<meta http-equiv="refresh" content="1; url=/nojavascript/"> Negative Exponents ( Read ) | Algebra | CK-12 Foundation

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# Zero and Negative Exponents

How can you use the quotient rules for exponents to understand the meaning of a zero or negative exponent?

### Guidance

##### Zero Exponent

Recall that $\boxed{\frac{a^m}{a^n}=a^{m-n}}$ . If $m = n$ , then the following would be true:

$\frac{a^m}{a^n}&=a^{{\color{red}m-n}}=a^{\color{red}0}\\\frac{3^3}{3^3} &= 3^{{\color{red}3-3}}=3^{\color{red}0}$

However, any quantity divided by itself is equal to one. Therefore, $\frac{3^3}{3^3}=1$ which means $3^{\color{red}0}={\color{red}1}$ . This is true in general:

$\boxed{a^{\color{red}0}=1 \ \text{if} \ a \neq 0.}$

Note that if $a=0, \ 0^{\color{red}0}$ is not defined.

##### Negative Exponents

$4^2 \times 4^{-2}=4^{\color{red}2+(-2)}=4^{\color{red}0}={\color{red}1}$

Therefore:

$& 4^2 \times 4^{-2}=1\\& \frac{4^2 \times 4^{-2}}{4^2}=\frac{1}{4^2} && \text{Divide both sides by} \ 4^2.\\& \frac{\cancel{4^2} \times 4^{-2}}{\cancel{4^2}}=\frac{1}{4^2} && \text{Simplify the equation.}\\& \boxed{4^{{\color{red}-2}}=\frac{1}{4^{\color{red}2}}}$

This is true in general and creates the following laws for negative exponents:

• $\boxed{a^{{\color{red}-m}}=\frac{1}{a^{\color{red}m}}}$
• $\boxed{\frac{1}{a^{{\color{red}-m}}}=a^{\color{red}m}}$

These laws for negative exponents can be expressed in many ways:

• If a term has a negative exponent, write it as 1 over the term with a positive exponent. For example: $a^{\color{red}-m}=\frac{1}{a^{\color{red}m}}$ and $\frac{1}{a^{\color{red}-m}}=a^{\color{red}m}$
• If a term has a negative exponent, write the reciprocal with a positive exponent. For example: $\left(\frac{2}{3}\right)^{{\color{red}-2}}=\left(\frac{3}{2}\right)^{\color{red}2}$ and $a^{{\color{red}-m}}=\frac{a^{-m}}{1}=\frac{1}{a^{\color{red}m}}$
• If the term is a factor in the numerator with a negative exponent, write it in the denominator with a positive exponent. For example: $3x^{{\color{red}-3}}y=\frac{3y}{x^{\color{red}3}}$ and $a^{{\color{red}-m}}b^n=\frac{1}{a^{\color{red}m}}(b^n)=\frac{b^n}{a^{\color{red}m}}$
• If the term is a factor in the denominator with a negative exponent, write it in the numerator with a positive exponent. For example: $\frac{2x^3}{x^{-2}}=2x^3(x^2)$ and $\frac{b^n}{a^{{\color{red}-m}}}=b^n \left(\frac{a^{{\color{red}m}}}{1}\right)=b^na^{\color{red}m}$

These ways for understanding negative exponents provide shortcuts for arriving at solutions without doing tedious calculations. The results will be the same.

#### Example A

Evaluate the following using the laws of exponents.

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

Solution:

There are two methods that can be used to evaluate the expression.

