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# 6.2: More Laws of Exponents

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Introduction

In this lesson you will learn more laws of exponents. These laws will include algebraic expressions involving zero exponents and negative exponents.

In addition to learning these laws, you will also learn use them, in conjunction with the laws of exponents you’ve previously learned, to perform various operations with exponential expressions.

Objectives

The lesson objectives for More Laws of Exponents:

• Understanding the Zero Exponent
• Negative Exponents
• Integrating all the Laws of Exponents

## Zero and Negative Exponents

Introduction

The term $a^n$ has been defined if $a \varepsilon R$ and $n \varepsilon N$, but what meaning does $a^n$ have if $n=0$ or if $n \varepsilon I$? These possibilities must be closely examined. The second law of exponents that dealt with dividing two powers was:

$\boxed{\frac{a^m}{a^n}=a^{m-n}}$

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 $3^{\color{red}0}={\color{red}1}$

Likewise,

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

However if $a=0, \ 0^{\color{red}0}$ is not defined. The reason is based on the fact that division by zero is undefined.

$0^{\color{red}0} &= 0^{{\color{red}1-1}}\\\frac{0^1}{0^1} &= \frac{0}{0} \ \text{{\color{red}which is undefined}}.$

This concept will be used determine if the other laws of exponents are true for integral exponents.

Watch This

Guidance

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

The first law of exponents $\boxed{a^m \times a^n=a^{m+n}}$ must also be true if $a^0 = 1$ and $a \neq 0$.

$a^m \times a^0 &= a^{{\color{red}m+0}}=a^{\color{red}m}\\a^m \times a^n &= a^{\color{red}m} \times {\color{red}a^0}=a^m \times {\color{red}1}=a^{\color{red}m}$

This first law of exponents must also be true for 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 law 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 $\frac{b^n}{a^{{\color{red}-m}}}=b^n \left(\frac{a^{{\color{red}m}}}{a}\right)=b^na^{\color{red}m}$

These expressions for understanding negative exponents all use the rules $\boxed{a^{{\color{red}-m}}=\frac{1}{a^{\color{red}m}}}$ and $\boxed{\frac{1}{a^{{\color{red}-m}}}=a^{\color{red}m}}$ and they provide shortcuts for arriving at the solution without doing all of the tedious calculations. The results will be the same.

With these definitions, it can be shown that the five laws of exponents are true for any integer exponents. For example, the law for raising a product to a power $\boxed{(ab)^n=a^nb^n}$ is true.

$& (2 \times 5)^{-4}=\frac{1}{(2 \times 5)^{\color{red}4}}=\frac{1}{2^{\color{red}4}} \times \frac{1}{5^{\color{red}4}}=2^{{\color{red}-4}} \times 5^{{\color{red}-4}}\\& \boxed{(2 \times 5)^{-4}=2^{-4} \times 5^{-4}}$

You should be able to use these definitions to prove the remaining laws of exponents.

The following table will summarize the laws of exponents for integral exponents.

Laws of Exponents for Integral Exponents

If $a \varepsilon R, a \neq 0$ and $m, n \varepsilon I$, then

1. $a^m \times a^n=a^{m+n}$

2. $\frac{a^m}{a^n}=a^{m-n} \ (\text{if} \ m>n, a \neq 0)$

3. $(a^m)^n=a^{mn}$

4. $(ab)^n=a^nb^n$

5. $\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n} \ (b \neq 0)$

6. $a^0=1 \ (a \neq 0)$

7. $a^{-m}=\frac{1}{a^m}$

Example A

Evaluate the following using the laws of exponents.

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

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

Method 1: Apply the law of exponents $\boxed{a^{-m}=\frac{1}{a^m}}$

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

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

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

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

Example C

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

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 first law of exponents and work with the terms in exponential form.

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

Whichever method is used, the result is the same.

