Exponents Rules - Examples, Exercises and Solutions

Question Types:
Applying Combined Exponents Rules: Factoring Out the Greatest Common Factor (GCF)Applying Combined Exponents Rules: More than one unknownApplying Combined Exponents Rules: Worded problemsApplying Combined Exponents Rules: Using variablesApplying Combined Exponents Rules: Variable in the exponent of the powerApplying Combined Exponents Rules: Complete the equationApplying Combined Exponents Rules: Using laws of exponents with parametersApplying Combined Exponents Rules: FactorizationApplying Combined Exponents Rules: Multiplying Exponents with the same baseApplying Combined Exponents Rules: TrinomialApplying Combined Exponents Rules: converting Negative Exponents to Positive ExponentsApplying Combined Exponents Rules: Number of termsApplying Combined Exponents Rules: Presenting powers with negative exponents as fractionsApplying Combined Exponents Rules: Two VariablesApplying Combined Exponents Rules: Identify the greater valueApplying Combined Exponents Rules: Presenting powers in the denominator as powers with negative exponentsApplying Combined Exponents Rules: BinomialApplying Combined Exponents Rules: Variables in the exponent of the powerApplying Combined Exponents Rules: Single VariableApplying Combined Exponents Rules: Using the laws of exponentsApplying Combined Exponents Rules: Variable in the base of the powerApplying Combined Exponents Rules: Presenting powers with negative exponents as fractionsApplying Combined Exponents Rules: Applying the formulaApplying Combined Exponents Rules: Calculating powers with negative exponentsApplying Combined Exponents Rules: MonomialApplying Combined Exponents Rules: A power lawApplying Combined Exponents Rules: Using multiple rules

Definition of Exponentiation

Exponents are a way to write the multiplication of a term by itself several times in a shortened form.

The number that is multiplied by itself is called the base, while the number of times the base is multiplied is called the exponent.

A - Exponentiation

an=aaa a^n=a\cdot a\cdot a ... (n times)

For example:

5555=54 5\cdot5\cdot5\cdot5=5^4

5 5 is the base, while 4 4 is the exponent.

In this case, the number 5 5 is multiplied by itself 4 4 times and, therefore, it is expressed as 5 5 to the fourth power or 5 5 to the power of 4 4 .

Practice Exponents Rules

Examples with solutions for Exponents Rules

Exercise #1

2423= \frac{2^4}{2^3}=

Video Solution

Step-by-Step Solution

Let's keep in mind that the numerator and denominator of the fraction have terms with the same base, therefore we use the property of powers to divide between terms with the same base:

bmbn=bmn \frac{b^m}{b^n}=b^{m-n} We apply it in the problem:

2423=243=21 \frac{2^4}{2^3}=2^{4-3}=2^1 Remember that any number raised to the 1st power is equal to the number itself, meaning that:

b1=b b^1=b Therefore, in the problem we obtain:

21=2 2^1=2 Therefore, the correct answer is option a.

Answer

2 2

Exercise #2

9993= \frac{9^9}{9^3}=

Video Solution

Step-by-Step Solution

Note that in the fraction and its denominator, there are terms with the same base, so we will use the law of exponents for division between terms with the same base:

bmbn=bmn \frac{b^m}{b^n}=b^{m-n} Let's apply it to the problem:

9993=993=96 \frac{9^9}{9^3}=9^{9-3}=9^6 Therefore, the correct answer is b.

Answer

96 9^6

Exercise #3

(35)4= (3^5)^4=

Video Solution

Step-by-Step Solution

To solve the exercise we use the power property:(an)m=anm (a^n)^m=a^{n\cdot m}

We use the property with our exercise and solve:

(35)4=35×4=320 (3^5)^4=3^{5\times4}=3^{20}

Answer

320 3^{20}

Exercise #4

(62)13= (6^2)^{13}=

Video Solution

Step-by-Step Solution

We use the formula:

(an)m=an×m (a^n)^m=a^{n\times m}

Therefore, we obtain:

62×13=626 6^{2\times13}=6^{26}

Answer

626 6^{26}

Exercise #5

50= 5^0=

Video Solution

Step-by-Step Solution

We use the power property:

X0=1 X^0=1 We apply it to the problem:

50=1 5^0=1 Therefore, the correct answer is C.

Answer

1 1

Exercise #6

1120=? 112^0=\text{?}

Video Solution

Step-by-Step Solution

We use the zero exponent rule.

X0=1 X^0=1 We obtain

1120=1 112^0=1 Therefore, the correct answer is option C.

