Square Root Quotient Property - Examples, Exercises and Solutions

Let's use the definition of root as a power:

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

$$when we remember that in a square root (also called "root to the power of 2") we don't write the root's power and:

$n=2$

meaning:

$\sqrt{a}=\sqrt[2]{a}=a^{\frac{1}{2}}$

We will convert therefore all the roots in the problem to powers:

$\frac{\sqrt{36}}{\sqrt{9}}=\frac{36^{\frac{1}{2}}}{9^{\frac{1}{2}}}$

Now let's recall the power law for a fraction in parentheses:

$\frac{a^n}{c^n}= \big(\frac{a}{c}\big)^n$

But in the opposite direction,

Note that both the numerator and denominator in the last expression we got are raised to the same power, therefore we can write the expression using the above power law as a fraction in parentheses raised to a power:
$\frac{36^{\frac{1}{2}}}{9^{\frac{1}{2}}}=\big(\frac{36}{9}\big)^{\frac{1}{2}}$

We emphasize that we could do this only because both the numerator and denominator of the fraction were raised to the same power,

Let's summarize our solution steps so far we got that:

$\frac{\sqrt{36}}{\sqrt{9}}=\frac{36^{\frac{1}{2}}}{9^{\frac{1}{2}}} =\big(\frac{36}{9}\big)^{\frac{1}{2}}$

Now let's calculate (by reducing the fraction) the expression inside the parentheses:

$\big(\frac{36}{9}\big)^{\frac{1}{2}} =4^\frac{1}{2}$

and we'll return to the root form using the definition of root as a power mentioned above, but in the opposite direction:

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

Let's apply this definition to the expression we got:

$4^\frac{1}{2}=\sqrt[2]{4}\ =\sqrt{4}=2$

where in the last step we calculated the numerical value of the root of 4,

Let's summarize the solution steps, we got that:

$\frac{\sqrt{36}}{\sqrt{9}}=\big(\frac{36}{9}\big)^{\frac{1}{2}} =\sqrt{4}=2$

Therefore the correct answer is answer B.

Let's simplify the expression, first we'll reduce the fraction under the square root:

$\sqrt{\frac{2}{4}}= \\ \sqrt{\frac{1}{2}}=$

We'll use two exponent laws:

A. Definition of root as a power:

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

B. The power law for powers applied to terms in parentheses:

$\big(\frac{a}{b}\big)^n=\frac{a^n}{b^n}$

Let's return to the expression we received, first we'll use the law mentioned in A and convert the square root to a power:

$\sqrt{\frac{1}{2}}=\\ \big(\frac{1}{2}\big)^{\frac{1}{2}}=$

We'll continue and apply the power law mentioned in B, meaning- we'll apply the power separately to the numerator and denominator, in the next step we'll remember that raising the number 1 to any power will always give the result 1, and in the fraction's denominator we'll return to the root notation, again, using the power law mentioned in A (in the opposite direction):

$\big(\frac{1}{2}\big)^{\frac{1}{2}}= \\ \frac{1^{\frac{1}{2}}}{2^{\frac{1}{2}}}=\\ \boxed{\frac{1}{\sqrt{2}}}\\$ Let's summarize the simplification of the given expression:

$\sqrt{\frac{2}{4}}= \\ \sqrt{\frac{1}{2}}= \\ \frac{1^{\frac{1}{2}}}{2^{\frac{1}{2}}}=\\ \boxed{\frac{1}{\sqrt{2}}}\\$Therefore, the correct answer is answer D.

Let's use the definition of root as a power:

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

when we remember that in a square root (also called "root to the power of 2") we don't write the root's power and:

$n=2$

meaning:

$\sqrt{a}=\sqrt[2]{a}=a^{\frac{1}{2}}$

Let's return to the problem and convert using the root definition we mentioned above the roots in the problem:

$\frac{\sqrt{25x^2}}{\sqrt{x^2}}=\frac{(25x^2)^{\frac{1}{2}}}{(x^{2)^{\frac{1}{2}}}}$

Now let's recall two laws of exponents:

a. The law of exponents for a power applied to a product in parentheses:

$(a\cdot b)^n=a^n\cdot b^n$

b. The law of exponents for a power of a power:

