# 2. Fractional Exponents

**Fractional exponents** can be used instead of using the radical sign (√). We use fractional exponents because often they are more convenient, and it can make algebraic operations easier to follow.

## Fractional Exponent Laws

The ** n-th root** of a number can be written using the power `1/n`, as follows:

`a^(1/n)=root(n)a`

**Meaning:** The *n*-th root of *a* when multiplied by itself *n* times, gives us *a*.

a^{1/n}×a^{1/n}×a^{1/n}× ... ×a^{1/n}=a[Multiply

ntimes]

### Example 1

The cube root of `8` is `2` (since `2^3=8`).

We can write the cube root of 8 as:

`8^(1//3)`

or

`root(3)8`

The following 3 numbers are equivalent:

`8^{1/3}=root(3)8=2`

### Example 2

The square root of a number can be written using the radical sign (√) or with exponent 1/2.

The following are equivalent:

`sqrt(100)=100^(1/2)=10`

### Example 3

The 4-th root of `625` can be written as either:

625

^{1/4}

or equivalently, as

`root(4)625`

Its value is `5`, since `5^4= 625`.

So we could write:

`625^(1/4) = root(4)625 = 5`

**Definitions**

### Radicand

The number under the radical is called the **radicand** (in Example 3, the number `625` is the radicand).

### Order/Index of the radical

The number indicating the root being taken is called the **order** (or *index*) of the radical (in Example 3, the order is `4`).

These definitions are here so you know what your textbook is talking about.

## Raising the *n*-th root to the Power *m*

If we need to raise the *n*-th root of a number to the power *m* (say), we can write this as:

`a^(m/n)=(root(n)a)^m`

This experssion means we need to "take the *n*-th root of the number *a*, then raise the result to the power *m*". With fractional exponents, we would write this as:

`(a^(1//n))^m`

Actually, we get the same final answer if we do it in the other order, "raise *a* to the power *m*, then find the *n*-th root of the result". That is,

`(a^m)^(1//n)`

But the first one is usually easier to do becuase finding the *n*-th root first gives us a smaller number, which is then easy to raise to the power *m*.

### Example 4

Evaluate `8^(2/3)`

Answer

`8^(2/3)=(root(3)8)^2=(2)^2=4`

First, we found the cube root of `8` and the answer was `2`.

Then we raised this result to the power `2`, giving `4`.

Easy to understand math videos:

MathTutorDVD.com

### Example 5

Simplify `(8a^2b^4)^(1/3)`

Answer

`(8a^2b^4)^(1/3)=(8)^(1/3)(a^2)^(1/3)(b^4)^(1/3)`

`=2a^(2/3)b^(4/3)`

In the first line, we used this rule from the last section:

(

a^{m})^{n}=a^{mn}

That is, we took each item inside the brackets and raised them to the power `1/3`. We can do this because each term is multiplied inside the bracket (if they were added or subtracted, we could not do this).

When we expand this out, the only thing we can do is to find the cube root of `8`, which is `2`, and then just write the *a* and *b* parts with fractional powers.

Easy to understand math videos:

MathTutorDVD.com

### Example 6

Simplify `a^(3text(/)4)a^(4text(/)5)`

Answer

`a^(3/4)a^(4/5)=a^(3/4+4/5)=a^(31/20)`

Please support IntMath!

### Example 7

Simplify

`((4^(-3/2)x^(2/3)y^(-7/4))/(2^(3/2)x^(-1/3)y^(3/4)))^(2/3)`

Answer

`((4^(-3/2)x^(2/3)y^(-7/4))/(2^(3/2)x^(-1/3)y^(3/4)))^(2/3)`

Don't panic when you see this one!

In the first step, we move the top expressions with negative exponents to the bottom, and the bottom ones with negative exponents to the top.

`=((x^(2/3)x^(1/3))/(2^(3/2)4^(3/2)y^(3/4)y^(7/4)))^(2/3)`

Then we multiply terms with the same base (the *x* and *y* terms), by adding their indices.
We can collect the `2` and `4` because they both have power `3/2`.

`=((x^(2/3+1/3))/((2xx4)^(3/2)y^(3/4+7/4)))^(2/3)`

Next, we raise everything to the power `2/3`, since that was the power outside the bracket.

`=(x/(8^(3/2)y^(10/4)))^(2/3)`

The final step is to tidy up the expression.

`=x^(2/3)/(8^((3/2xx2/3))y^((10/4xx2/3)))`

`=x^(2/3)/(8y^(5/3))`

Whew!

Easy to understand math videos:

MathTutorDVD.com

### Exercises

**Question 1:** Evaluate `5^(1//2)5^(3//2)`

Answer

Each item in this question has the same base (`5`), so we need to add the indices, as follows:

`5^(1text(/)2)5^(3text(/)2) = 5^(1text(/)2+3text(/)2) =5^(4text(/)2) =5^2 =25 `

Get the Daily Math Tweet!

IntMath on Twitter

**Question 2:** Evaluate `(1000^(1text(/)3))/(400^(-1text(/)2))`

Answer

On the top, the cube root of `1000` is `10`.

On the bottom of the fraction, we have a negative index. We need to recall the following 2 properties of fractions and negative indices:

`1/a=a^(-1)`

If the number on the bottom has a negative index, for example negative `1`, then this is the result:

`1/x^(-1)=1/(1text(/)x)=1/(1text(/)x)xxx/x=x`

So `400^(−1//2` on the bottom becomes `400^(1//2` on the top.

And `400^(1//2) =sqrt(400)= 20`.

Here's the complete answer:

`1000^(1text(/)3)/(400^(-1text(/)2)) =10 xx 400^(1 text(/)2)` `=10 xx 20` `=200`

### Search IntMath, blog and Forum

### Online Algebra Solver

This algebra solver can solve a wide range of math problems.

Go to: Online algebra solver

### Algebra Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand algebra lessons on DVD. See samples before you commit.

More info: Algebra videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!