Laws Of Exponents

Question 1

a small number that is written on the upper right of another number (or variable) which is called the base

Answer 1

Exponent

Question 2

Example 1:Compute 35

Answer 2

Solution:The exponent of 5 tells us that 3 is multiplied by itself five times.

Therefore, 35= 243

Question 3

Example 1:What is the value of -94?

Answer 3

Solution:-94is an example of Case 1. To compute for -94, we need to calculate 94first. Afterward, multiply the result by -1:

6561

Question 4

Example 2:What is the value of (- 9)4?

Answer 4

Solution:Since – 9 is inside the parentheses, it indicates that – 9 is what is being raised to the power of 4. Thus, we need to multiply – 9 to itself four times.

6561

Question 5

Example 1:Write k5in expanded form.

Answer 5

Solution:The exponent in k5tells us that the variable k is being multiplied by itself 5 times. Thus, the expanded form of k5is:

k5= k∙k∙k∙k∙k

Question 6

Example 2:Expressu∙u∙u∙u∙u∙u in exponential form.

Answer 6

Solution:Note that the variableuis used six times. Hence, we must use an exponent of 6. Thus:

u∙u∙u∙u∙u∙u= u6

Question 7

Example 1:Write -3x5in expanded form.

Answer 7

Solution:The variablexis the only one raised to the power of 5. Thus, only the variablexis the base of exponent 5 in the given, and -3 is not included.

Thus, -3x5= -3(x∙x∙x∙x∙x)

Question 8

Example 2:Write (-3x)5in expanded form.

Answer 8

Solution:The existence of the parentheses indicates that both the -3 and x in -3xare raised to the power of 5. Thus, -3x is the base of exponent 5.

In other words, (-3x)5= -3x∙–3x∙ -3x∙–3x∙ -3x

Question 9

Suppose we want to multiplyx2byx4. Note thatx2andx4have the same base.How can we multiply them?

Answer 9

One possible method is to expandx2andx4:

x2= x∙x

x4= x∙x∙x∙x

Multiplying the expanded values:

      x2∙x4

       (x∙x)∙( x∙x∙x∙x)

Note that we can express the product of the expanded values into exponential form:

(x∙x)∙( x∙x∙x∙x) =x∙x∙x∙x∙x∙x =x6

Therefore,x2∙x4=x6

Question 10

Example 1:Compute for 2⁴∙2²

Answer 10

Solution:We have expressions with the same bases (i.e., 2) being multiplied together. Thus, we can apply the product rule.

Let us copy the common base first:

Then, add the exponents:

Thus, using the product rule:2⁴∙2²= 2⁶

Question 11

Example 2:Multiply b⁵by b³

Answer 11

Solution:Since we have the same bases being multiplied together, we can apply the product rule:

Let us copy the common base first:

Then, add the exponents:

Therefore, using the product rule:b⁵∙b³=b⁸

Question 12

Example 3:Multiply a³b²by a²b⁴

Answer 12

Solution:We have two bases involved here, the variablesaandb.

Thus, we need to apply the product rule, each foraandb:

Let us copy the common bases first:

Add the exponents for the common bases.

Hence,a³b²∙a²b⁴=a⁵b⁶

Question 13

Example 4:Multiply (x + 5)⁶by (x + 5)³

Answer 13

Solution:In this case, the common base isx + 5.Hence, we can apply the product rule.

Let us copy the common base first:

Add the exponents:

Therefore,(x + 5)⁶∙(x + 5)³= (x + 5)⁹

Question 14

Example 5:Compute for a(a²)

Answer 14

Solution: If two variables are written together with the other one enclosed in parentheses, it implies that the variables are being multiplied. Since we have a common base in the given (which isa), we can apply the product rule here.

Note that if a number or a variable has no exponent written above it, it implies that the exponent is 1.

Let us copy the common base first:

Add the exponents:

Therefore,a(a²) = a³

Question 15

Example 1:Compute for x⁷÷x³

Answer 15

Solution:Since we are dividing exponential expressions with the same base, we can apply the quotient rule.

Let us copy the common base first:

Subtract the exponents:

Therefore,x⁷÷x³= x⁴

Question 16

Example 2:Simplify x⁹⁄x⁴

Answer 16

Solution:x9⁄x4also means x9÷ x4. Since we are dividing exponential expressions with the same base, we can apply the quotient rule.