$(a^m)^n=a^{m\cdot n}$

Let's apply these laws to the numerator and denominator of the fraction in the expression we got in the last step:

$\frac{(25x^2)^{\frac{1}{2}}}{(x^{2)^{\frac{1}{2}}}}=\frac{25^{\frac{1}{2}}\cdot(x^2)^{\frac{1}{2}}}{(x^2)^{\frac{1}{2}}}=\frac{25^{\frac{1}{2}}x^{2\cdot\frac{1}{2}}}{x^{2\cdot\frac{1}{2}}}$

$$where in the first stage we applied the above law of exponents mentioned in a' and applied the power to both factors of the product in parentheses in the fraction's numerator, we did this carefully using parentheses since one of the factors in the parentheses is already raised to a power, in the second stage we applied the second law of exponents mentioned in b' to the second factor in the product in the fraction's numerator and similarly to the factor in the fraction's denominator,

Let's simplify the expression we got:

$\frac{25^{\frac{1}{2}}x^{2\cdot\frac{1}{2}}}{x^{2\cdot\frac{1}{2}}}=\frac{\sqrt{25}x^{\frac{2}{2}}}{x^{\frac{2}{2}}}=\frac{5x^1}{x^1}$

where in the first stage we converted back the fraction's power to a root, for the first factor in the product, this was done using the definition of root as a power mentioned at the beginning of the solution, but in the opposite direction, then we calculated the numerical value of the root,

Additionally - we calculated the product of the power of the second factor in the product in the fraction's numerator in the expression we got and similarly for the factor in the fraction's denominator, then we simplified the resulting fraction for that factor.

Let's complete the calculation and simplify the resulting fraction:

$\frac{5x^1}{x^1} =\frac{5\not{x}}{\not{x}}=5$

Let's summarize the solution steps so far, we got that:

$\frac{\sqrt{25x^2}}{\sqrt{x^2}}=\frac{(25x^2)^{\frac{1}{2}}}{(x^{2)^{\frac{1}{2}}}} =\frac{5x^1}{x^1} =5$

Therefore the correct answer is answer c.

Let's use the definition of root as a power:

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

when we remember that in a square root (also called "root to the power of 2") we don't write the root's power and:

$n=2$

meaning:

$\sqrt{a}=\sqrt[2]{a}=a^{\frac{1}{2}}$

Let's return to the problem and convert using the root definition we mentioned above the root in the fraction's numerator in the problem:

$\frac{\sqrt{49x^2}}{x}=\frac{(49x^2)^{\frac{1}{2}}}{x}$

Now let's recall two laws of exponents:

a. The law of exponents for a power applied to a product in parentheses:

$(a\cdot b)^n=a^n\cdot b^n$

b. The law of exponents for a power of a power:

$(a^m)^n=a^{m\cdot n}$

Let's apply these laws to the fraction's numerator in the expression we got in the last step:

$\frac{(49x^2)^{\frac{1}{2}}}{x}=\frac{49^{\frac{1}{2}}\cdot(x^2)^{\frac{1}{2}}}{x}=\frac{49^{\frac{1}{2}}x^{2\cdot\frac{1}{2}}}{x}$

$$where in the first stage we applied the above-mentioned law of exponents noted in a' and applied the power to both factors of the product in parentheses in the fraction's numerator, we did this carefully using parentheses since one of the factors in the parentheses is already raised to a power, in the second stage we applied the second law of exponents mentioned in b' to the second factor in the product,

Let's simplify the expression we got:

$\frac{49^{\frac{1}{2}}x^{2\cdot\frac{1}{2}}}{x}=\frac{\sqrt{49}x^{\frac{2}{2}}}{x}=\frac{7x^1}{x}$

where in the first stage we converted back the fraction's power to a root, for the first factor in the product, using the definition of root as a power mentioned at the beginning of the solution, but in the opposite direction,

Additionally- we calculated the product in the exponent of the second factor in the product in the fraction's numerator in the expression we got, then we simplified the resulting fraction in that exponent for that factor.

