Rules of Roots Combined - Examples, Exercises and Solutions

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

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

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

We use the formula:

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

and therefore we obtain:

$(a^5)^7=a^{5\times7}=a^{35}$

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

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

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:

$$$\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.

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

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

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:

$$$\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.

To simplify the given expression, we will use the following three laws of exponents:

a. Definition of root as an exponent:

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

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

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

c. Law of exponents for an exponent raised to an exponent:

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

We'll start by converting the fourth root to an exponent using the law of exponents mentioned in a.:

$\sqrt{25x^4}= \\ \downarrow\\ (25x^4)^{\frac{1}{2}}=$

We'll continue, using the law of exponents mentioned in b. and apply the exponent to each factor in the parentheses:

$(25x^4)^{\frac{1}{2}}= \\ 25^{\frac{1}{2}}\cdot(x^4)^{{\frac{1}{2}}}$

We'll continue, using the law of exponents mentioned in c. and perform the exponent applied to the term with an exponent in parentheses (the second factor in the multiplication):

$25^{\frac{1}{2}}\cdot(x^4)^{{\frac{1}{2}}} = \\ 25^{\frac{1}{2}}\cdot x^{4\cdot\frac{1}{2}}=\\ 25^{\frac{1}{2}}\cdot x^{2}=\\ \sqrt{25}\cdot x^2=\\ \boxed{5x^2}$

In the final steps, we first converted the power of one-half applied to the first factor in the multiplication back to the fourth root form, again, according to the definition of root as an exponent mentioned in a. (in the reverse direction) and then calculated the known fourth root of 25.

Therefore, the correct answer is answer a.

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

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

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:

$$$\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.

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.

We use the zero exponent rule.

$X^0=1$We obtain

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

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

We use the property with our exercise and solve:

$(3^5)^4=3^{5\times4}=3^{20}$

We use the formula:

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

Therefore, we obtain:

$6^{2\times13}=6^{26}$

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:

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

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

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

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

Using the quotient rule for exponents: $\frac{a^m}{a^n} = a^{m-n}$.

Here, we have $\frac{3^5}{3^2} = 3^{5-2}$

Simplifying, we get $3^3$

Using the quotient rule for exponents: $\frac{a^m}{a^n} = a^{m-n}$.

Here, we have $\frac{5^6}{5^4} = 5^{6-4}$. Simplifying, we get $5^2$.

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:

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

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

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:

$a^n\times a^m=a^{n+m}$

We are going to focus on the term a:

$a^3\times a^2=a^{3+2}=a^5$

We are going to focus on the term b:

$b^4\times b^5=b^{4+5}=b^9$

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

$a^5\times b^9$