Method 1: Apply the negative exponent rule $\boxed{a^{-m}=\frac{1}{a^m}}$

$& \left(\frac{3}{4}\right)^{-2}=\frac{1}{{\color{red}\left(\frac{3}{4}\right)^2}} && \text{Write the expression with a positive exponent by applying} \ \boxed{a^{-m}=\frac{1}{a^m}}.\\& \frac{1}{\left(\frac{3}{4}\right)^2}=\frac{1}{{\color{red}\frac{3^2}{4^2}}} && \text{Apply the law of exponents for raising a quotient to a power.} \ \boxed{\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}}\\& \frac{1}{\frac{3^2}{4^2}}=\frac{1}{{\color{red}\frac{9}{16}}} && \text{Evaluate the powers.}\\& \frac{1}{\frac{9}{16}}=1 \div \frac{9}{16} && \text{Divide}\\& 1 \div \frac{9}{16}=1 \times \frac{16}{9}={\color{red}\frac{16}{9}}\\& \boxed{\left(\frac{3}{4}\right)^{-2}=\frac{16}{9}}$

Method 2: Apply the shortcut and write the reciprocal with a positive exponent.

$& \left(\frac{3}{4}\right)^{-2}={\color{red}\left(\frac{4}{3}\right)^2} && \text{Write the reciprocal with a positive exponent.}\\& \left(\frac{4}{3}\right)^2={\color{red}\frac{4^2}{3^2}} && \text{Apply the law of exponents for raising a quotient to a power.} && \boxed{\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}}\\& \frac{4^2}{3^2}={\color{red}\frac{16}{9}} && \text{Simplify.}\\& \boxed{\left(\frac{3}{4}\right)^{-2}=\frac{16}{9}}$

Applying the shortcut facilitates the process for obtaining the solution.

#### Example B

State the following using only positive exponents: (If possible, use shortcuts)

i) $y^{-6}$

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

iii) $\frac{x^5}{y^{-4}}$

iv) $a^2 \times a^{-5}$

Solutions:

i)

$& y^{-6} && \text{Write the expression with a positive exponent by applying} && \boxed{a^{-m}=\frac{1}{a^m}}.\\& \boxed{y^{-6}=\frac{1}{y^6}}$

ii)

$& \left(\frac{a}{b}\right)^{-3} && \text{Write the reciprocal with a positive exponent.}\\& \left(\frac{a}{b}\right)^{-3}={\color{red}\left(\frac{b}{a}\right)^3} && \text{Apply the law of exponents for raising a quotient to a power.} && \boxed{\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}}\\& \left(\frac{b}{a}\right)^3={\color{red}\frac{b^3}{a^3}}\\& \boxed{\left(\frac{a}{b}\right)^{-3}=\frac{b^3}{a^3}}$

iii)

$& \frac{x^5}{y^{-4}} && \text{Apply the negative exponent rule.} \ \boxed{\frac{1}{a^{{\color{red}-m}}}=a^{\color{red}m}}\\& \frac{x^5}{y^{-4}}=x^5 \left(\frac{y^{\color{red}4}}{1}\right) && \text{Simplify}.\\& \boxed{\frac{x^5}{y^{-4}}=x^5 y^4}$

iv)

$& a^2 \times a^{-5} && \text{Apply the product rule for exponents} \ \boxed{a^m \times a^n=a^{m+n}}.\\& a^2 \times a^{-5}=a^{{\color{red}2+(-5)}} && \text{Simplify}.\\& a^{2+(-5)}=a^{{\color{red}-3}} && \text{Write the expression with a positive exponent by applying} \ \boxed{a^{-m}=\frac{1}{a^m}}.\\& a^{-3}={\color{red}\frac{1}{a^3}}\\& \boxed{a^2 \times a^{-5}=\frac{1}{a^3}}$

#### Example C

Evaluate the following: $\frac{7^{-2}+7^{-1}}{7^{-3}+7^{-4}}$

Solution:

There are two methods that can be used to evaluate the problem.

Method 1: Work with the terms in the problem in exponential form.