Let's finish the calculation and simplify the resulting fraction:

$\frac{7x^1}{x}=\frac{7\not{x}}{\not{x}}=7$

Let's summarize the solution steps so far, we got that:

$\frac{\sqrt{49x^2}}{x}=\frac{(49x^2)^{\frac{1}{2}}}{x}=\frac{49^{\frac{1}{2}}x^{2\cdot\frac{1}{2}}}{x} =7$

Therefore the correct answer is answer c.

Let's use the definition of root as a power:

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

when we remember that in a square root (also called "root to the power of 2") we don't write the root's power and:

$n=2$

meaning:

$\sqrt{a}=\sqrt[2]{a}=a^{\frac{1}{2}}$

Let's return to the problem and convert using the root definition we mentioned above the root in the numerator of the fraction in the problem:

$\frac{\sqrt{x^4}}{x}=\frac{(x^4)^{\frac{1}{2}}}{x}$

Now let's remember the power law for power of power:

$(a^m)^n=a^{m\cdot n}$

Let's apply this law to the numerator of the fraction in the expression we got in the last step:

$\frac{(x^4)^{\frac{1}{2}}}{x}=\frac{x^{4\cdot\frac{1}{2}}}{x}=\frac{x^\frac{4}{2}}{x}$

where in the first step we applied the above power law and in the second step we performed the multiplication in the power exponent of the numerator term,

Let's continue and simplify the expression we got, first we'll reduce the fraction with the power exponent in the numerator term and then we'll use the power law for division between terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$

to simplify the fraction in the complete expression:

$\frac{x^\frac{4}{2}}{x}=\frac{x^2}{x}=x^{2-1}=x$

Let's summarize the solution steps, we got that:

$$$\frac{\sqrt{x^4}}{x}=\frac{(x^4)^{\frac{1}{2}}}{x}=\frac{x^2}{x}=x$

Therefore the correct answer is answer A.

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

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

Introduction:

We will address the following two laws of exponents:

a. Definition of root as an exponent:

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

b. The law of exponents for an exponent applied to terms in parentheses:

$(\frac{a}{b})^n=\frac{a^n}{b^n}$

Note:

By combining the two laws of exponents mentioned in a (in the first and third stages below) and b (in the second stage below), we can derive another new rule:

$\sqrt[n]{\frac{a}{b}}=\\ (\frac{a}{b})^{\frac{1}{n}}=\\ \frac{a^{\frac{1}{n}}}{ b^{\frac{1}{n}}}=\\ \frac{\sqrt[n]{a}}{ \sqrt[n]{ b}}\\ \downarrow\\ \boxed{ \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{ \sqrt[n]{ b}}}$

And specifically for the fourth root we get:

$\boxed{ \sqrt{\frac{a}{ b}}=\frac{\sqrt{a}}{\sqrt{ b}} }$

Therefore, we can proceed with solving the problem:

$\frac{\sqrt{64}}{\sqrt{4}}=2x$Let's start by simplifying the expression on the left side, using the new rule we received in the introduction:

$\boxed{ \sqrt{\frac{a}{ b}}=\frac{\sqrt{a}}{\sqrt{ b}} }$

(But in the opposite direction, meaning we'll insert the product of roots as a product of terms under the same root) Then we'll perform the multiplication under the root:

$\frac{\sqrt{64}}{\sqrt{4}}=2x \\ \sqrt{\frac{64}{4}}=2x \\ \sqrt{16}=2x \\ 4=2x \\$In the final stage, we used the known fourth root of the number 16,

After simplifying the expression on the left side, to isolate the unknown, we'll divide both sides of the equation by its coefficient:

$4=2x\hspace{6pt}\text{/}:2 \\ 2=x \\ \downarrow\\ \boxed{x=2}$

Let's summarize the solution of the equation:

$\frac{\sqrt{64}}{\sqrt{4}}=2x \\ \sqrt{16}=2x \\ 4=2x \\ \downarrow\\ \boxed{x=2}$$$

Therefore, the correct answer is answer b.

