## Definition of Exponentiation

Exponents are a way to write the multiplication of a term by itself several times in a shortened form.

The number that is multiplied by itself is called the base, while the number of times the base is multiplied is called the exponent.

$a^n=a\cdot a\cdot a$... (n times)

For example:

$5\cdot5\cdot5\cdot5=5^4$

$5$ is the base, while $4$ is the exponent.

In this case, the number $5$ is multiplied by itself $4$ times and, therefore, it is expressed as $5$ to the fourth power or $5$ to the power of $4$.

## Examples with solutions for Exponents Rules

### 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

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

### 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 #3

$(3^5)^4=$

### 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 #4

$5^0=$

### Step-by-Step Solution

We use the power property:

$X^0=1$We apply it to the problem:

$5^0=1$Therefore, the correct answer is C.

$1$

### Exercise #5

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

### 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 #6

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

### 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 #7

$(3\times4\times5)^4=$

### Step-by-Step Solution

We use the power law for multiplication within parentheses:

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

$(3\cdot4\cdot5)^4=3^4\cdot4^4\cdot5^4$Therefore, the correct answer is option b.

Note:

From the formula of the power property mentioned above, we understand that it refers not only to two terms of the multiplication within parentheses, but also for multiple terms within parentheses.

$3^44^45^4$

### Exercise #8

$(3\times2\times4\times6)^{-4}=$

### Step-by-Step Solution

We begin by using the power rule for parentheses.

$(z\cdot t)^n=z^n\cdot t^n$That is, the power applied to a product inside parentheses is applied to each of the terms within when the parentheses are opened,

We apply the above rule to the given problem:

$(3\cdot2\cdot4\cdot6)^{-4}=3^{-4}\cdot2^{-4}\cdot4^{-4}\cdot6^{-4}$Therefore, the correct answer is option d.

Note:

According to the formula of the power property inside parentheses mentioned above, it might seem as though it refers to only two terms of the product inside of the parentheses, but in reality, it is also valid for the power over a multiplication of many terms inside parentheses, as was seen above.

A good exercise is to demonstrate that if the previous property is valid for a power over a product of two terms inside parentheses (as formulated above), then it is also valid for a power over several terms of the product inside parentheses (for example - three terms, etc.).

$3^{-4}\times2^{-4}\times4^{-4}\times6^{-4}$

### Exercise #9

$(4\times7\times3)^2=$

### Step-by-Step Solution

We use the power law for multiplication within parentheses:

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

$(4\cdot7\cdot3)^2=4^2\cdot7^2\cdot3^2$Therefore, the correct answer is option a.

Note:

From the formula of the power property mentioned above, we understand that we can apply it not only to the multiplication of two terms within parentheses, but is also for multiple terms within parentheses.

$4^2\times7^2\times3^2$

### Exercise #10

$(4^2)^3+(g^3)^4=$

### Step-by-Step Solution

We use the formula:

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

$(4^2)^3+(g^3)^4=4^{2\times3}+g^{3\times4}=4^6+g^{12}$

$4^6+g^{12}$

### Exercise #11

$(5\cdot x\cdot3)^3=$

### Step-by-Step Solution

We use the formula:

$(a\times b)^n=a^nb^n$

$(5\times x\times3)^3=(15x)^3$

$(15x)^3=(15\times x)^3$

$15^3x^3$

$15^3\cdot x^3$

### Exercise #12

Solve the exercise:

$(a^5)^7=$

### Step-by-Step Solution

We use the formula:

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

and therefore we obtain:

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

$a^{35}$

### Exercise #13

$(a^4)^6=$

### Step-by-Step Solution

We use the formula

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

Therefore, we obtain:

$a^{4\times6}=a^{24}$

$a^{24}$

### Exercise #14

$(a\times b\times c\times4)^7=$

### Step-by-Step Solution

We use the formula:

$(a\times b)^x=a^xb^x$

Therefore, we obtain:

$a^7b^7c^74^7$

$a^7\times b^7\times c^7\times4^7$

### Exercise #15

$(a\cdot5\cdot6\cdot y)^5=$

### Step-by-Step Solution

We use the formula:

$(a\times b)^x=a^xb^x$

Therefore, we obtain:

$(a\times5\times6\times y)^5=(a\times30\times y)^5$

$a^530^5y^5$

$a^5\cdot30^5\cdot y^5$