Numerator:

$& 7^{-2}=\frac{1}{7^2} \ \text{and} \ 7^{-1}=\frac{1}{7} && \text{Apply the definition} \ a^{-m}=\frac{1}{a^m}\\& \frac{1}{7^2}+\frac{1}{7} && \text{A common denominator is needed to add the fractions.}\\& \frac{1}{7^2}+\frac{1}{7} {\color{red}\left(\frac{7}{7}\right)} && \text{Multiply} \ \frac{1}{7} \ \text{by} \ \frac{7}{7} \ \text{to obtain the common denominator of} \ 7^2\\& \frac{1}{7^2}+\frac{{\color{red}7}}{7^2}=\frac{1+7}{7^2}={\color{red}\frac{8}{7^2}} && \text{Add the fractions.}$

Denominator:

$& 7^{-3}=\frac{1}{7^3} \ \text{and} \ 7^{-4}=\frac{1}{7^4} && \text{Apply the definition} \ a^{-m}=\frac{1}{a^m}\\& \frac{1}{7^3}+\frac{1}{7^4} && \text{A common denominator is needed to add the fractions.}\\& {\color{red}\left(\frac{7}{7}\right)} \frac{1}{7^3}+\frac{1}{7^4} && \text{Multiply} \ \frac{1}{7^3} \ \text{by} \ \frac{7}{7} \ \text{to obtain the common denominator of} \ 7^4\\& \frac{{\color{red}7}}{7^4}+\frac{1}{7^4}=\frac{1+{\color{red}7}}{7^4}={\color{red}\frac{8}{7^4}} && \text{Add the fractions.}$

Numerator and Denominator:

$& \frac{8}{7^2} \div \frac{8}{7^4} && \text{Divide the numerator by the denominator.}\\& \frac{8}{7^2} \times \frac{7^4}{8} && \text{Multiply by the reciprocal.}\\& \frac{\cancel{8}}{7^2} \times \frac{7^4}{\cancel{8}}=\frac{7^4}{7^2}=7^{\color{red}2}={\color{red}49} && \text{Simplify.}\\& \boxed{\frac{7^{-2}+7^{-1}}{7^{-3}+7^{-4}}=49}$

Method 2: Multiply the numerator and the denominator by $7^4$ . This will change all negative exponents to positive exponents. Apply the product rule for exponents and work with the terms in exponential form.

$& \frac{7^{-2}+7^{-1}}{7^{-3}+7^{-4}}\\& {\color{red}\left(\frac{7^4}{7^4}\right)} \frac{7^{-2}+7^{-1}}{7^{-3}+7^{-4}} && \text{Apply the distributive property with the product rule for exponents.}\\& \frac{7^{\color{red}2}+7^{\color{red}3}}{7^{\color{red}1}+7^{\color{red}0}} && \text{Evaluate the numerator and the denominator.}\\& \frac{49+343}{7+1}=\frac{392}{8}={\color{red}49}\\& \boxed{\frac{7^{-2}+7^{-1}}{7^{-3}+7^{-4}}=49}$

Whichever method is used, the result is the same.

#### Concept Problem Revisited

By the quotient rule for exponents, $\frac{x^m}{x^m}=x^{m-m}=x^0$ . Since anything divided by itself is equal to 1 (besides 0), $\frac{x^m}{x^m}=1$ . Therefore, $x^0=1$ as long as $x\neq 0$ .

Also by the quotient rule for exponents, $\frac{x^2}{x^5}=x^{2-5}=x^{-3}$ . If you were to expand and reduce the original expression you would have $\frac{x^2}{x^5}=\frac{x\cdot x}{x\cdot x \cdot x\cdot x \cdot x}=\frac{1}{x^3}$ . Therefore, $x^{-3}=\frac{1}{x^3}$ . This generalizes to $x^{-a}=\frac{1}{x^a}$ .

### Vocabulary

Base
In an algebraic expression, the base is the variable, number, product or quotient, to which the exponent refers. Some examples are: In the expression $2^5$ , ‘2’ is the base. In the expression $(-3y)^4$ , ‘ $-3y$ ’ is the base.
Exponent
In an algebraic expression, the exponent is the number to the upper right of the base that tells how many times to multiply the base times itself. Some examples are:
In the expression $2^5$ , ‘5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: $2^5=2 \times 2 \times 2 \times 2 \times 2$ .
In the expression $(-3y)^4$ , ‘4’ is the exponent. It means to multiply $-3y$ times itself 4 times as shown here: $(-3y)^4=-3y \times -3y \times -3y \times -3y$ .
Laws of Exponents
The laws of exponents are the algebra rules and formulas that tell us the operation to perform on the exponents when dealing with exponential expressions.