Let's use the definition of root as a power:

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

We'll apply this definition and convert the roots in the problem:

$\frac{\sqrt[4]{2}}{\sqrt[5]{2}}= \frac{2^{\frac{1}{4}}}{2^{\frac{1}{5}}}$

Now let's recall the law of powers for division with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$

Let's apply this law to our problem:

$\frac{2^{\frac{1}{4}}}{2^{\frac{1}{5}}}=2^{\frac{1}{4}-\frac{1}{5}}$

Next, for convenience, we'll handle the expression in the power numerator from the last step separately and calculate the value of the fraction:

$\frac{1}{4}-\frac{1}{5}=\frac{5\cdot1-4\cdot1}{20}=\\ \frac{5-4}{20}=\frac{1}{20}$

In the first step, we combined the two fractions into one fraction line, by expanding to the common denominator of 20 and performing subtraction (in the first fraction on the left we expanded both numerator and denominator by 5, and in the second fraction we expanded both numerator and denominator by 4), in the following steps we simplified the resulting expression,

Let's return to the problem and consider the result of the subtraction operation between the fractions we just performed, we get:

$2^{\frac{1}{4}-\frac{1}{5}}=2^{\frac{1}{20}}$

Let's summarize the solution steps, we found that:

$\frac{\sqrt[4]{2}}{\sqrt[5]{2}}=2^{\frac{1}{4}-\frac{1}{5}}=2^{\frac{1}{20}}$

Therefore, the correct answer is answer C.

Introduction:

We will address the following three laws of exponents:

a. Definition of root as an exponent:

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

b. The law of exponents for exponents applied to multiplication of terms in parentheses:

$(a\cdot b)^n=a^n\cdot b^n$

c. The law of exponents for exponents applied to division of terms in parentheses:

$(\frac{a}{b})^n=\frac{a^n}{b^n}$

Note:

(1). By combining the two laws of exponents mentioned in a (in the first and third steps later) and b (in the second step later), we can obtain a new rule:

$\sqrt[n]{a\cdot b}=\\ (a\cdot b)^{\frac{1}{n}}=\\ a^{\frac{1}{n}}\cdot b^{\frac{1}{n}}=\\ \sqrt[n]{a}\cdot \sqrt[n]{ b}\\ \downarrow\\ \boxed{\sqrt[n]{a\cdot b}=\sqrt[n]{a}\cdot \sqrt[n]{ b}}$

And specifically for the fourth root we get:

$\boxed{ \sqrt{a\cdot b}=\sqrt{a}\cdot \sqrt{ b}}$

(2). Similarly, note that by combining the two laws of exponents mentioned in a (in the first and third steps later) and c (in the second step later), we can obtain another new rule:

$\sqrt[n]{\frac{a}{b}}=\\ (\frac{a}{b})^{\frac{1}{n}}=\\ \frac{a^{\frac{1}{n}}}{ b^{\frac{1}{n}}}=\\ \frac{\sqrt[n]{a}}{ \sqrt[n]{ b}}\\ \downarrow\\ \boxed{ \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{ \sqrt[n]{ b}}}$

And specifically for the fourth root we get:

$\boxed{ \sqrt{\frac{a}{ b}}=\frac{\sqrt{a}}{\sqrt{ b}} }$

Therefore, in solving the problem, meaning - in simplifying the given expression, we will use the two new rules we received in the introduction:

(1).

$\boxed{ \sqrt{a\cdot b}=\sqrt{a}\cdot \sqrt{ b}}$

(2).

$\boxed{ \sqrt{\frac{a}{ b}}=\frac{\sqrt{a}}{\sqrt{ b}} }$

We'll start by simplifying the expression in the numerator using the rule we received in the introduction (1) (but in the opposite direction, meaning we'll insert the multiplication of roots as a multiplication of terms under the same root) Then we'll perform the multiplication under the root in the numerator:

$\frac{\sqrt{20}\cdot\sqrt{4}}{\sqrt{5}}= \\ \frac{\sqrt{20\cdot4}}{\sqrt{5}}= \\ \frac{\sqrt{80}}{\sqrt{5}}= \\$We'll continue and simplify the fraction, using the rule we received in the introduction (2) (but in the opposite direction, meaning we'll insert the division of roots as a division of terms under the same root) Then we'll reduce the fraction under the root:

$\frac{\sqrt{80}}{\sqrt{5}}= \\ \\ \sqrt{\frac{80}{5}}=\\ \sqrt{16}=\\ \boxed{4}$

In the final stage, after reducing the fraction under the root, we used the known fourth root of the number 16.

