# Fundamental Rules of Exponents

By Kathleen Knowles, 03 Apr 2021

Knowing the fundamental rules of exponents will help you simplify mathematical equations and statements so you can arrive at a solution with relative ease. As with everything, it may take some practice to get the hang of working with exponents. But if you take your time, don't allow yourself to be overwhelmed by variables, and memorize each rule, you'll be a master in no time.

## Know Your Exponents

Before getting into the theory of exponents, let's first look at a few notations and related terminology.

Let's take the term `x^n`

.

We read this as "*x* raised to the power of *n*," which means we multiply *x* n number of times.

* x^n=x⋅x⋅x*...n times.

Here, *x* is any real number and *n* is a positive integer. We refer to *x* as "base," and *n* as "power" or "exponent."

Another example, we read `9^5`

as "9 raised to the power of 5" and it is equal to 9⋅9⋅9⋅9⋅9.

`9^5`

= 9⋅9⋅9⋅9⋅9 = 59,049

## Rules of Exponents

There are eight fundamental rules of exponents.

### 1. The One Rule

A number to the 1 power is equal to the number itself.

* x^n = n* where

*n = 1*

We multiply *x* one time, which is equal to the number itself.

`x^1`

= x

**Examples:**

`9^1`

= 9

`01 = 0`

`-1^1 = -1`

### 2. The Zero Rule

Any number raised to the power of 0 (except zero) is equal to 1.

`x^0 = 1`

**Examples:**

`9^0 = 1`

`1^0 = 1`

Note: `0^0`

is undefined.

### 3. The Product Rule

When multiplying, you can add two exponents with the same base.

`x^a⋅x^b = x^a+b`

As the bases of both the terms are the same, we can add their powers.

**Examples:**

`9^5⋅9^7 = 9^(5+7) = 9^12`

`8^-2⋅8^10 = 8^(-2+10) = 8^8`

In special cases, we can apply this rule even when there are the same bases but with different signs (positive/negative). We just need to rewrite the terms in accordance with our rule.

**For example:**

Simplify: `(-9)^5⋅(9)^7`

We can write `(-9)^5`

as `(-1⋅9)^5`

which is equal to `(-1)^5⋅(9)^5`

*.*

This implies, `(-9)^5⋅(9)^7 = (-1)^5⋅(9)^5⋅(9)^7 = (-1)^5⋅(9)^5+7 = (-1)^5⋅(9)^12 = -9^12`

*.*

### 4. The Quotient Rule

By subtracting the exponents, we can divide two powers with the same base.

`x^a/x^b = x^(a-b)`

Here the bases of both the terms under division are the same, so we subtract the exponents.

**Examples:**

`9^5/9^7 = 9^(5-7) = 9^-2`

`2^2/2^2 = 2^(2-2) = 2^0 = 1`

### 5. The Power Rule

To raise a power to another power, we multiply the exponents.

`(x^m)^n = x^(m⋅n)`

We raise the inner term to an exponent and so we multiply both the exponents.

**Examples:**

`(9^5)^7 = 9^(5⋅7) = 9^35`

`(8^2)^2 = 8^(2⋅2) = 8^4`

### 6. The Negative Exponent

Any non-zero number raised to a negative power equals its reciprocal raised to the opposite positive power.

`x^-n = 1/x^n`

*x* is raised to a negative exponent and we can write it as *x*'s reciprocal with the opposite sign of *n*.

**Examples:**

`9^-5 = 1/9^5`

`1^-1 = 1/1^1 = 1`

### 7. The Power of a Fraction Rule

When there is a power to a fraction, we can rewrite the term as both the numerator and denominator raised to the power of the exponent.

`(x/y)^n = x^n/x^y`

**Examples:**

`(2/5)^5 = 2^5/5^5`

`(1/2)^2 = 1^2/2^2 = 1/2^2`

### 8. A Fractional Exponent

When you have a fractional exponent, the numerator is the power and the denominator is the root (square root for denominator = 2, cube root for denominator = 3, and so on).

`b^n/m = (^m √b)^n`

## Examples

As we studied the fundamental laws of exponents, we shall now implement these rules in the following examples. Observe the statement or equation carefully before rushing in to apply the rules. You may be able to solve what seems like a five-step problem in just two steps. This is especially true when dealing with exponents.

**Problem 1**

Simplify: `((x/y)^n)^(1/n)`

**Solution **

From the power rule, we get `(x/y)^(n⋅(1/n))`

which is equal to `(x/y)^1 = x/y`

.

### Problem 2

Simplify: `((x^a/x^b)x^c)^-d`

**Solution**

From the negative exponent rule, we get `1/((x^a/x^b)x^c)^d`

.

From the quotient rule, we get `1/((x^a-b)x^c)^d`

.

From the product rule, we get `1/(x^a-b+c)^d`

.

From the power rule, we get `1/x^((a-b+c)^d)`

which equals to `1/x^(ad-bd+cd)`

.

Hence, `1/((x^a/x^b)x^c)^-d = 1/(x^(ad-bd+cd))`

### Problem 3

Simplify: `(((x⋅x⋅x⋅x⋅x)^y)^-1)⋅(((x⋅x⋅x⋅x⋅x)^y)^1)`

**Solution**

Look carefully. We can simplify this in two steps without even simplifying the individual terms under multiplication. Upon observation, we find that the bases of the powers are equal and we can apply the rule of product.

`(((x⋅x⋅x⋅x⋅x)^y)^-1)⋅(((x⋅x⋅x⋅x⋅x)^y)^1) = ((x⋅x⋅x⋅x⋅x)^y)^-1+1 = ((x⋅x⋅x⋅x⋅x)^y)^0`

Applying the zero rule we get, `((x⋅x⋅x⋅x⋅x)^y)^0 = 1`

## Conclusion

Our aim is not just to solve problems or equations. Our aim is to solve them efficiently and effectively. It is true that we can solve or handle the statements without using these rules but that process is highly ineffective, complex, and time-consuming. So, we use these rules of exponents to solve and simplify equations and statements efficiently.

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