By the end of this section, you should be able to:

1. Solve problems involving rational exponents

Wow... rational exponents... that sounds hard, right? So far, we've only dealt with **integral exponents**, that is, positive and negative whole-number exponents.

Well, if you look back to Lesson #4: Types of Numbers, we actually define the rational number. Do you remember what a rational number is?

A rational number is a number that can be expressed as a fraction of two integers!

For example, \(\frac{1}{2}\)is a rational number. \(\pi\), which cannot be expressed as a fraction (it's the ratio of a circle's cirumference to its diameter!), is an **irrational number**.

Returning from that tangent, a rational exponent, therefore, is when we have a fraction as an exponent! Spooky, right? Not really, in fact, you've met rational exponents before!

For instance, if I asked you the square root of 4, you might instantly recognize that it's 2; \(\sqrt{4}=2\). What you may not know, however, is that we can rewrite the square root as \(4^{\frac{1}{2}} = 2\). Four, raised to the power of \(\frac{1}{2}\), is 2! What about a cube root? \(\sqrt[3]{8}\), for example, denotes the cube root of 8. You might know that the cube root of eight is two (\(\sqrt[3]{8} = 2\)), but did you know that we can rewrite this relationship as \(8^{1/3} = 2\)? Again, eight, raised to the power of \(\frac{1}{3}\), is 2!

Are you spotting a pattern? Let me show you:

\(a^{1/n} = \sqrt[n]{a}\). The form to the right of the equals sign is called the **radical form**, while the form to the left is the exponent form.

Let's dive into some examples, shall we?

Provide a real number solution for \(\sqrt{-256}\)

If you've been following along, in Lesson 16: Complex Numbers, we discussed complex numbers, and how \(i = \sqrt{-1}\). Right now, you might be tempted to factor out the \(\sqrt{-1}\)... STOP! The question is only asking for *real number* solutions, and that's a complex number solution! How about real number solutions? Are there any? No? Well then say that!

There are no real number solutions!

Alright, that one was silly. How about another?

Provide a real number solution for \((-1024)^{\frac{1}{5}}\)

Remember that we're using an alternative form here, but it's really the same as the radical.

Well, as you might find using a calculator, \(4^5 = 1024\), so \((-1024)^{\frac{1}{5}} = -4\).

You might be wondering, why don't we have to factor out the \(i = \sqrt{-1}\). Well, that's because the \(n\) value of our power is an odd number. As you might know, the product of an even number of negative numbers is always even, so there's no way that the even root of a negative number could ever be positive (e.g., \(-2 \times -2 = +4\)).

In this case, however, the product of an odd number of negative numbers is always negative (e.g., \(-2 \times -2 \times -2 = -8\)). Therefore, the odd root of a negative number will be a negative number!

Evaluate \((2^3)^{\frac{1}{3}}\)

Well, as you might remember from Lesson #17: Laws of Exponents, the exponent of an exponent is the product of the exponents. Confusing, right? Well, it's just saying that we need to multiply the powers. In this case, \(3 \times \frac{1}{3} = 1\), so our result is \(2^1 = 2\)

Evaluate \((25x^4)^{\frac{-1}{2}}\)

Whoa, a negative fraction! Remember from Lesson 17 that, to get rid of the negative exponent, we need to rewrite it as an inverse;

\(=(\frac{1}{25x^4})^{\frac{1}{2}}\)

Now, let's apply that \(\frac{1}{2}\) exponent to the numerator and denominator:

\(=(\frac{1^{\frac{1}{2}}}{(25x^4)^\frac{1}{2}})\)

Now, let's simplify:

\(=\frac{1}{5x^2}\)

Dun dun duuun! We got it! One more:

Evaluate \(2^\frac{1}{2} \times 2^\frac{3}{4}\)

Remember from Lesson 17 that when two exponents have the same base and are multiplied, the exponent of the product is the sum of the exponents. In other words, we need to add the powers:

\(=\frac{1}{2}+\frac{3}{4}\)

\(=\frac{2}{4}+\frac{3}{4}\)

\(=\frac{5}{4}\)

Therefore, our answer is \(2^\frac{5}{4}\)!

With that, the lesson's done. Time for some practice problems!

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