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2. 2 Rational Exponents
In section 1.2, we explored the rules of exponents when those exponents
were integers. The same rules will apply, however, if we have powers
which are rational numbers (fractions with integer numerator and
denominator). This section will review those rules, but also will
introduce the meaning of a rational exponent.
The key to understanding rational exponents is simply a definition:
๐
๐
= ( โ
๐
๐
๐
๐
Notice that the denominator becomes the index of the radical and the
numerator becomes the power.
Either form will give the same answer when you simplify, but I believe
the first form is easier most of the time (since it is easier to take a root of
a smaller number), so you will see this one used more often in this text.
Sometimes it is more convenient to bring the power inside of the root,
but only when there are leftovers inside of the radical.
Examples
Simplify each of the following rational exponent expressions.
2
3
= ( โ
3
2
2
So 8 to the
2
3
power is just 4. Notice that we first took the cube root
of 8 and then we squared our result. Working from the inside out
makes it easy because we can focus on one thing at a time. We
could have also used the second form and brought the exponent
inside before taking the root: โ
2
3
3
= 4. Either way, we
will get 4.
1
2
= ( โ 16
2
1
You can see here that the power of
1
2
just means square root.
5
2
= ( โ
5
5
Here we take the square root of 4 first and then put it to the fifth
power. Donโt forget that 2
5
= 2 ยท 2 ยท 2 ยท 2 ยท 2 = 8 ยท 4 = 32. You
can group the 2โs together in whichever way you like since the 5
together rather than to multiply five 2โs together, so be careful to
pay attention to what you are doing. The more practice you get, the
less likely you are to make such mistakes.
2
3
= ( โ 125
3
2
2
If you did not know that โ 125
3
= 5 , then you could simply factor
125 using a factor tree. You will recognize quickly that 5 divides
125 (the divisibility rule for 5 says that if a number ends in a 0 or a
5 that it is divisible by 5). Then 125 = 5 ยท 25 and the rest is
easyโฆ.
4
3
= ( โ
3
4
3
4
4
We can pull the negative out of the cube root, but we still have to
apply the fourth power to it, which gives us positive 81.
3
4 = ( โ
4
3
Not a real number
This is not a real number because we cannot take an even root of a
negative (there is no real number that when put to the fourth power
will yield a negative number).
The exponent rules still apply when the exponents are rational
numbers. We will now go through each of these rules and provide an
example.
We will assume that all variables represent positive real numbers for
the examples in this section.
Rule Example
๐
๐
๐+๐
2
3 ยท ๐ฅ
4
3 = ๐ฅ
2
3
4
3 = ๐ฅ
6
3 = ๐ฅ
2
๐
๐
๐
๐
๐โ๐
๐ฆ
5
9
๐ฆ
1
4
5
9
โ
1
4
= ๐ฆ
20
36
โ
9
36
= ๐ฆ
11
36
๐
๐
๐๐
4
3
)
2
3
4
3
ยท
2
3
= ๐ฅ
8
9
Donโt forget that Rule 1 and Rule 3 often get confused, so make sure to
note whether you will be adding your powers or multiplying your
powers. In rule 3, we have a single base while in Rule 1, we have two of
the same base. This might help you to remember.
๐
๐
๐
9
1
5 )
2
3
2
3
(๐ฅ
9
2
3 ( ๐ฆ
1
5 )
2
3
3
2
9
1
ยท
2
3
๐ฆ
1
5
ยท
2
3
= 9 ๐ฅ
6
2
15
๐
๐
๐
๐
๐
๐
๐
๐ฅ
36 ๐ฆ
8
1
2
๐ฅ
1
2
( 36 ๐ฆ
8
)
1
2
๐ฅ
1
2
36
1
2 ( ๐ฆ
8
)
1
2
๐ฅ
1
2
6 ๐ฆ
4
โ๐
1
๐
๐
โ
2
3
=
1
8
2
3
1
( โ
8
3
)
2
1
4
0
4
7
๐ฆ
3
5
๐ง)
0
The next set of examples will integrate these rules. A good rule of thumb
is to work from the outside in when you have parentheses involved.
Look at the outermost exponent first and if it is negative, move that
piece across the fraction bar using rule 6. When all of the outermost
exponents are positive, then distribute exponents over multiplication
using rules 3 and 4. Then combine exponents using rules 1 and 2.
Examples
Simplify each of the following rational exponent expressions.
Assume all variables represent non-negative real numbers.
4
1
2
= 25
1
2 ( ๐
4
1
2
= โ 25 ๐
2
2
For this problem, the outermost exponent is positive, so we can
begin by distributing the exponent.
6
โ 2
1
3
= โโ 27
3
6
1
3 ( ๐ฆ
โ 2
1
3
= โ 3 ๐ฅ
2
โ
2
3
= โ
3 ๐ฅ
2
๐ฆ
2
3
Note that you could have approached this problem differently by
working on the inside first (moving ๐ฆ
โ 2
down) and then applying
the outer exponent. You will get the same answer. We will always
5. [(๐ฅ + 2 )
1
8
]
8
This problem reminds us that radicals and powers โundoโ each
other since inside we have the 8th root and outside we have the
power 8. It is easier to just use rule 3 and multiply the powers
together to get a power of 1.
2
3
(๐ฆ
โ
2
3
1
2
) = ๐ฆ
2
3
ยท ๐ฆ
โ
2
3
2
3
ยท ๐ฆ
1
2
= ๐ฆ
0
7
6
= 1 + ๐ฆ
7
6
The answer to this one could also be written with radicals as
7
6
6