Let's summarize the simplification process of the expression in the problem:

$\frac{\sqrt{20}\cdot\sqrt{4}}{\sqrt{5}}= \\ \frac{\sqrt{80}}{\sqrt{5}}= \\ \sqrt{16}=\\ \boxed{4}$

Therefore, the correct answer is answer B.

Introduction:

We will address the following three laws of exponents:

a. Definition of root as an exponent:

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

b. The law of exponents for exponents applied to multiplication of terms in parentheses:

$(a\cdot b)^n=a^n\cdot b^n$

c. The law of exponents for exponents applied to division of terms in parentheses:

$(\frac{a}{b})^n=\frac{a^n}{b^n}$

Note:

d. By combining the two laws of exponents mentioned in a' (in the first and third steps later) and b' (in the second step later), we can obtain a new rule:

$\sqrt[n]{a\cdot b}=\\ (a\cdot b)^{\frac{1}{n}}=\\ a^{\frac{1}{n}}\cdot b^{\frac{1}{n}}=\\ \sqrt[n]{a}\cdot \sqrt[n]{ b}\\ \downarrow\\ \boxed{\sqrt[n]{a\cdot b}=\sqrt[n]{a}\cdot \sqrt[n]{ b}}$

And specifically for the fourth root we get:

$\boxed{ \sqrt{a\cdot b}=\sqrt{a}\cdot \sqrt{ b}}$

e. Similarly, note that by combining the two laws of exponents mentioned in a' (in the first and third steps later) and c' (in the second step later), we can obtain another new rule:

$\sqrt[n]{\frac{a}{b}}=\\ (\frac{a}{b})^{\frac{1}{n}}=\\ \frac{a^{\frac{1}{n}}}{ b^{\frac{1}{n}}}=\\ \frac{\sqrt[n]{a}}{ \sqrt[n]{ b}}\\ \downarrow\\ \boxed{ \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{ \sqrt[n]{ b}}}$

And specifically for the fourth root we get:

$\boxed{ \sqrt{\frac{a}{ b}}=\frac{\sqrt{a}}{\sqrt{ b}} }$

Therefore, in solving the problem, that is - in simplifying the given expression, we will use the two new rules we received in the introduction:

(1).

$\boxed{ \sqrt{a\cdot b}=\sqrt{a}\cdot \sqrt{ b}}$$$(2).

$\boxed{ \sqrt{\frac{a}{ b}}=\frac{\sqrt{a}}{\sqrt{ b}} }$

We'll start by simplifying the expression in the numerator using the rule we received in the introduction (1) (but in the opposite direction, meaning we'll insert the multiplication of roots as a multiplication of terms under the same root) then we'll perform the multiplication under the root in the numerator:

$\frac{\sqrt{70}\cdot\sqrt{10}}{\sqrt{7}}= \\ \frac{\sqrt{70\cdot10}}{\sqrt{7}}= \\ \frac{\sqrt{700}}{\sqrt{7}}= \\$We'll continue and simplify the fraction, using the rule we received in the introduction (2) (but in the opposite direction, meaning we'll insert the division of roots as a division of terms under the same root) then we'll reduce the fraction under the root:

$\frac{\sqrt{700}}{\sqrt{7}}= \\ \sqrt{\frac{700}{7}}=\\ \sqrt{100}=\\ \boxed{10}$

In the final stage, after reducing the fraction under the root, we used the known fourth root of the number 100.

Let's summarize the process of simplifying the expression in the problem:

$\frac{\sqrt{70}\cdot\sqrt{10}}{\sqrt{7}}= \\ \frac{\sqrt{700}}{\sqrt{7}}= \\ \sqrt{100}=\\ \boxed{10}$

Therefore, the correct answer is answer a'.

