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# Negative Exponents

## Any value to the zero power equals 1, convert negative exponents

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Practice Negative Exponents
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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 $x^{-a}=\frac{1}{x^a}$ .

### 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.

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

### Explore More

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

### Vocabulary Language: English

Negative Exponent Property

Negative Exponent Property

The negative exponent property states that $\frac{1}{a^m} = a^{-m}$ and $\frac{1}{a^{-m}} = a^m$ for $a \neq 0$.
quotient rule

quotient rule

In calculus, the quotient rule states that if $f$ and $g$ are differentiable functions at $x$ and $g(x) \ne 0$, then $\frac {d}{dx}\left [ \frac{f(x)}{g(x)} \right ]= \frac {g(x) \frac {d}{dx}\left [{f(x)} \right ] - f(x) \frac{d}{dx} \left [{g(x)} \right ]}{\left [{g(x)} \right ]^2}$.
Zero Exponent Property

Zero Exponent Property

The zero exponent property says that for all $a \neq 0$, $a^0 = 1$.