Answer

1

Exercise #7

Simplify the expression:

a3a2b4b5= a^3\cdot a^2\cdot b^4\cdot b^5=

Video Solution

Step-by-Step Solution

In the exercise of multiplying powers, we will add up all the powers of the same product, in this case the terms a, b

We use the formula:

an×am=an+m a^n\times a^m=a^{n+m}

We are going to focus on the term a:

a3×a2=a3+2=a5 a^3\times a^2=a^{3+2}=a^5

We are going to focus on the term b:

b4×b5=b4+5=b9 b^4\times b^5=b^{4+5}=b^9

Therefore, the exercise that will be obtained after simplification is:

a5×b9 a^5\times b^9

Answer

a5b9 a^5\cdot b^9

Exercise #8

k2t4k6t2= k^2\cdot t^4\cdot k^6\cdot t^2=

Video Solution

Step-by-Step Solution

Using the power property to multiply terms with identical bases:

aman=am+n a^m\cdot a^n=a^{m+n} It is important to note that this law is only valid for terms with identical bases,

We notice that in the problem there are two types of terms. First, for the sake of order, we will use the substitution property to rearrange the expression so that the two terms with the same base are grouped together. The, we will proceed to solve:

k2t4k6t2=k2k6t4t2 k^2t^4k^6t^2=k^2k^6t^4t^2 Next, we apply the power property to each different type of term separately,

k2k6t4t2=k2+6t4+2=k8t6 k^2k^6t^4t^2=k^{2+6}t^{4+2}=k^8t^6 We apply the property separately - for the terms whose bases arek k and for the terms whose bases aret t We add the powers in the exponent when we multiply all the terms with the same base.

The correct answer then is option b.

Answer

k8t6 k^8\cdot t^6

Exercise #9

ababa2 a\cdot b\cdot a\cdot b\cdot a^2

Video Solution

Step-by-Step Solution

We use the power property to multiply terms with identical bases:

aman=am+n a^m\cdot a^n=a^{m+n} It is important to note that this property is only valid for terms with identical bases,

We return to the problem

We notice that in the problem there are two types of terms with different bases. First, for the sake of order, we will use the substitution property of multiplication to rearrange the expression so that the two terms with the same base are grouped together. Then, we will proceed to work:

ababa2=aaa2bb a\cdot b\operatorname{\cdot}a\operatorname{\cdot}b\operatorname{\cdot}a^2=a\cdot a\cdot a^2\cdot b\cdot b Next, we apply the power property for each type of term separately,

aaa2bb=a1+1+2b1+1=a4b2 a\cdot a\cdot a^2\cdot b\cdot b=a^{1+1+2}\cdot b^{1+1}=a^4\cdot b^2

We apply the power property separately - for the terms whose bases area a and then for the terms whose bases areb b and we add the exponents and simplify the terms.

Therefore, the correct answer is option c.

Note:

We use the fact that:

a=a1 a=a^1 and the same for b b .

Answer

a4b2 a^4\cdot b^2

Exercise #10

Solve the exercise:

(a5)7= (a^5)^7=

Video Solution

Step-by-Step Solution

We use the formula:

(am)n=am×n (a^m)^n=a^{m\times n}

and therefore we obtain:

(a5)7=a5×7=a35 (a^5)^7=a^{5\times7}=a^{35}

Answer

a35 a^{35}

Exercise #11

(3×4×5)4= (3\times4\times5)^4=

Video Solution

Step-by-Step Solution

We use the power law for multiplication within parentheses:

(xy)n=xnyn (x\cdot y)^n=x^n\cdot y^n We apply it to the problem:

(345)4=344454 (3\cdot4\cdot5)^4=3^4\cdot4^4\cdot5^4 Therefore, the correct answer is option b.

Note:

From the formula of the power property mentioned above, we understand that it refers not only to two terms of the multiplication within parentheses, but also for multiple terms within parentheses.

Answer

34×44×54 3^4\times4^4\times5^4

Exercise #12

(4×7×3)2= (4\times7\times3)^2=

Video Solution

Step-by-Step Solution

We use the power law for multiplication within parentheses:

(xy)n=xnyn (x\cdot y)^n=x^n\cdot y^n We apply it to the problem:

(473)2=427232 (4\cdot7\cdot3)^2=4^2\cdot7^2\cdot3^2 Therefore, the correct answer is option a.

Note:

From the formula of the power property mentioned above, we understand that we can apply it not only to the multiplication of two terms within parentheses, but is also for multiple terms within parentheses.

Answer

42×72×32 4^2\times7^2\times3^2

Exercise #13

(y×7×3)4= (y\times7\times3)^4=

Video Solution

Step-by-Step Solution

We use the power law for multiplication within parentheses:

(xy)n=xnyn (x\cdot y)^n=x^n\cdot y^n We apply it in the problem:

(y73)4=y47434 (y\cdot7\cdot3)^4=y^4\cdot7^4\cdot3^4 Therefore, the correct answer is option a.

Note:

From the formula of the power property mentioned above, we can understand that it applies not only to two terms within parentheses, but also for multiple terms within parentheses.

Answer

y4×74×34 y^4\times7^4\times3^4

Exercise #14

(42)3+(g3)4= (4^2)^3+(g^3)^4=

Video Solution

Step-by-Step Solution

We use the formula:

(am)n=am×n (a^m)^n=a^{m\times n}

(42)3+(g3)4=42×3+g3×4=46+g12 (4^2)^3+(g^3)^4=4^{2\times3}+g^{3\times4}=4^6+g^{12}

Answer

46+g12 4^6+g^{12}

Exercise #15

(4274)2= (\frac{4^2}{7^4})^2=

Video Solution

Step-by-Step Solution

(4274)2=42×274×2=4478 (\frac{4^2}{7^4})^2=\frac{4^{2\times2}}{7^{4\times2}}=\frac{4^4}{7^8}

Answer

4478 \frac{4^4}{7^8}