Example D

Evaluate the following using the laws of exponents:

$& 2^{-1} \times 5^{-2} \times 3^0 \times 4^2 && \text{Write} \ 4 \ \text{as a power of} \ 2 \ \text{and apply the law} \ \boxed{(a^m)^n=a^{mn}}.\\& 4^2=({\color{red}2^2})^{\color{red}2}=2^{\color{red}4} && \text{Replace} \ 4^2 \ \text{with} \ 2^4 \ \text{in the problem and apply the law} \ \boxed{a^m \times a^n=a^{m+n}}.\\& 2^{-1} \times 5^{-2} \times 3^0 \times {\color{red}2^4}\\& 2^{-1{\color{red}+4}} \times 5^{-2} \times 3^0\\& 2^{\color{red}3} \times 5^{-2} \times 3^0 && \text{For} \ 5^{-2} \ \text{apply the law} \ \boxed{a^{-m}=\frac{1}{a^m}}.\\& 2^3 \times {\color{red}\frac{1}{5^2}} \times 3^0 && \text{Evaluate the terms and apply the law} \ \boxed{a^0=1}.\\& {\color{red}8} \times {\color{red}\frac{1}{25}} \times {\color{red}1} && \text{Simplify.}\\& {\color{red}\frac{8}{25}}\\& \boxed{2^{-1} \times 5^{-2} \times 3^0 \times 4^2 = \frac{8}{25}}$

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. $(x^2 y^{-1} -1)^2$
3. Use the laws of exponents to evaluate the following: $[5^{-4} \times (25)^3]^2$

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

2. $& (x^2y^{-1}-1)^2 && \text{Begin by expanding the binomial and using the FOIL method. Remember}\\& && \text{to apply the law of exponents} \ \boxed{a^m \times a^n=a^{m+n}}\\& (x^2y^{-1}-1)(x^2y^{-1}-1)= (x^{{\color{red}2+2}} y^{{\color{red}-1+(-1)}} -1x^2y^{-1}-1x^2y^{-1}+1) && \text{Simplify.}\\& (x^{{\color{red}2+2}}y^{{\color{red}-1+(-1)}} -1x^2y^{-1}-1x^2y^{-1}+1)=(x^{\color{red}4}y^{{\color{red}-2}}-{\color{red}2}x^2y^{-1}+1) && \text{Apply the law of exponents} \ \boxed{a^{-m}=\frac{1}{a^m}}.\\& (x^4y^{-2}-2x^2y^{-1}+1)=\left(\frac{x^4}{{\color{red}y^2}}-\frac{2x^2}{{\color{red}y}}+1 \right)\\& \boxed{(x^2y^{-1}-1)^2=\left(\frac{x^4}{y^2}-\frac{2x^2}{y}+1\right)}$

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

Summary

In this lesson you have learned more laws of exponents. You have learned the laws of exponents that involve exponents that belong to the integer number system. The two laws that were presented were $\boxed{a^0=1}$ if $a \neq 0$ and $\boxed{a^{{\color{red}-m}}=\frac{1}{a^{\color{red}m}} \ \text{or} \ \frac{1}{a^{{\color{red}-m}}}=a^{\color{red}m}}$. These laws were used with the laws that you had previously learned to evaluate exponential expressions, to simplify exponential expressions and to perform indicated operations with exponential expressions.

Problem Set

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. $7^{-4} \times 7^4$

Simplify the following:

1. $(2^{-1}-2^{-2})^2$
2. $(4^0 + 4^{-1})^{-1}$
3. $(3^{-1}-2^{-1})^{-2}$
4. $(x^{-1}+y^{-1})^2$
5. $\left(\frac{1}{6^{0}}-\frac{1}{6^{-1}}\right)^{-1}$

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. $(x^2-1)(x^{-2}+2)$
4. $m^4(m^2+m-5m^{-2})$
5. $\frac{x^{-2}-y^{-2}}{x^{-1}+y^{-1}}$

Evaluate each of...

$& -\left(\frac{2}{3}\right)^0\\& \boxed{-\left(\frac{2}{3}\right)^0=-1}$

$& (-3)^{-3}\\& (-3)^{-3} = \frac{1}{(-3)^3}\\& \frac{1}{(-3)^3}=\frac{1}{-27}\\& \frac{1}{-27}=-\frac{1}{27}\\& \boxed{(-3)^{-3}=-\frac{1}{27}}$

$& 7^{-4} \times 7^4\\& 7^{-4} \times 7^4=7^{-4+4}\\& 7^{-4+4}=7^0\\& 7^0=1\\& \boxed{7^{-4} \times 7^4=1}$

Simplify the following...

