5.4 Fractional exponents

On the homework, we demonstrated the rule of negative exponents by building a table. Now, we’re going to demonstrate it another way—by using the rules of exponents.

• A

  According to the rules of exponents , $\frac{7^3}{7^5}$ $7^{\mathrm{[\; ]}}$ .

• B

  But if you write it out and cancel the excess 7s, then $\frac{7^3}{7^5}$ = ——.

• C

  Therefore, since $\frac{7^3}{7^5}$ can only be one thing, we conclude that these two things must be equal: write that equation!

Now, we’re going to approach fractional exponents the same way. Based on our rules of exponents , $9^{\left(\frac{1}{2}\right)}^2$ =

So, what does that tell us about $9^{\left(\frac{1}{2}\right)}$ ? Well, it is some number that when you square it, you get _______ (*same answer you gave for number 2). So therefore, $9^{\left(\frac{1}{2}\right)}$ itself must be:

Using the same logic, what is $16^{\left(\frac{1}{2}\right)}$ ?

What is $25^{\left(\frac{1}{2}\right)}$ ?

What is $x^{\left(\frac{1}{2}\right)}$ ?

Construct a similar argument to show that $8^{\left(\frac{1}{2}\right)}=2$ .

What is $27^{\left(\frac{1}{3}\right)}$ ?

What is $-1^{\left(\frac{1}{3}\right)}$ ?

What is $x^{\left(\frac{1}{3}\right)}$ ?

What would you expect $x^{\left(\frac{1}{5}\right)}$ to be?

What is $25^{\left(\frac{-1}{2}\right)}$ ? (You have to combine the rules for negative and fractional exponents here!)

Check your answer to #12 on your calculator. Did it come out the way you expected?

Using the rules of exponents, $8^{\left(\frac{1}{3}\right)}^2=8^{\mathrm{[\; ]}}$ .

$8^{\left(\frac{2}{3}\right)}=$

Construct a similar argument to show what $16^{\left(\frac{3}{4}\right)}$ should be.

Check $16^{\left(\frac{3}{4}\right)}$ on your calculator. Did it come out the way you predicted?

$8^{\left(\frac{-1}{2}\right)}=$

$8^{\left(\frac{-2}{3}\right)}=$

$10^{-4}$

$2^{\left(\frac{-3}{4}\right)}$

$x^{\left(\frac{a}{b}\right)}$

$x^{\left(\frac{\mathrm{-a}}{b}\right)}$