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

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

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

Fifth property:
$\sqrt[n]{\sqrt[m]{a}}=\sqrt[n\times m]{a}$

## Examples with solutions for Rules of Roots Combined

### Exercise #1

$112^0=\text{?}$

### Step-by-Step Solution

We use the zero exponent rule.

$X^0=1$We obtain

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

1

### Exercise #2

$\frac{1}{12^3}=\text{?}$

### Step-by-Step Solution

To begin with, we must remind ourselves of the Negative Exponent rule:

$a^{-n}=\frac{1}{a^n}$We apply it to the given expression :

$\frac{1}{12^3}=12^{-3}$Therefore, the correct answer is option A.

$12^{-3}$

### Exercise #3

$\frac{1}{2^9}=\text{?}$

### Step-by-Step Solution

We use the power property for a negative exponent:

$a^{-n}=\frac{1}{a^n}$We apply it to the given expression:

$\frac{1}{2^9}=2^{-9}$

Therefore, the correct answer is option A.

$2^{-9}$

### Exercise #4

$\frac{2^4}{2^3}=$

### 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:

$\frac{b^m}{b^n}=b^{m-n}$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.

$2$

### Exercise #5

$(9\times2\times5)^3=$

### Step-by-Step Solution

We use the law of exponents for a power that is applied to parentheses in which terms are multiplied:

$(x\cdot y)^n=x^n\cdot y^n$We apply the rule to the problem:

$(9\cdot2\cdot5)^3=9^3\cdot2^3\cdot5^3$When we apply the power within parentheses to the product of the terms we do so separately and maintain the multiplication,

Therefore, the correct answer is option B.

$9^3\times2^3\times5^3$

### Exercise #6

$\frac{81}{3^2}=$

### Step-by-Step Solution

First, we recognize that 81 is a power of the number 3, which means that:

$3^4=81$We replace in the problem:

$\frac{81}{3^2}=\frac{3^4}{3^2}$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:

$\frac{3^4}{3^2}=3^{4-2}=3^2$Therefore, the correct answer is option b.

$3^2$

### Exercise #7

$(3^5)^4=$

### Step-by-Step Solution

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:

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

$3^{20}$

### Exercise #8

$(6^2)^{13}=$

### Step-by-Step Solution

We use the formula:

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

Therefore, we obtain:

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

$6^{26}$

### Exercise #9

$7^{-24}=\text{?}$

### Step-by-Step Solution

Using the rules of negative exponents: how to raise a number to a negative exponent:

$a^{-n}=\frac{1}{a^n}$We apply it to the problem:

$7^{-24}=\frac{1}{7^{24}}$Therefore, the correct answer is option D.

$\frac{1}{7^{24}}$

### Exercise #10

$\frac{1}{8^3}=\text{?}$

### Step-by-Step Solution

We use the negative exponent rule.

$b^{-n}=\frac{1}{b^n}$

We apply it to the problem in the opposite sense.:

$\frac{1}{8^3}=8^{-3}$

Therefore, the correct answer is option A.

$8^{-3}$

### Exercise #11

$\frac{9^9}{9^3}=$

### 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:

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

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

$9^6$

### Exercise #12

$(\frac{2}{6})^3=$

### Step-by-Step Solution

We use the formula:

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

$(\frac{2}{6})^3=(\frac{2}{2\times3})^3$

We simplify:

$(\frac{1}{3})^3=\frac{1^3}{3^3}$

$\frac{1\times1\times1}{3\times3\times3}=\frac{1}{27}$

$\frac{1}{27}$

### Exercise #13

$19^{-2}=\text{?}$

### Step-by-Step Solution

In order to solve the exercise, we use the negative exponent rule.

$a^{-n}=\frac{1}{a^n}$

We apply the rule to the given exercise:

$19^{-2}=\frac{1}{19^2}$

We can then continue and calculate the exponent.

$\frac{1}{19^2}=\frac{1}{361}$

$\frac{1}{361}$

### Exercise #14

$(\frac{1}{4})^{-1}$

### Step-by-Step Solution

We use the power property for a negative exponent:

$a^{-n}=\frac{1}{a^n}$We will write the fraction in parentheses as a negative power with the help of the previously mentioned power:

$\frac{1}{4}=\frac{1}{4^1}=4^{-1}$We return to the problem, where we obtained:

$\big(\frac{1}{4}\big)^{-1}=(4^{-1})^{-1}$We continue and use the power property of an exponent raised to another exponent:

$(a^m)^n=a^{m\cdot n}$And we apply it in the problem:

$(4^{-1})^{-1}=4^{-1\cdot-1}=4^1=4$Therefore, the correct answer is option d.

$4$

### Exercise #15

$4^{-1}=\text{?}$

### Step-by-Step Solution

We begin by using the power rule of negative exponents.

$a^{-n}=\frac{1}{a^n}$We then apply it to the problem:

$4^{-1}=\frac{1}{4^1}=\frac{1}{4}$We can therefore deduce that the correct answer is option B.

$\frac{1}{4}$