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

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 #2

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 #3

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 #4

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 #5

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 #6

(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 #7

(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 #8

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 #9

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 #10

Solve the following exercise:

103= \sqrt{10}\cdot\sqrt{3}=

Video Solution

Step-by-Step Solution

To simplify the given expression, we use two laws of exponents:

A. Defining the root as an exponent:

an=a1n \sqrt[n]{a}=a^{\frac{1}{n}} B. The law of exponents for dividing powers with the same base (in the opposite direction):

xnyn=(xy)n x^n\cdot y^n =(x\cdot y)^n

Let's start by using the law of exponents shown in A:

103=1012312= \sqrt{10}\cdot\sqrt{3}= \\ \downarrow\\ 10^{\frac{1}{2}}\cdot3^{\frac{1}{2}}= We continue, since we have a multiplication between two terms with equal exponents, we can use the law of exponents shown in B and combine them under the same base which is raised to the same exponent:

1012312=(103)12=3012=30 10^{\frac{1}{2}}\cdot3^{\frac{1}{2}}= \\ (10\cdot3)^{\frac{1}{2}}=\\ 30^{\frac{1}{2}}=\\ \boxed{\sqrt{30}} In the last steps, we performed the multiplication of the bases and used the definition of the root as an exponent shown earlier in A (in the opposite direction) to return to the root notation.

Therefore, the correct answer is B.

Answer

30 \sqrt{30}

Exercise #11

Solve the following exercise:

10025= \sqrt{100}\cdot\sqrt{25}=

Video Solution

Step-by-Step Solution

We can simplify the expression without using the laws of exponents, because the expression has known square roots, so let's simplify the expression and then perform the multiplication:

10025=105=50 \sqrt{100}\cdot\sqrt{25}=\\ 10\cdot5=\\ \boxed{50} Therefore, the correct answer is answer D.

Answer

50 50

Exercise #12

Solve the following exercise:

254= \sqrt{25}\cdot\sqrt{4}=

Video Solution

Step-by-Step Solution

We can simplify the expression directly without using the laws of exponents, since the expression has known square roots, so let's simplify the expression and then perform the multiplication:

254=52=10 \sqrt{25}\cdot\sqrt{4}=\\ 5\cdot2=\\ \boxed{10} Therefore, the correct answer is answer C.

Answer

10 10

Exercise #13

Solve the following exercise:

94= \sqrt{9}\cdot\sqrt{4}=

Video Solution

Step-by-Step Solution

We can simplify the expression without using the laws of exponents, since the expression has known square roots, so let's simplify the expression and then perform the multiplication:

94=32=6 \sqrt{9}\cdot\sqrt{4}=\\ 3\cdot2=\\ \boxed{6} Therefore, the correct answer is answer B.

Answer

6 6

Exercise #14

Solve the following exercise:

22525= \sqrt{\frac{225}{25}}=

Video Solution

Step-by-Step Solution

Let's simplify the expression. First, we'll reduce the fraction under the square root, then we'll calculate the result of the root:

22525=93 \sqrt{\frac{225}{25}}= \\ \sqrt{9}\\ \boxed{3} Therefore, the correct answer is option B.

Answer

3

Exercise #15

Solve the following exercise:

25= \sqrt{2}\cdot\sqrt{5}=

Video Solution

Step-by-Step Solution

In order to simplify the given expression we use two laws of exponents:

A. Defining the root as an exponent:

an=a1n \sqrt[n]{a}=a^{\frac{1}{n}} B. The law of exponents for dividing powers with the same bases (in the opposite direction):

xnyn=(xy)n x^n\cdot y^n =(x\cdot y)^n

Let's start by changing the square roots to exponents using the law of exponents shown in A:

25=212512= \sqrt{2}\cdot\sqrt{5}= \\ \downarrow\\ 2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}= We continue: since we are multiplying two terms with equal exponents we can use the law of exponents shown in B and combine them together as the same base raised to the same power:

212512=(25)12=1012=10 2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}= \\ (2\cdot5)^{\frac{1}{2}}=\\ 10^{\frac{1}{2}}=\\ \boxed{\sqrt{10}} In the last steps wemultiplied the bases and then used the definition of the root as an exponent shown earlier in A (in the opposite direction) to return to the root notation.

Therefore, the correct answer is answer B.

Answer

10 \sqrt{10}

Topics learned in later sections

  1. Square Root Rules