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# 8.3: Zero, Negative, and Fractional Exponents

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## Learning Objectives

• Simplify expressions with negative exponents.
• Simplify expressions with zero exponents.
• Simplify expression with fractional exponents.
• Evaluate exponential expressions.

## Introduction

The product and quotient rules for exponents lead to many interesting concepts. For example, so far we’ve mostly just considered positive, whole numbers as exponents, but you might be wondering what happens when the exponent isn’t a positive whole number. What does it mean to raise something to the power of zero, or -1, or $\frac{1}{2}$? In this lesson, we’ll find out.

## Simplify Expressions With Negative Exponents

When we learned the quotient rule for exponents $\left ( \frac{x^n}{x^m}=x^{(n-m)} \right )$, we saw that it applies even when the exponent in the denominator is bigger than the one in the numerator. Canceling out the factors in the numerator and denominator leaves the leftover factors in the denominator, and subtracting the exponents leaves a negative number. So negative exponents simply represent fractions with exponents in the denominator. This can be summarized in a rule:

Negative Power Rule for Exponents: $x^{-n} = \frac{1}{x^n}$, where $x \neq 0$

Negative exponents can be applied to products and quotients also. Here’s an example of a negative exponent being applied to a product:

$& (x^3y)^{-2} = x^{-6}y^{-2} && \text{using the power rule}\\& x^{-6}y^{-2} = \frac{1}{x^6} \cdot \frac{1}{y^2} = \frac{1}{x^6y^2} && \text{using the negative power rule separately on each variable}$

And here’s one applied to a quotient:

$& \left ( \frac{a}{b} \right )^{-3} = \frac{a^{-3}}{b^{-3}} && \text{using the power rule for quotients}\\& \frac{a^{-3}}{b^{-3}} = \frac{a^{-3}}{1} \cdot \frac{1}{b^{-3}} = \frac{1}{a^3} \cdot \frac{b^3}{1} && \text{using the negative power rule on each variable separately}\\& \frac{1}{a^3} \cdot \frac{b^3}{1} = \frac{b^3}{a^3} && \text{simplifying the division of fractions}\\& \frac{b^3}{a^3} = \left ( \frac{b}{a} \right )^3 && \text{using the power rule for quotients in reverse.}$

That last step wasn’t really necessary, but putting the answer in that form shows us something useful: $\left ( \frac{a}{b} \right )^{-3}$ is equal to $\left ( \frac{b}{a} \right )^3$. This is an example of a rule we can apply more generally:

Negative Power Rule for Fractions: $\left ( \frac{x}{y} \right )^{-n} = \left ( \frac{y}{x} \right )^n$, where $x \neq 0, y \neq 0$

This rule can be useful when you want to write out an expression without using fractions.

Example 1

Write the following expressions without fractions.

a) $\frac{1}{x}$

b) $\frac{2}{x^2}$

c) $\frac{x^2}{y^3}$

d) $\frac{3}{xy}$

Solution

a) $\frac{1}{x} = x^{-1}$

b) $\frac{2}{x^2} = 2x^{-2}$

c) $\frac{x^2}{y^3} = x^2y^{-3}$

d) $\frac{3}{xy} = 3x^{-1}y^{-1}$

Example 2

Simplify the following expressions and write them without fractions.

a) $\frac{4a^2b^3}{2a^5b}$

b) $\left ( \frac{x}{3y^2} \right )^3 \cdot \frac{x^2y}{4}$

Solution

a) Reduce the numbers and apply the quotient rule to each variable separately:

$\frac{4a^2b^3}{2a^5b} = 2 \cdot a^{2-5} \cdot b^{3-1} = 2a^{-3}b^2$

b) Apply the power rule for quotients first:

$\left ( \frac{2x}{y^2} \right )^3 \cdot \frac{x^2y}{4} = \frac{8x^3}{y^6} \cdot \frac{x^2y}{4}$

Then simplify the numbers, and use the product rule on the $x$’s and the quotient rule on the $y$’s:

$\frac{8x^3}{y^6} \cdot \frac{x^2y}{4} = 2 \cdot x^{3+2} \cdot y^{1-6} = 2x^5y^{-5}$

You can also use the negative power rule the other way around if you want to write an expression without negative exponents.

Example 3

Write the following expressions without negative exponents.

a) $3x^{-3}$

b) $a^2b^{-3}c^{-1}$

c) $4x^{-1}y^3$

d) $\frac{2x^{-2}}{y^{-3}}$

Solution

a) $3x^{-3} = \frac{3}{x^3}$

b) $a^2b^{-3}c^{-1} = \frac{a^2}{b^3c}$

c) $4x^{-1}y^3 = \frac{4y^3}{x}$

d) $\frac{2x^{-2}}{y^{-3}} = \frac{2y^3}{x^2}$

Example 4

Simplify the following expressions and write the answers without negative powers.

a) $\left ( \frac{ab^{-2}}{b^3} \right )^2$

b) $\frac{x^{-3}y^2}{x^2y^{-2}}$

Solution

a) Apply the quotient rule inside the parentheses: $\left ( \frac{ab^{-2}}{b^3} \right )^2 = (ab^{-5})^2$