### Guided Practice

1. Use the laws of exponents to simplify the following: $(-3x^2)^3 (9x^4y)^{-2}$

2. Rewrite the following using only positive exponents. $(x^2 y^{-1})^2$

3. Use the laws of exponents to evaluate the following: $[5^{-4} \times (25)^3]^2$

1.

$& (-3x^2)^3(9x^4y)^{-2} && \text{Apply the laws of exponents} \ \boxed{(a^m)^n=a^{mn}} \ \text{and} \ \boxed{a^{-m}=\frac{1}{a^m}}\\& (-3x^2)^3 (9x^4y)^{-2}=(-3^{\color{red}3}x^{\color{red}6}) \left(\frac{1}{(9x^4y)^2}\right) && \text{Simplify and apply} \ \boxed{(ab)^n=a^nb^n}\\ & (-3^3x^6) \left(\frac{1}{(9x^4y)^2}\right)={\color{red}-27}x^6 \left(\frac{1}{(9^{\color{red}2} x^{\color{red}8} y^{\color{red}2})}\right) && \text{Simplify}.\\& -27x^6 \left(\frac{1}{(9^2x^8y^2)}\right)=\frac{-27x^6}{{\color{red}81}x^8y^2} && \text{Simplify and apply the quotient rule for exponents } \boxed{\frac{a^m}{a^n}=a^{m-n}}.\\& \frac{-27x^6}{81x^8y^2}={\color{red}-\frac{1x^{-2}}{3y^2}} && \text{Apply the negative exponent rule} \ \boxed{a^{-m}=\frac{1}{a^m}}\\& \boxed{(-3x^2)^3 (9x^4y)^{-2}=-\frac{1}{3x^2y^2}}$

2.

$(x^2 y^{-1})^2 &= x^4y^{-2}\\&=\frac{x^4}{y^2}$

3.

$& [5^{-4} \times (25)^3]^2 && \text{Try to do this one by applying the laws of exponents.}\\& [5^{-4} \times (25)^3]^2=[5^{-4} \times ({\color{red}5^2})^3]^2\\& [5^{-4} \times ({\color{red}5^2})^3]^2=[5^{-4} \times 5^{\color{red}6}]^2\\& [5^{-4} \times 5^{\color{red}6}]^2=(5^{\color{red}2})^2\\& (5^{\color{red}2})^2=5^{\color{red}4}\\& 5^4={\color{red}625}\\& \boxed{[5^{-4} \times (25)^3]^2=5^4=625}$

### Practice

Evaluate each of the following expressions:

1. $-\left(\frac{2}{3}\right)^0$
2. $\left(-\frac{2}{5}\right)^{-2}$
3. $(-3)^{-3}$
4. $6 \times \left(\frac{1}{2}\right)^{-2}$
5. $7^{-4} \times 7^4$

Rewrite the following using positive exponents only. Simplify where possible.

1. $(4wx^{-2}y^3z^{-4})^3$
2. $\frac{a^2b^3c^{-2}}{d^{-2}bc^{-6}}$
3. $x^{-2}(x^2-1)$
4. $m^4(m^2+m-5m^{-2})$
5. $\frac{x^{-2}y^{-2}}{x^{-1}y^{-1}}$
6. $\left(\frac{x^{-2}}{y^4}\right)^3\left(\frac{y^{-4}}{x^6}\right)^{-7}$
7. $\frac{(x^{-2}y^4)^2}{(x^5y^{-3})^4}$
8. $\frac{(3xy^2)^3}{(3x^2y)^4}$
9. $\left(\frac{x^2y^{-25}z^5}{-12.4x^3y}\right)^0$
10. $\left(\frac{x^{-2}}{y^3}\right)^5\left(\frac{y^{-2}}{x^4}\right)^{-3}$