# Power Functions and Exponents

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

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### 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

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

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

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.