Then apply the power rule: $(ab^{-5})^2 = a^2b^{-10} = \frac{a^2}{b^{10}}$

b) Apply the quotient rule to each variable separately: $\frac{x^{-3}y^2}{x^2y^{-2}} = x^{-3-2}y^{2-(-2)} = x^{-5}y^4 = \frac{y^4}{x^5}$

## Simplify Expressions with Exponents of Zero

Let’s look again at the quotient rule for exponents $\left ( \frac{x^n}{x^m} = x^{(n-m)}\right )$ and consider what happens when $n = m$. For example, what happens when we divide $x^4$ by $x^4$? Applying the quotient rule tells us that $\frac{x^4}{x^4} = x^{(4-4)}=x^0$—so what does that zero mean?

Well, we first discovered the quotient rule by considering how the factors of $x$ cancel in such a fraction. Let’s do that again with our example of $x^4$ divided by $x^4$:

$\frac{x^4}{x^4} = \frac{\cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot \cancel{x}}{\cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot \cancel{x}} = 1$

So $x^0=1 !$ You can see that this works for any value of the exponent, not just 4:

$\frac{x^n}{x^n} = x^{(n-n)} = x^0$

Since there is the same number of $x$’s in the numerator as in the denominator, they cancel each other out and we get $x^0 = 1$. This rule applies for all expressions:

Zero Rule for Exponents: $x^0=1$, where $x \neq 0$

## Simplify Expressions With Fractional Exponents

So far we’ve only looked at expressions where the exponents are positive and negative integers. The rules we’ve learned work exactly the same if the powers are fractions or irrational numbers—but what does a fractional exponent even mean? Let’s see if we can figure that out by using the rules we already know.

Suppose we have an expression like $9^{\frac{1}{2}}$—how can we relate this expression to one that we already know how to work with? For example, how could we turn it into an expression that doesn’t have any fractional exponents?

Well, the power rule tells us that if we raise an exponential expression to a power, we can multiply the exponents. For example, if we raise $9^ \frac{1}{2}$ to the power of 2, we get $\left(9^\frac{1}{2}\right)^2=9^{2 \cdot \frac{1}{2}}=9^1=9$.

So if $9^{\frac{1}{2}}$ squared equals 9, what does $9^{\frac{1}{2}}$ itself equal? Well, 3 is the number whose square is 9 (that is, it’s the square root of 9), so $9^{\frac{1}{2}}$ must equal 3. And that’s true for all numbers and variables: a number raised to the power of $\frac{1}{2}$ is just the square root of the number. We can write that as $\sqrt{x}=x^{\frac{1}{2}}$, and then we can see that’s true because $\left( \sqrt{x} \right)^2=x$ just as $\left( x^{\frac{1}{2}} \right)^2=x$.

Similarly, a number to the power of $\frac{1}{3}$ is just the cube root of the number, and so on. In general, $x^{\frac{1}{n}}=\sqrt[n]{x}$. And when we raise a number to a power and then take the root of it, we still get a fractional exponent; for example, $\sqrt[3]{x^4}=\left(x^4\right)^{\frac{1}{3}}=x^{\frac{4}{3}}$. In general, the rule is as follows:

Rule for Fractional Exponents: $\sqrt[m]{a^n}=a^{\frac{n}{m}}$ and $\left( \sqrt[m]{a}\right)^n=a^{\frac{n}{m}}$

We’ll examine roots and radicals in detail in a later chapter. In this section, we’ll focus on how exponent rules apply to fractional exponents.

Example 5

Simplify the following expressions.

a) $a^{\frac{1}{2}} \cdot a^{\frac{1}{3}}$

b) $\left(a^{\frac{1}{3}}\right)^2$

c) $\frac{a^{\frac{5}{2}}}{a^{\frac{1}{2}}}$

d) $\left(\frac{x^2}{y^3}\right)^{\frac{1}{3}}$

Solution

a) Apply the product rule: $a^\frac{1}{2} \cdot a^\frac{1}{3}=a^{\frac{1}{2}+\frac{1}{3}}=a^{\frac{5}{6}}$

b) Apply the power rule: $\left(a^{\frac{1}{3}}\right)^2=a^{\frac{2}{3}}$

c) Apply the quotient rule: $\frac{a^{\frac{5}{2}}}{a^{\frac{1}{2}}}=a^{\frac{5}{2}-\frac{1}{2}}=a^{\frac{4}{2}}=a^2$

d) Apply the power rule for quotients: $\left(\frac{x^2}{y^3}\right)^{\frac{1}{3}}=\frac{x^{\frac{2}{3}}}{y}$

## Evaluate Exponential Expressions

When evaluating expressions we must keep in mind the order of operations. You must remember PEMDAS:

1. Evaluate inside the Parentheses.
2. Evaluate Exponents.
3. Perform Multiplication and Division operations from left to right.
4. Perform Addition and Subtraction operations from left to right.