$& (2^{-1}-2^{-2})^2\\& (2^{-1}-2^{-2})^2=\left(\frac{1}{2}-\frac{1}{2^2}\right)^2\\& \left(\frac{1}{2}-\frac{1}{2^2}\right)^2=\left(\frac{1}{2}-\frac{1}{4}\right)^2\\& \left(\frac{1}{2}-\frac{1}{4}\right)^2=\left(\frac{2}{4}-\frac{1}{4}\right)^2\\& \left(\frac{2}{4}-\frac{1}{4}\right)^2=\left(\frac{1}{4}\right)^2\\& \left(\frac{1}{4}\right)^2=\frac{1^2}{4^2}\\& \frac{1^2}{4^2}=\frac{1}{16}\\& \boxed{(2^{-1}-2^{-2})^2=\frac{1}{16}}$

$& (3^{-1}-2^{-1})^{-2}\\& (3^{-1}-2^{-1})^{-2}=\left(\frac{1}{3}-\frac{1}{2}\right)^{-2}\\& \left(\frac{1}{3}-\frac{1}{2}\right)^{-2}=\left(\frac{2}{6}-\frac{3}{6}\right)^{-2}\\& \left(\frac{2}{6}-\frac{3}{6}\right)^{-2}=\left(-\frac{1}{6}\right)^{-2}\\& \left(-\frac{1}{6}\right)^{-2}=\left(-\frac{6}{1}\right)^2\\& \left(-\frac{6}{1}\right)^2=(-6)^2\\& (-6)^2=36\\& \boxed{(3^{-1}-2^{-1})^{-2}=36}$

$& \left(\frac{1}{6^0}-\frac{1}{6^{-1}}\right)^{-1}\\& \left(\frac{1}{6^0}-\frac{1}{6^{-1}}\right)^{-1}=\left(\frac{1}{1}-\frac{6}{1}\right)^{-1}\\& \left(\frac{1}{1}-\frac{6}{1}\right)^{-1}=(-5)^{-1}\\& (-5)^{-1}=\frac{1}{-5}\\& \frac{1}{-5}=-\frac{1}{5}\\& \boxed{\left(\frac{1}{6^0}-\frac{1}{6^{-1}}\right)^{-1}=-\frac{1}{5}}$

Rewrite the following...

$& (4wx^{-2}y^3z^{-4})^3\\& (4wx^{-2}y^3z^{-4})^3=4^{1 \times 3}w^{1 \times 3}x^{-2 \times 3}y^{3 \times 3}z^{-4 \times 3}\\& 4^{1 \times 3}w^{1 \times 3}x^{-2 \times 3}y^{3 \times 3}z^{-4 \times 3}=4^3w^3x^{-6}y^9z^{-12}\\& 4^3w^3x^{-6}y^9z^{-12}=64w^3x^{-6}y^9z^{-12}\\& 64w^3x^{-6}y^9z^{-12}=\frac{64w^3y^9}{x^6z^{12}}\\& \boxed{(4wx^{-2}y^3z^{-4})^3=\frac{64w^3y^9}{x^6z^{12}}}$

$& (x^2-1)(x^{-2}+2)\\& (x^2-1)(x^{-2}+2)=x^{2-2}+2x^2-1x^{-2}-2\\& x^{2-2}+2x^2-1x^{-2}-2=x^0+2x^2-\frac{1}{x^2}-2\\& x^0+2x^2-\frac{1}{x^2}-2=1+2x^2-\frac{1}{x^2}-2\\& 1+2x^2-\frac{1}{x^2}-2=2x^2-\frac{1}{x^2}-1\\& \boxed{(x^2-1)(x^{-2}+2)=2x^2-\frac{1}{x^2}-1}$