Introduction:

We will address the following three laws of exponents:

a. Definition of root as an exponent:

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

b. The law of exponents for an exponent applied to a product in parentheses:

$(a\cdot b)^n=a^n\cdot b^n$

c. The law of exponents for an exponent applied to a quotient in parentheses:

$(\frac{a}{b})^n=\frac{a^n}{b^n}$

Note:

(1). By combining the two laws of exponents mentioned in a (in the first and third steps later) and b (in the second step later), we can obtain a new rule:

$\sqrt[n]{a\cdot b}=\\ (a\cdot b)^{\frac{1}{n}}=\\ a^{\frac{1}{n}}\cdot b^{\frac{1}{n}}=\\ \sqrt[n]{a}\cdot \sqrt[n]{ b}\\ \downarrow\\ \boxed{\sqrt[n]{a\cdot b}=\sqrt[n]{a}\cdot \sqrt[n]{ b}}$

And specifically for the fourth root we get:

$\boxed{ \sqrt{a\cdot b}=\sqrt{a}\cdot \sqrt{ b}}$

(2). Similarly, note that by combining the two laws of exponents mentioned in a (in the first and third steps later) and c (in the second step later), we can obtain another new rule:

$\sqrt[n]{\frac{a}{b}}=\\ (\frac{a}{b})^{\frac{1}{n}}=\\ \frac{a^{\frac{1}{n}}}{ b^{\frac{1}{n}}}=\\ \frac{\sqrt[n]{a}}{ \sqrt[n]{ b}}\\ \downarrow\\ \boxed{ \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{ \sqrt[n]{ b}}}$

And specifically for the fourth root we get:

$\boxed{ \sqrt{\frac{a}{ b}}=\frac{\sqrt{a}}{\sqrt{ b}} }$

Therefore, in solving the problem, meaning - in simplifying the given expression, we will use the two new rules we received in the introduction:

(1).

$\boxed{ \sqrt{a\cdot b}=\sqrt{a}\cdot \sqrt{ b}}$

(2).

$\boxed{ \sqrt{\frac{a}{ b}}=\frac{\sqrt{a}}{\sqrt{ b}} }$

We will start by simplifying the expression in the numerator using the rule we received in the introduction (1) (but in the opposite direction, meaning we will insert the product of roots as a product of terms under the same root) Then we will perform the multiplication under the root in the numerator:

$\frac{\sqrt{4}\cdot\sqrt{5}}{\sqrt{10}}= \\ \frac{\sqrt{4\cdot5}}{\sqrt{10}}= \\ \frac{\sqrt{20}}{\sqrt{10}}= \\$We will continue and simplify the fraction, using the rule we received in the introduction (2) (but in the opposite direction, meaning we will insert the quotient of roots as a quotient of terms under the same root) Then we will reduce the fraction under the root:

$\frac{\sqrt{20}}{\sqrt{10}}= \\ \sqrt{\frac{20}{10}}=\\ \boxed{\sqrt{2}}$

Let's summarize the process of simplifying the expression in the problem:

$\frac{\sqrt{4}\cdot\sqrt{5}}{\sqrt{10}}= \\ \frac{\sqrt{20}}{\sqrt{10}}= \\ \boxed{\sqrt{2}}$

Therefore, the correct answer is answer c.

Introduction:

We will address the following two laws of exponents:

a. The definition of root as an exponent:

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

b. The law of exponents for an exponent applied to terms in parentheses:

$(\frac{a}{b})^n=\frac{a^n}{b^n}$

Note:

By combining these two laws of exponents mentioned in a (in the first and third steps below) and b (in the second step below), we can derive another new rule:

$\sqrt[n]{\frac{a}{b}}=\\ (\frac{a}{b})^{\frac{1}{n}}=\\ \frac{a^{\frac{1}{n}}}{ b^{\frac{1}{n}}}=\\ \frac{\sqrt[n]{a}}{ \sqrt[n]{ b}}\\ \downarrow\\ \boxed{ \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{ \sqrt[n]{ b}}}$

Therefore, in solving the problem, meaning - simplifying the given expression, we will use the new rule we received in the introduction:

$\boxed{ \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{ \sqrt[n]{ b}}}$

We'll start by simplifying the expression using the rule we received in the introduction (but in the opposite direction, meaning we'll insert the product of roots as a product of terms under the same root) then we'll perform the multiplication under the root and finally we'll perform the fifth root operation:

$\frac{\sqrt[4]{128}}{\sqrt[4]{8}}= \\ \sqrt[4]{\frac{128}{8}}=\\ \sqrt[4]{16}=\\ \boxed{2}$

Therefore, the correct answer is answer B.