Let us start by copying the common base:

Finally, subtract the exponents:

Therefore, x⁹/x⁴= x⁵

Question 17

Example 3:Simplify p⁸q²⁄p⁶q

Answer 17

Solution:We have two bases involved: the variablespandq. Thus, we will use the quotient rule for the variablespandq.

Copying the common bases:

Finally, subtract the exponents for each of the common bases.

Hence,p8q2⁄p6q= p²q

Question 18

Example 4:Divide 1 000 000 000 by 1 000 000

Answer 18

Solution:Note that we can express1 000 000 000 as 109. On the other hand, we can express 1 000 000 as 106. Therefore, we can answer the problem by dividing 109by 106.

Since we have a common base (which is 10), we can apply the quotient rule:

Let us copy the common base first:

Then, subtract the exponents:

Therefore, the answer is 103or 1000.

Tip:We can express a multiple of 10 into exponential form quickly by counting the number of zeros it has. For example, 1 000 000 000 has 9 zeros. Thus, if we express 1 000 000 000 in exponential form, we can determine the exponent to be used based on the number of zeros it has. Therefore 1 000 000 000 = 10⁹

Question 19

Example 1:Simplify (k⁴)²

Answer 19

Solution:Notice that the entirek4is raised to 2. Applying the power rule, we can combine the exponents into one by multiplying them. Thus,

(k4)2= k4×2= k8

Therefore,(k4)2= k⁸

Question 20

Example 2:What is the value of (32)3?

Answer 20

Solution:Applying the product rule:

(32)3= 32×3= 36

Now, all we need to do next is expand 36:

3 ∙ 3 ∙ 3 ∙ 3 ∙ 3 ∙ 3 = 729

Therefore,(32)3= 729

Question 21

Example 3:Simplify (a⁵)²

Answer 21

Solution:Applying the product rule:

(a5)2= a5×2= a10

Therefore, (a5)2= a¹⁰

Question 22

Example 4:Simplify (8y⁵)²

Answer 22

Solution:Note that the base of the exponent 2 is 8y5. This means that we need to apply the power rule both for 8 and y5:

(8y5)2= 82y5×2= 82y10

Since 82= 8 ∙ 8 = 64, then 82y10= 64y10

Therefore, (8y5)2= 64y¹⁰

Question 23

If an expression has more than one variable multiplied together and raised to a certain power, we can simplify that expression using the

Answer 23

Power of a product rule

Question 24

Example 1:Simplify (xy)²

Answer 24

Solution:The given expression has two variables multiplied together (which isxy)and raised to the power of 2. This means that we can simplify it using the power of a product rule.

The power of the product rule allows us to “distribute” the exponent to each variable:

(xy)2 = (x2)(y2)

Therefore,(xy)2= x²y²

Question 25

Example 2: Simplify (a⁴b³)³

Answer 25

Solution: Let us apply the power of the product rule to simplify the given expression. We start by “distributing” 3 to each of the variables: (a4b3)3 = (a4)3 (b3)3  To further simplify the expression, we can apply the product rule to each variable:  (a4)3 (b3)3 = a4 × 3 b3 × 3  = a12b9 Hence, (a4b3)3 =  a¹²b⁹

Question 26

Example 3: Simplify (4a³b²)²

Answer 26

Solution: Let us apply the power of the product rule to simplify the given expression. Note that we should also raise 4 to the power of 2: (4a3b2)2 = (4)2 (a3)2 (b2)2 Now, we apply the product rule to each variable: (4)2(a6)(b4) Lastly, since 42 = 16: 16a6b4 Therefore, (4a3b2)2 = 16a⁶b⁴

Question 27

if variables are divided together and raised to a certain power, we can simplify the expression using the

Answer 27

Power of the quotient rule

Question 28

Example 1: Apply the power of the quotient rule to (x⁄y)²

Answer 28

Solution: Through the power of the quotient rule, we can distribute the exponent to the variables involved in the division process: Therefore, using the power of the quotient rule, (x⁄y)2 = x²⁄y²

Question 29

Example 2: Simplify (a⁴⁄b²)²

Answer 29

Solution: Since we have two variables (a4 and b2) divided together and raised to a certain power, we can apply the power of the quotient rule: Notice that we can simplify the expression further using the product rule: Therefore, (a4⁄b2)2 = a⁸⁄b⁴

Question 30

The Zero-Exponent Rule states that any nonzero base raised to 

Answer 30

0

Question 31

Example 1: Suppose that m ≠ 0, what is the value of m⁰?

Answer 31

Solution: By the zero-exponent rule, m⁰ = 1.

Question 32

Example 2: What is the value of 109⁰?