1. $\frac{x^{-2}-y^{-2}}{x^{-1}+y^{-1}}$

$& \text{Numerator}:&& \text{Denominator}\\& x^{-2}-y^{-2} && x^{-1}-y^{-1}\\& \frac{1}{x^2}-\frac{1}{y^2} && \frac{1}{x}-\frac{1}{y}\\& \frac{y^2}{y^2} \left(\frac{1}{x^2}\right)-\frac{x^2}{x^2} \left(\frac{1}{y^2}\right) && \frac{y}{y} \left(\frac{1}{x}\right)-\frac{x}{x} \left(\frac{1}{y}\right)\\& \frac{y^2-x^2}{x^2y^2} && \frac{y-x}{xy}\\& \frac{(y+x)(y-x)}{(xy)(xy)} && \frac{(y+x)(y-x)}{(xy)(xy)}$

$& \text{Numerator} \div \text{Denominator}\\& \frac{(y+x)(y-x)}{(xy)(xy)} \div \frac{(y+x)}{(xy)}\\& \frac{\cancel{(y+x)}(y-x)}{(xy)\cancel{(xy)}} \cdot \frac{\cancel{(xy)}}{\cancel{(y+x)}}\\& \frac{y-x}{xy}\\& \boxed{\frac{x^{-2}-y^{-2}}{x^{-1}+y^{-1}}=\frac{y-x}{xy}}$

## Rational Exponents

Introduction

In this lesson you will learn about rational exponents. So far, $a^n$ where $a \neq 0$ has been defined where $n \varepsilon N$ and where $n \varepsilon I$. Now you will learn that it is possible to define $a^n$ where the exponent is a rational number, such as $3^{\frac{1}{2}}$ or $8^{\frac{2}{3}}$.

In addition to learning this law of exponents, you will also learn use them, in conjunction with the laws of exponents you’ve previously learned, to perform various operations with exponential expressions.

Objectives

The lesson objectives for More Laws of Exponents:

• Rational exponents.
• Simplifying and Evaluating expressions with rational exponents.
• Integrating all the Laws of Exponents.

Introduction

In the law of exponents for raising a power to a power $\boxed{(a^m)^n=a^{mn}}$, $m$ can be substituted for the rational number $\frac{1}{n}$ to give $\left(a^{{\color{red}\frac{1}{n}}}\right)^n=a^{\frac{1}{\cancel{n}} \times \cancel{n}}=a^{\color{red}1}=a$. Therefore, $\boxed{\left(a^{\frac{1}{n}}\right)^n=a}$. If $a \ge 0$, the $n$th root of both sides of the equation can be taken to give:

$& \left(a^{\frac{1}{n}}\right)^n=a\\& \sqrt[{\color{red}n}]{\left(a^{\frac{1}{n}}\right)^n}=\sqrt[{\color{red}n}]{a}\\& \boxed{a^{{\color{red}\frac{1}{n}}}=\sqrt[{\color{red}n}]{a}} \ \text{and} \ n \varepsilon N.$

If ‘$n$’ is an even number, then the value of ‘$a$’ must be greater than or equal to zero. If ‘$n$’ is an odd number, then the value of ‘$a$’ can be any real number.

$& a^m \times a^{\frac{1}{n}} \rightarrow m, n \varepsilon N\\& a^m \times a^{\frac{1}{n}}=a^{\frac{m}{n}}\\& a^{\frac{m}{n}}=(a^m)^{{\color{red}\frac{1}{n}}} \quad \text{or} \quad a^{\frac{m}{n}}=\left(a^{{\color{red}\frac{1}{n}}}\right)^m\\& a^{\frac{m}{n}}=\sqrt[{\color{red}n}]{a^{\color{red}m}} \qquad \quad \ a^{\frac{m}{n}}=(\sqrt[{\color{red}n}]{a})^{\color{red}m}$

Using this concept, the following law of exponents can be written for rational exponents.

$\boxed{a^{\frac{m}{n}}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m \ m, n \varepsilon N}$

Watch This

Guidance

To determine the value of $4^{\frac{3}{2}}$, there are several methods that can be applied.

These methods refer back to the concepts presented in this lesson.

i) $4^{\frac{3}{2}}$

$& 4^{\frac{3}{2}}=(\sqrt{4})^3$

‘3’ is the exponent and ‘2’ is the index. Remember the index tells what root to find. The ‘2’ is understood and is never written when the operation is to take the square root of a number or term.