Example 6

Evaluate the following expressions.

a) $5^0$

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

c) $16^{\frac{1}{2}}$

d) $8^{-\frac{1}{3}}$

Solution

a) $5^0=1$ A number raised to the power 0 is always 1.

b) $\left(\frac{2}{3}\right)^3=\frac{2^3}{3^3}=\frac{8}{27}$

c) $16^{\frac{1}{2}}=\sqrt{16}=4$ Remember that an exponent of $\frac{1}{2}$ means taking the square root.

d) $8^{-\frac{1}{3}}=\frac{1}{8^{\frac{1}{3}}}=\frac{1}{\sqrt[3]{8}}=\frac{1}{2}$ Remember that an exponent of $\frac{1}{3}$ means taking the cube root.

Example 7

Evaluate the following expressions.

a) $3 \cdot 5^2-10 \cdot 5+1$

b) $\frac{2 \cdot 4^2-3 \cdot 5^2}{3^2-2^2}$

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

Solution

a) Evaluate the exponent: $3 \cdot 5^2 - 10 \cdot 5+1=3 \cdot 25-10 \cdot 5+1$

Perform multiplications from left to right: $3 \cdot 25-10 \cdot 5+1=75-50+1$

Perform additions and subtractions from left to right: $75-50+1=26$

b) Treat the expressions in the numerator and denominator of the fraction like they are in parentheses: $\frac{(2 \cdot 4^2-3 \cdot 5^2)}{(3^2-2^2)}=\frac{(2 \cdot 16-3 \cdot 25)}{(9-4)}=\frac{(32-75)}{5}=\frac{-43}{5}$

c) $\left(\frac{3^3}{2^2}\right)^{-2} \cdot \frac{3}{4}=\left(\frac{2^2}{3^3}\right)^2 \cdot \frac{3}{4}=\frac{2^4}{3^6} \cdot \frac{3}{4}=\frac{2^4}{3^6} \cdot \frac{3}{2^2}=\frac{2^2}{3^5}=\frac{4}{243}$

Example 8

Evaluate the following expressions for $x = 2, y = - 1, z = 3$.

a) $2x^2-3y^3+4z$

b) $(x^2-y^2)^2$

c) $\left(\frac{3x^2y^5}{4z}\right)^{-2}$

Solution

a) $2x^2-3y^3+4z=2 \cdot 2^2-3 \cdot (-1)^3+4 \cdot 3=2 \cdot 4-3 \cdot (-1)+4 \cdot 3=8+3+12=23$

b) $(x^2-y^2)^2=(2^2 - (-1)^2)^2=(4-1)^2=3^2=9$

c) $\left( \frac{3x^2y^5}{4z}\right)^{-2}=\left( \frac{3 \cdot 2^2 \cdot (-1)^5}{4 \cdot 3}\right)^{-2}=\left(\frac{3 \cdot 4 \cdot (-1)}{12}\right)^{-2}=\left(\frac{-12}{12}\right)^{-2}=\left(\frac{-1}{1}\right)^{-2}=\left(\frac{1}{-1}\right)^2=(-1)^2=1$

## Review Questions

Simplify the following expressions in such a way that there aren't any negative exponents in the answer.

1. $x^{-1}y^2$
2. $x^{-4}$
3. $\frac{x^{-3}}{x^{-7}}$
4. $\frac{x^{-3}y^{-5}}{z^{-7}}$
5. $(x^\frac{1}{2} y^\frac{-2}{3})(x^2y^\frac{1}{3})$
6. $\left ( \frac{a}{b} \right )^{-2}$
7. $(3a^{-2}b^2c^3)^3$
8. $x^{-3} \cdot x^3$

Simplify the following expressions in such a way that there aren't any fractions in the answer.

1. $\frac{a^{-3}(a^5)}{a^{-6}}$
2. $\frac{5x^6y^2}{x^8y}$
3. $\frac{(4ab^6)^3}{(ab)^5}$
4. $\left ( \frac{3x}{y^\frac{1}{3}} \right )^3$
5. $\frac{3x^2y^\frac{3}{2}}{xy^\frac{1}{2}}$
6. $\frac{(3x^3)(4x^4)}{(2y)^2}$
7. $\frac{a^{-2}b^{-3}}{c^{-1}}$
8. $\frac{x^\frac{1}{2}y^\frac{5}{2}}{x^\frac{3}{2} y^\frac{3}{2}}$

Evaluate the following expressions to a single number.

1. $3^{-2}$
2. $(6.2)^0$
3. $8^{-4} \cdot 8^6$
4. $\left (16^\frac{1}{2} \right )^3$
5. $x^2 \cdot 4x^3 \cdot y^4 \cdot 4y^2$, if $x = 2$ and $y = -1$
6. $a^4(b^2)^3 + 2ab$, if $a = -2$ and $b = 1$
7. $5x^2 - 2y^3 + 3z$, if $x = 3, y = 2,$ and $z = 4$
8. $\left ( \frac{a^2}{b^3} \right )^{-2}$, if $a = 5$ and $b = 3$
9. $\left ( \frac{x^{-2}}{y^4} \right )^\frac{1}{2}$, if $x=-3$ and $y=2$

Feb 22, 2012

Aug 22, 2014