# Power Functions and Exponents

In this document, we discuss power functions and how exponents affect them.

**Power Functions And Exponents**

Power functions use rational numbers as exponents. There are several properties that define their behavior. We will go over these properties in some example problems.

**Example 1**

Simplify \( 16^{3/4}\)

You can look at this as saying the 4th root of 16 cubed.

\[ \sqrt[4]{16^3}\]

The fourth root of 16 = 2.

\[ (2)^3 = 8 \]

**Example 2**

Simplify: \( \frac{4^{⅓}} {4^{⅚}} \)

Put the numerator and denominator into one term and subtract the exponents.

\[=4^{ {⅓ - ⅚} } \]

Subtract the exponents.

\[ = 4^{ {-½}} \]

A negative exponent is just an inverse root. Let us move it to the denominator.

\[ = \frac{ 1}{\sqrt{4}} \]

Now, just simplify.

\[ = \frac{1}{2} \]

**Example 3**

Rewrite \( \sqrt{x} \) as a positive rational exponent.

The square root of something has an exponent of \( 1/2 \).

This evaluates :

\[ x^{1/2} \]

**Example 4**

Rewrite \( \sqrt[3] x^{2} \) as a positive rational exponent

We have two things to consider with this example.

\( x^2 \) and \( \sqrt{3} \)

\( x^2 \) is in the correct form already. \( \sqrt{3} \) is \( 1/3 \) as an exponent.

So we have \( (x^{2}) ^{1/3} \)

Multiple the exponents together.

We now get : \( x^{2/3} \)

These are what we call power functions. They have a positive exponent. The domain is all real numbers.

**Example 5 **

Here is a slightly more complicated version. It has a rational exponent.

Solve: \( 2x^{5/2} - 7 = 23 \)

We want to solve this equation. This means finding a value for \(x\).

Start by adding \(7\) to both sides. We get: \( 2x^{5/2} = 30 \)

Now we want to get rid of the coefficient in front of the \(x\) variable.

To do that, we need to divide both sides of the equation by \(2\).

This gives us: \( x^{5/2} = 15 \).

Now we can square each side: \( x^5 = 225 \)

The next step is to take the fifth root of each side to give us whole numbers to work with. This gives us: \( x = 225^{1/5} \).

A calculator makes this step easy and quick. You should get a value of: \( 2.95 \)

**Example 6**

Solving an equation with a square root

Solve \( x = \sqrt{15-2x} \)

To get rid of square roots, we can square each side.

\[ x^2 = \sqrt{15-2x}^2 \]

Now we can simplify some

\[ x^2 = 15 - 2x \]

Add \( 2x \) to both sides

\[ x^2 + 2x = 15 \]

Subtract 15 from both sides

\[ x^2 + 2x - 15 = 0 \]

We now have a proper equation. It is time to factor.

\[ (x + 5) (x - 3) = 0 \]

\[ x = -5, x = 3 \]

Next, we substitute the values back into the original equation to see if they work.

Specifically, we want to do this anytime squaring has been used to make sure the values match.

\[ -5 \rightarrow \sqrt{15-2(-5)} = 5 \]

Our input does not match our output so that solution does not work.

\[ 3 \rightarrow \sqrt{15-2(3)} = 3 \]

That input of 3 gives an output of 3 so that is the only correct solution.

**Example 7**

Solving an equation with a cube root

\[ \sqrt[3]{2x + 5} -2 =1 \]

Add 2 to each side

\[ \sqrt[3]{2x + 5} = 3 \]

Cube each side

\[ \sqrt[3]{2x + 5}^3 = 3^3 \]

Now simplify

\[ 2x + 5 = 27 \]

Subtract 5 from each side

\[ 2x = 22 \]

Divide both sides by 2

\[ x = 11 \]

**Conclusion**

In this section, we talked about power functions and how to solve different versions of them.