$& (\sqrt{4})^3=({\color{red}2})^3 && \text{The square root of} \ 4 \ \text{is} \ 2.\\& (2)^3= {\color{red}8}$

Therefore, $\boxed{4^{\frac{3}{2}}=8}$.

ii) $4^{\frac{3}{2}}$

$& 4^{\frac{3}{2}}=\sqrt{4^3} && \text{The first step is to evaluate} \ 4^3.\\& \sqrt{4^3}=\sqrt{{\color{red}64}} && \text{Take the square root of} \ 64.\\& \sqrt{64}={\color{red}8}$

Therefore, $\boxed{4^{\frac{3}{2}}=8}$.

iii) $4^{\frac{3}{2}}$

$& 4^{\frac{3}{2}}=\left(4^{\color{red}\frac{1}{2}}\right)^{\color{red}3} && \text{Write the power as} \ \left(a^{\frac{1}{n}}\right)^m.\\& \left(4^{\color{red}\frac{1}{2}}\right)^{\color{red}3}=\left({\color{red}\sqrt{4}}\right)^{3} && 4^{\frac{1}{2}} \ \text{means to take the square root of} \ 4.\\& \left(\sqrt{4}\right)^3={\color{red}2}^3 && \text{Evaluate} \ 2^3.\\& 2^3={\color{red}8}$

Therefore, $\boxed{4^{\frac{3}{2}}=8}$.

iv) $& 4^{\frac{3}{2}} && \text{Write} \ 4 \ \text{as a power of} \ 2.\\& 4^{\frac{3}{2}}=(2^2)^\frac{3}{2} && \text{Raise the power to a power by multiplying the exponents.}\\& (2^2)^{\frac{3}{2}}=2^{{\color{red}2 \times \frac{3}{2}}} && \text{Multiply the exponents.}\\& 2^{\cancel{2} \times \frac{3}{\cancel{2}}}=2^{\color{red}3} && \text{Evaluate} \ 2^3.\\& 2^3={\color{red}8}$

Therefore, $\boxed{4^{\frac{3}{2}}=8}$.

The following table will summarize the laws of exponents for rational exponents.

Laws of Exponents for Rational Exponents

If $a \varepsilon R, a \ge 0$ and $m, n \varepsilon Q$, then

1. $a^m \times a^n=a^{m+n}$
2. $\frac{a^m}{a^n}=a^{m-n} \ (\text{if} \ m > n, a \neq 0)$
3. $(a^m)^n=a^{mn}$
4. $(ab)^n=a^nb^n$
5. $\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n} \ (b \neq 0)$
6. $a^0=1 \ (a \neq 0)$
7. $a^{-m}=\frac{1}{a^m}$
8. $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m$

Example A

Simplify the following:

$(125)^{-\frac{2}{3}}$

$& (125)^{-\frac{2}{3}} && \text{Apply the law of exponents for negative exponents} \ \boxed{a^{-m}=\frac{1}{a^m}}.\\& \frac{1}{125^{{\color{red}\frac{2}{3}}}} && \text{Apply the law of exponents for rational exponents} \ \boxed{a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left( \sqrt[n]{a}\right)^{m} \ m,n \varepsilon N.}\\& \frac{1}{\left(\sqrt[{\color{red}3}]{125}\right)^{\color{red}2}}$

The cube root of 125 is ‘5’.

$\frac{1}{{\color{red}5}^2}$

Evaluate the denominator.