Answer 32

Solution: By the zero-exponent rule, 109⁰ = 1

Question 33

Example 3: Simplify 15x⁰

Answer 33

Solution: We know that by the zero-exponent rule, x0 = 1. Take note that x0 is multiplied by 15 in 15x0. Since x0 = 1: Hence, 15x⁰ = 15

Question 34

Example 4: Simplify a⁰b²c

Answer 34

Solution: By the zero-exponent rule, a0 = 1. Since a0 is multiplied to b2c in the given expression a0b2c: Hence,  a0b2c = b²c

Question 35

Example 1: What is the value of 5^-3?

Answer 35

Solution: Using the negative exponent rule, we can express 5-3 as 1⁄53. Note that we can expand 53 as 5 x 5 x 5 and obtain 125. Therefore, 5^-3 = 1⁄125

Question 36

Example 2: Express y^-1 as an expression without a negative exponent.

Answer 36

Solution: Using the negative exponent rule, we can express y-1 as 1⁄y which has no negative exponent involved. Thus, the answer is 1⁄y.

Question 37

Example 3: Express a^2b^-3c without a negative exponent.

Answer 37

Solution: We can apply the negative exponent rule to express a2b-3c without a negative exponent. However, since b– 3 is the only base raised to a negative exponent, then we can only apply the negative exponent rule to b-3 and it is the only base that we are going to put in the denominator. Therefore, a2b-3c can be written without a negative exponent as a^2c⁄b^3

Question 38

a^m ∙ a^n = a^m + n

Answer 38

Product rule

Question 39

a^m⁄a^n= a^m – n  

Answer 39

Quotient rule

Question 40

(a^m)^n = a^mn

Answer 40

Power rule

Question 41

(a b)^p  = a^p b^p

Answer 41

Power of a Product Rule

Question 42

(a⁄b)^m = a^m⁄b^m, where b ≠ 0

Answer 42

Power of a Quotient Rule

Question 43

a^0 = 1

Answer 43

Zero Exponent Rule

Question 44

a^-m = 1⁄a^m

Answer 44

Negative Exponent Rule

Question 45

Example 1: Simplify 5p^0q^-2 

Answer 45

Solution: We can simplify the given expression by making all of its exponents positive. Let us start by applying the zero-exponent rule: 5p0q-2 5(1)q-2 (since p0 = 1) 5q-2 We can then remove the negative exponent using the negative exponent Rule: 5q-2 5⁄q2 Thus, the answer is 5⁄q^2

Question 46

Example 2: Simplify (a^4⁄a^2)^2

Answer 46

Solution 1: Note that we can distribute the exponent 2 which is outside the parentheses to the bases that are inside the parentheses using the power of the quotient rule: (a4⁄a2)2 = a4 × 2⁄a2 × 2 = a8⁄a4   Since we are dividing the same bases, we can apply the quotient rule: a8⁄a4  = a8 – 4 = a4 Therefore, (a4⁄a2)2 = a4 You can also simplify the given expression using the alternative solution below. Solution 2: This time, let us start applying the quotient rule since we are dividing the same bases: (a4⁄a2)2  = (a4 – 2)2 = (a2)2 Notice that we can apply now the power rule since (a2)2 is an expression raised to an exponent then raised to another exponent.         (a2)2 = a2 × 2 = a^4

Question 47

Example 3: Simplify [(x + y)^2(x + y)^3]^-1

Answer 47

Solution: We can start by making the negative exponent positive. To do this, put the base into the denominator (negative exponent rule). The base in the given expression is the entire (x + y)2(x + y)3 We can simplify the expression further using the product rule since we are multiplying the same bases: Therefore, the answer is 1⁄(x + y)^5

Question 48

1) Compute 3a^2b^3•ab^2 a) 3a^3b^5 b) 3ab^5 c) 3a^2b^5 d) 3ab

Question 49

3) Express a^-1b^-3c^2 as an expression without negative exponent a)𝑎𝑐^2/𝑏^3 b)𝑎𝑏^3/𝑐^2 c) ab^3c^2 d)𝑐^2/𝑎𝑏^3

Question 50

4) Express (k + m)^2(k + m)^-1 as an expression without any negative exponent a) (k + m)^2 b) (k + m)^-1 c) k + m d) k^2 + m^2

Question 51

5) Apply the Laws of Exponents to compute for the value of (3 • 10^2)(3 • 10^3)

Answer 51

a) 900 000 b) 9 000 000 c) 90 000 d) 9 000