Fractional Exponents

Fractional exponents are related to roots or radicals.

If n is a positive integer, then

$$a^{1/n} \quad\hbox{is the $n^{\rm th}$ root of a}.$$

If a is positive, it is the positive number b such that

$$b^n = a.$$

If a is negative, then:

1. If n is odd, $a^{1/n}$ is the negative number b such that

$$b^n = a.$$

2. If n is even, $a^{1/n}$ is undefined.

$a^{1/n}$ is also written $\root n \of a$ .


Example. Compute the exact values of:

(a) $9^{1/2}$ .

(b) $125^{1/3}$ .

(c) $-64^{1/3}$ .

(d) $(-16)^{1/4}$ .

(e) $-16^{1/4}$ .

$$9^{1/2} = 3, \quad\hbox{since}\quad 3^2 = 9.$$

$9^{1/2}$ is the same as $\sqrt{9}$ . Note that $\sqrt{9}$ is not "$\pm 3$ ".

$$125^{1/3} = 5, \quad\hbox{since}\quad 5^3 = 125.$$

$$-64^{1/3} = -4, \quad\hbox{since}\quad (-4)^3 = -64.$$

$$(-16)^{1/4} \quad\hbox{is undefined}.$$

$$-16^{1/4} = -(16^{1/4}) = -2.$$

In the last example, exponentiation takes precedence over negation.


If m and n are positive integers,

$$a^{m/n} \quad\hbox{means}\quad {\root n \of {a^m}}.$$

This makes sense, since

$${\root n \of {a^m}} = (a^m)^{1/n},$$

and this should equal $a^{m/n}$ if the rule for multiplying exponents is to hold in this case.

Equivalently,

$$a^{m/n} = ({\root n \of a})^m.$$

In other words, you can do the root and the power in either order.

${\root n \of {a^m}}$ involves an $n^{\rm th}$ root, so it may be positive, negative, or undefined.


Example. Compute the exact values of:

(a) $8^{5/3}$ .

(b) $64^{5/6}$ .

(c) $(-27)^{2/3}$ .

(d) $36^{-3/2}$ .

(e) $(-125)^{-4/3}$ .

$$8^{5/3} = ({\root 3 \of 8})^5 = 2^5 = 32.$$

$$64^{5/6} = (\root 6 \of {64})^5 = 2^5 = 32.$$

$$(-27)^{2/3} = (\root 3 \of {-27})^2 = (-3)^2 = 9.$$

$$36^{-3/2} = \dfrac{1}{36^{3/2}} = \dfrac{1}{(\sqrt{36})^3} = \dfrac{1}{6^3} = \dfrac{1}{216}.$$

$$(-125)^{-4/3} = \dfrac{1}{(-125)^{4/3}} = \dfrac{1}{(\root 3 \of {-125})^4} = \dfrac{1}{(-5)^4} = \dfrac{1}{625}.\quad\halmos$$


Example. Use a calculator to approximate $5^{1/3}$ and $(-10)^{1/7}$ .

$$5^{1/3} \approx 1.70998.$$

Roots of negative numbers can present a problem; some calculators will return a complex number, or give an error message. You can fix things by figuring out the sign of the result beforehand. Then make the base positive for your calculator and fix the sign at the end.

For example, $(-10)^{1/7}$ is an odd root of a negative number, so it's negative. Knowing this, I use the calculator to compute $10^{1/7}$ :

$$10^{1/7} \approx 1.38950.$$

Therefore,

$$(-10)^{1/7} \approx -1.38950.\quad\halmos$$


Example. Is "$(-1)^{2/6}$ " undefined, since -1 is negative and the $6^{\rm th}$ root is an even root?

Before considering a fractional exponent, the fraction should be reduced to lowest terms: $\dfrac{2}{6} = \dfrac{1}{3}$ .

$$(-1)^{1/3} = -1, \quad\hbox{since}\quad (-1)^3 = -1.\quad\halmos$$


The rules I gave earlier for working with integer exponents work with fractional exponents --- with certain exceptions for even roots.

$$\eqalign{ x^0 & = 1 \cr x^a x^b & = x^{a + b} \cr \dfrac{x^a}{x^b} & = x^{a - b} \cr (x^a)^b & = x^{a b} \cr (x y)^a & = x^a y^a \cr \left(\dfrac{x}{y}\right)^a & = \dfrac{x^a}{y^a} \cr}$$


Example. Simplify $(x^3 y^{15})^{1/3}$ .

$$(x^3 y^{15})^{1/3} = (x^3)^{1/3}(y^{15})^{1/3} = x^1 y^5 = x y^5.\quad\halmos$$


Example. Simplify $(x^{3/7} y^{9/11})^{1/3}$ .

$$(x^{3/7} y^{9/11})^{1/3} = (x^{3/7})^{1/3}(y^{9/11})^{1/3} = x^{1/7} y^{3/11}.\quad\halmos$$


However,

$$(x^2)^{1/2} \quad\hbox{is not the same as}\quad x.$$

Why? $(x^2)^{1/2} =
   \sqrt{x^2}$ , and $\sqrt{\hbox{junk}}$ is always nonnegative. But x could be negative: For example,

$$\sqrt{(-3)^2} = \sqrt{9} = 2, \quad\hbox{but}\quad x = -3.$$

In this case, $\sqrt{x^2}
   \ne x$ .