$& {\color{red}\frac{1}{25}}\\& \boxed{(125)^{-\frac{2}{3}}=\frac{1}{25}}\\& \boxed{a^{-m}=\frac{1}{a^m}}$

Example B

Simplify the following:

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

$& (2a^2b^4)^{\frac{3}{2}} && \text{Apply the law of exponents for raising a power to a power} \ \boxed{(a^m)^n=a^{mn}}.\\& (2a^2b^4)^{\frac{3}{2}}=2^{{\color{red}1 \times \frac{3}{2}}}(a^2)^{{\color{red}\frac{3}{2}}}(b^4)^{{\color{red}\frac{3}{2}}} && \text{Simplify the expression}.\\& 2^{{\color{red}1 \times \frac{3}{2}}} (a^2)^{{\color{red}\frac{3}{2}}} (b^4)^{{\color{red}\frac{3}{2}}}=2^{{\color{red}\frac{3}{2}}}(a)^{{\color{red}2 \times \frac{3}{2}}}(b)^{{\color{red}4 \times \frac{3}{2}}} && \text{Simplify. Apply the rule for rational exponents} \ \boxed{a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m \ m,n \varepsilon N}.\\& 2^{\color{red}\frac{3}{2}}(a)^{{\color{red}\cancel{2} \times \frac{3}{\cancel{2}}}} (b)^{{\color{red}\overset{2}{\cancel{4}} \times \frac{3}{\cancel{2}}}}=\sqrt{2^{{\color{red}3}}} (a)^{\color{red}3}(b)^{\color{red}6} && \text{Simplify}.\\& \sqrt{2^{\color{red}3}}(a)^{\color{red}3}(b)^{\color{red}6}=\sqrt{{\color{red}8}}a^3b^6\\& \sqrt{{\color{red}8}}a^3b^6={\color{red}2} \sqrt{{\color{red}2}} a^3b^6\\& \boxed{(2a^2b^4)^{\frac{3}{2}}=2 \sqrt{2}a^3b^6}$

Example C

The rational exponents represent the exponent and the index of the base. The numerator is the exponent and the denominator is the index.

a) State the following using radicals:

i) $2^{\frac{3}{8}}$

ii) $7^{-\frac{1}{5}}$

iii) $3^{\frac{3}{4}}$

b) State the following using exponents:

i) $\sqrt[3]{7^2}$

ii) $\frac{1}{\left(\sqrt[4]{5}\right)^3}$

iii) $\left(\sqrt[5]{a}\right)^2$

a) i) $& 2^{\frac{3}{8}}\\& \boxed{2^{\frac{3}{8}}=\sqrt[{\color{red}8}]{2^{\color{red}3}}=\sqrt[{\color{red}8}]{8}}$

ii) $& 7^{-\frac{1}{5}}\\& 7^{-\frac{1}{5}}=\frac{1}{7^{\frac{1}{5}}}\\& \boxed{7^{-\frac{1}{5}}=\frac{1}{\sqrt[{\color{red}5}]{7}}}$

iii) $& 3^{\frac{3}{4}}\\& \boxed{3^{\frac{3}{4}}=\sqrt[{\color{red}4}]{3^{\color{red}3}}=\sqrt[{\color{red}4}]{27}}$

b) i) $& \sqrt[3]{7^2}\\& \boxed{\sqrt[3]{7^2}=7^{\color{red}\frac{2}{3}}}$

ii) $& \frac{1}{\left(\sqrt[4]{5}\right)^3}\\& \frac{1}{\left(\sqrt[4]{5}\right)^3}=\frac{1}{5^{\color{red}\frac{3}{4}}}\\& \frac{1}{5^{\color{red}\frac{3}{4}}}=5^{{\color{red}-\frac{3}{4}}}$

iii) $& \left(\sqrt[5]{a}\right)^2\\& \boxed{\left(\sqrt[5]{a}\right)^2=a^{\color{red}\frac{2}{5}}}$

Guided Practice

1. Use the laws of exponents to evaluate the following: $9^{\frac{3}{2}} \div 36^{-\frac{1}{2}}$
2. Simplify the following using the laws of exponents. $(20a^2b^3c^{-1})^{\frac{3}{2}}$
3. Use the laws of exponents to evaluate the following: $\frac{64^{\frac{2}{3}}}{216^{-\frac{1}{3}}}$

1. $9^{\frac{3}{2}} \div 36^{-\frac{1}{2}}$

Apply the law of exponents for rational exponents $\boxed{a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m \ m, n \varepsilon N}$ to $9^{\frac{3}{2}}$.

