Rules of Roots Combined - 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 fractionsRules of Roots Combined: Identify the greater valueApplying Combined Exponents Rules: Two VariablesApplying Combined Exponents Rules: Identify the greater valueApplying Combined Exponents Rules: Presenting powers in the denominator as powers with negative exponentsRules of Roots Combined: Same base and different indicatorApplying Combined Exponents Rules: BinomialApplying Combined Exponents Rules: Variables in the exponent of the powerRules of Roots Combined: Using variablesApplying Combined Exponents Rules: Single VariableRules of Roots Combined: Solving the equationRules of Roots Combined: Using multiple rulesApplying Combined Exponents Rules: Using the laws of exponentsRules of Roots Combined: Applying the formulaApplying 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 formulaRules of Roots Combined: Number of termsApplying Combined Exponents Rules: Calculating powers with negative exponentsApplying Combined Exponents Rules: MonomialApplying Combined Exponents Rules: A power lawApplying Combined Exponents Rules: Using multiple rules

Understanding the combination of powers and roots is important and necessary.

First property:
a=a12\sqrt a=a^{ 1 \over 2}
Second property:
amn=amn\sqrt[n]{a^m}=a^{\frac{m}{n}}
Third property:
(a×b)=a×b\sqrt{(a\times b)}=\sqrt{a}\times \sqrt{b}

Fourth property:
ab=ab\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}

Fifth property:  
amn=an×m\sqrt[n]{\sqrt[m]{a}}=\sqrt[n\times m]{a}

Suggested Topics to Practice in Advance

  1. Square root of a product
  2. Square root of a quotient
  3. Square Roots

Practice Rules of Roots Combined

Examples with solutions for Rules of Roots Combined

Exercise #1

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 #2

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 #3

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 #4

(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 #5

(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 #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

Choose the largest value

Video Solution

Step-by-Step Solution

Let's begin by calculating the numerical value of each of the roots in the given options:

25=516=49=3 \sqrt{25}=5\\ \sqrt{16}=4\\ \sqrt{9}=3\\ We can determine that:

5>4>3>1 Therefore, the correct answer is option A

Answer

25 \sqrt{25}

Exercise #8

Solve the following exercise:

301= \sqrt{30}\cdot\sqrt{1}=

Video Solution

Step-by-Step Solution

Let's start with a reminder of the definition of a root as a power:

an=a1n \sqrt[n]{a}=a^{\frac{1}{n}}

We will then use the fact that raising the number 1 to any power always yields the result 1, particularly raising it to the power of half of the square root (which we obtain by using the definition of root as a power mentioned earlier),

In other words:

301=3012=30112=301=30 \sqrt{30}\cdot\sqrt{1}= \\ \downarrow\\ \sqrt{30}\cdot\sqrt[2]{1}=\\ \sqrt{30}\cdot 1^{\frac{1}{2}}=\\ \sqrt{30} \cdot1=\\ \boxed{\sqrt{30}} Therefore, the correct answer is answer C.

Answer

30 \sqrt{30}

Exercise #9

Solve the following exercise:

161= \sqrt{16}\cdot\sqrt{1}=

Video Solution

Step-by-Step Solution

Let's start by recalling how to define a root as a power:

an=a1n \sqrt[n]{a}=a^{\frac{1}{n}}

Next, we will remember that raising 1 to any power will always yield the result 1, even the half power of the square root.

In other words:

161=1612=16112=161=16=4 \sqrt{16}\cdot\sqrt{1}= \\ \downarrow\\ \sqrt{16}\cdot\sqrt[2]{1}=\\ \sqrt{16}\cdot 1^{\frac{1}{2}}=\\ \sqrt{16} \cdot1=\\ \sqrt{16} =\\ \boxed{4} Therefore, the correct answer is answer D.

Answer

4 4

Exercise #10

Solve the following exercise:

12= \sqrt{1}\cdot\sqrt{2}=

Video Solution

Step-by-Step Solution

Let's start by recalling how to define a square root as a power:

an=a1n \sqrt[n]{a}=a^{\frac{1}{n}}

Next, we remember that raising 1 to any power always gives us 1, even the half power we got from converting the square root.

In other words:

12=122=1122=12=2 \sqrt{1} \cdot \sqrt{2}= \\ \downarrow\\ \sqrt[2]{1}\cdot \sqrt{2}=\\ 1^{\frac{1}{2}} \cdot\sqrt{2} =\\ 1\cdot\sqrt{2}=\\ \boxed{\sqrt{2}} Therefore, the correct answer is answer a.

Answer

2 \sqrt{2}

Exercise #11

Solve the exercise:

Y2+Y6Y5Y= Y^2+Y^6-Y^5\cdot Y=

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} We apply it in the problem:

Y2+Y6Y5Y=Y2+Y6Y5+1=Y2+Y6Y6=Y2 Y^2+Y^6-Y^5\cdot Y=Y^2+Y^6-Y^{5+1}=Y^2+Y^6-Y^6=Y^2 When we apply the previous property to the third expression from the left in the sum, and then simplify the total expression by adding like terms.

Therefore, the correct answer is option D.

Answer

Y2 Y^2

Exercise #12

Solve the exercise:

a2:a+a3a5= a^2:a+a^3\cdot a^5=

Video Solution

Step-by-Step Solution

First we rewrite the first expression on the left of the problem as a fraction:

a2a+a3a5 \frac{a^2}{a}+a^3\cdot a^5 Then we use two properties of exponentiation, to multiply and divide terms with identical bases:

1.

bmbn=bm+n b^m\cdot b^n=b^{m+n}

2.

bmbn=bmn \frac{b^m}{b^n}=b^{m-n}

Returning to the problem and applying the two properties of exponentiation mentioned earlier:

a2a+a3a5=a21+a3+5=a1+a8=a+a8 \frac{a^2}{a}+a^3\cdot a^5=a^{2-1}+a^{3+5}=a^1+a^8=a+a^8

Later on, keep in mind that we need to factor the expression we obtained in the last step by extracting the common factor,

Therefore, we extract from outside the parentheses the greatest common divisor to the two terms which are:

a a We obtain the expression:

a+a8=a(1+a7) a+a^8=a(1+a^7) when we use the property of exponentiation mentioned earlier in A.

a8=a1+7=a1a7=aa7 a^8=a^{1+7}=a^1\cdot a^7=a\cdot a^7

Summarizing the solution to the problem and all the steps, we obtained the following:

a2a+a3a5=a(1+a7) \frac{a^2}{a}+a^3\cdot a^5=a(1+a^{7)}

Therefore, the correct answer is option b.

Answer

a(1+a7) a(1+a^7)

Exercise #13

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 #14

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 #15

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

Topics learned in later sections

  1. Square Root Rules