Rules of Roots Combined: Solving the equation

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.

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:

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

Note:

By combining these two laws of exponents mentioned in a (in the first and third steps ahead) and b (in the second step ahead), 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}}$$$

$$Therefore, we will proceed with solving the problem as follows:

$\frac{x}{\sqrt{6}} = \sqrt{2}\cdot\sqrt{3}$$$

First, we'll eliminate the fraction line, which we'll do by multiplying both sides of the equation by the common denominator which is- $\sqrt{6}$:

$\frac{x}{\sqrt{6}} = \sqrt{2}\cdot\sqrt{3} \hspace{6pt}\text{/}\cdot\sqrt{6}\\ x=\sqrt{2}\cdot\sqrt{3}\cdot \sqrt{6}$

Let's continue and simplify the expression on the left side of the equation, using the rule we received in the introduction:

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

(which of course also applies to multiplication between numbers under a root), next we'll perform the multiplication under the root:

$x=\sqrt{2}\cdot\sqrt{3}\cdot\sqrt{6} \\ x=\sqrt{2\cdot3\cdot6} \\ x=\sqrt{36}\\ \boxed{x=6}$

In the final step, we used the known fourth root of the number 36,

Let's summarize the solution of the equation:

$\frac{x}{\sqrt{6}} = \sqrt{2}\cdot\sqrt{3} \hspace{6pt}\text{/}\cdot\sqrt{6}\\ x=\sqrt{2}\cdot\sqrt{3}\cdot \sqrt{6} \\ x=\sqrt{36}\\ \boxed{x=6}$

$$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.