In fact, if n is an even integer,

$$(x^n)^{1/n} = {\root n \of {x^n}} = |x|.$$

(Of course, I can drop the absolute values if I know x is nonnegative.)


Example. Simplify $(x^4)^{1/4}$ .

$$(x^4)^{1/4} = {\root 4 \of {x^4}} = |x|.$$

Note that since I didn't assume x was nonnegative, the answer is not "x".


Example. Assuming that x and y are nonnegative, simplify $\sqrt{16 x^4 y^6}$ .

$$\sqrt{16 x^4 y^6} = \sqrt{16} \sqrt{x^4} \sqrt{y^6} = 4 x^2 y^3.\quad\halmos$$


Example. Simplify ${\root 3 \of {8 x^6 y^3}}$ .

$${\root 3 \of {8 x^6 y^3}} = (8 x^6 y^3)^{1/3} = 8^{1/3} \cdot (x^6)^{1/3} \cdot (y^3)^{1/3} = 2 x^2 y.\quad\halmos$$


Example. Assuming that x and y are nonnegative, simplify $(-8
   x^6 y^{3/5})^{1/3}(x^{-5} y^{5/3})^{1/5}$ . Write your answer using positive powers.

$$(-8 x^6 y^{3/5})^{1/3}(x^{-5} y^{5/3})^{1/5} = (-8)^{1/3}(x^6)^{1/3}(y^{3/5})^{1/3}(x^{-5})^{1/5}(y^{5/3})^{1/5} =$$

$$(-2) x^2 y^{1/5} x^{-1} y^{1/3} = -2 x y^{8/15}.\quad\halmos$$


Example. Assuming that x and y are nonnegative, simplify $(4
   x^2 y^{-3})^{3/2} \cdot 3 x^{5/2} y^{1/3}$ . Write your answer using positive powers.

$$(4 x^2 y^{-3})^{3/2} \cdot 3 x^{5/2} y^{1/3} = 4^{3/2} (x^2)^{3/2} (y^{-3})^{3/2} \cdot 3 x^{5/2} y^{1/3} = 8 x^3 y^{-9/2} \cdot 3 x^{5/2} y^{1/3} =$$

$$24 x^{11/2} y^{-25/6} = \dfrac{24 x^{11/2}}{y^{25/6}}.\quad\halmos$$


Example. Assuming that x and y are nonnegative, simplify $\dfrac{(8 x^6 y^{3/2})^{1/3}}{6 (x^3 y^{21/2})^{1/6}}$ . Write your answer using positive powers.

$$\dfrac{(8 x^6 y^{3/2})^{1/3}}{6 (x^3 y^{21/2})^{1/6}} = \dfrac{8^{1/3} (x^6)^{1/3} (y^{3/2})^{1/3}}{6 (x^3)^{1/6} (y^{21/2})^{1/6}} = \dfrac{2 x^2 y^{1/2}}{6 x^{1/2} y^{7/4}} =$$

$$\dfrac{x^{3/2} y^{-5/4}}{3} = \dfrac{x^{3/2}}{3 y^{5/4}}.\quad\halmos$$


Example. Assuming that x and y are nonnegative, simplify $\dfrac{\sqrt{27 x^{14} y^{-4}}}{(x^{-6} y^3)^{1/3}}$ . Write your answer using positive powers.

$$\dfrac{\sqrt{27 x^{14} y^{-4}}}{(x^{-6} y^3)^{1/3}} = \dfrac{\sqrt{27} \cdot (x^{14})^{1/2} \cdot (y^{-4})^{1/2}}{(x^{-6})^{1/3} \cdot (y^3)^{1/3}} = \dfrac{3 \sqrt{3} \cdot x^7 \cdot y^{-2}}{x^{-2} \cdot y} = 3 \sqrt{3} x^9 y^{-3} = \dfrac{3 \sqrt{3} x^9}{y^3}.\quad\halmos$$


Example. Assuming that x, y, and z are nonnegative, simplify $\left(\dfrac{-32 x^9 y^{-3}}{z^{15}}\right)^{1/3}$ . Write your answer using positive powers.

$$\left(\dfrac{-32 x^9 y^{-3}}{z^{15}}\right)^{1/3} = \dfrac{(-32)^{1/3} (x^9)^{1/3} (y^{-3})^{1/3}}{(z^{15})^{1/3}} = \dfrac{-2 x^3 y^{-1}}{z^5} = \dfrac{-2 x^3}{y z^5}.\quad\halmos$$



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