10

81 h) 27_c9

i) y

3

w12 j)

p20

k24 k) c2d5 l) 1

1. Evaluate each using a rule of exponents.

a) x0 = b) 41 = c) (- 4)0 = d) (- 6)1 =

e) y1 = f) 2₃ 0 = g) 5₆ 1 = h) 01 =

2. Write each product as one base with one exponent. You may expand and count the number of factors or use the product rule.

a) x2 _{· x}6 _{= b) v}5 _{· v}9 _{= c) y}3 _{· y =}

d) c· c4 _{= e) p}5 _{· p}0 _{= f) k}0 _{· k}6 ₌

g) w · w0 _{= h) y}0 _{· y}0 _{= i) x}4 _{· x}10 ₌

j) c3 _{· c}3 _{= k) p}1 _{· p}1 _{= l) m}0 _{· q}4 ₌

3. Write each quotient as one base with one exponent. You may expand and count the number of factors or use the quotient rule.

a) x

9

x5 = b)

w6

w3 = c)

c5 c1 =

d) p

3

p0 = e)

y8

y = f) a

7

a7 =

g) m

12

m6 = h)

x3

x2 = i)

x x =

j) c

3

c3 = k)

y 10

y2 = l)

m 21

4. Use the Distributive Properties for Exponents to rewrite each of these. Simplify if possible.

a) (k·h)4 = b) (3m)4 = c) (-3·p)2 =

d) ( - 2x )3 _{= e) }y_x5 _{= f) }5_h0 ₌

g) p₆

1

= h) 2h₃ 

3

= i) -7p₈ 

2

=

j) -_2w1v

3

= k) (2p)5 = l) (8x)0 =

5. Use the Power Rule to rewrite each of these. Simplify if possible.

a) ( x2 ₎4 _{= b) ( r}5 ₎3 _{= c) ( 2k}3 ₎4 ₌

d) ( y4_b3 ₎2 _{= e) ( 4c}3 ₎3 _{= f) ( xw}2 ₎5 ₌

g) _p _

2

m3 2

= h) __xy4_

3

= i) __k22_

3

=

j) _m _

0

v1 4

= k) (x3w4)1 = l) (x6w)0 =

m) _4c_k9_ 1

= n) _m _

4

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