Apply the law of exponents for negative exponents $\boxed{a^{-m}=\frac{1}{a^m}}$ to $36^{-\frac{1}{2}}$.

$& 9^{\frac{3}{2}} \div 36^{-\frac{1}{2}}=\left({\color{red}\sqrt{9}}\right)^{\color{red}3} \div \frac{1}{36^{\color{red}\frac{1}{2}}} && \text{Find the square root of} \ 9. \ \text{Apply} \ \boxed{a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m \ m,n \varepsilon N} \ 36^{\frac{1}{2}}.\\& \left(\sqrt{9}\right)^3 \div \frac{1}{36^{\frac{1}{2}}}=({\color{red}3})^3 \div \frac{1}{{\color{red}\sqrt{36}}} && \text{Simplify}.\\& (3)^3 \div \frac{1}{\sqrt{36}}={\color{red}27} \div \frac{1}{\sqrt{6}} && \text{Perform the indicated operation of division.}\\& 27 \div \frac{1}{6}=27 \times \frac{6}{1}={\color{red}162}\\& \boxed{9^{\frac{3}{2}} \div 36^{-\frac{1}{2}}=162}$

2. $& (20a^2b^3c^{-1})^{\frac{3}{2}} && \text{Apply the law of exponents} \ \boxed{(ab)^n=a^nb^n}.\\& (20a^2b^3c^{-1})^{\frac{3}{2}}=20^{{\color{red}1 \times \frac{3}{2}}} (a)^{{\color{red}2 \times \frac{3}{2}}} (b)^{{\color{red}3 \times \frac{3}{2}}} (c)^{{\color{red}-1 \times \frac{3}{2}}} && \text{Simplify the exponents}.\\& 20^{1 \times \frac{3}{2}} (a)^{2 \times \frac{3}{2}} (b)^{3 \times \frac{3}{2}} (c)^{-1 \times \frac{3}{2}}=20^{\color{red}\frac{3}{2}} (a)^{\color{red}3} (b)^{\color{red}\frac{9}{2}} (c)^{\color{red}-\frac{3}{2}}$

Apply the law of exponents for rational exponents $\boxed{a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m m, n \varepsilon N}$ to $20^{\frac{3}{2}}$ and $b^{\frac{9}{2}}$. To $c^{-\frac{3}{2}}$, apply the law for negative exponents $\boxed{a^{-m}=\frac{1}{a^m}}$ and then the law $\boxed{a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m \ m,n \varepsilon N}$ for rational exponents.

$& 20^{\frac{3}{2}} (a)^3 (b)^{\frac{9}{2}} (c)^{-\frac{3}{2}}=\left({\color{red}\sqrt{20}}\right)^{\color{red}3} (a)^3 {\color{red}\sqrt{b^9}} \left({\color{red}\frac{1}{c^{\frac{3}{2}}}}\right) && \text{Simplify}.\\& 20^{\frac{3}{2}} (a)^3 (b)^{\frac{9}{2}} (c)^{-\frac{3}{2}}=\left({\color{red} 2 \sqrt{5}}\right)^{\color{red}3} {\color{red}(a^3) \left ( b^4 \sqrt{b} \right )} \left({\color{red}\frac{1}{\sqrt{c^3}}}\right) && \text{Simplify}\\& \left(2 \sqrt{5}\right)^3 (a^3) \left ( b^4 \sqrt{b}\right) \left(\frac{1}{\sqrt{c^3}}\right)={\color{red}8\sqrt{125}} a^3b^4 \sqrt{b} \frac{1}{{\color{red}c \sqrt{c}}}\\& 8\sqrt{125}a^3b^4 \sqrt{b}\frac{1}{c\sqrt{c}}={\color{red}40\sqrt{5}} a^3b^4 \sqrt{b} \left({\color{red}c \sqrt{c}}\right)^{{\color{red}-1}} && \text{Simplify}\\& \boxed{(20 a^2b^3c^{-1})^{\frac{3}{2}}=40 \sqrt{5}a^3b^4 \sqrt{b} \left(c\sqrt{c}\right)^{-1}}$

3. $\frac{64^{\frac{2}{3}}}{216^{-\frac{1}{3}}}$ Try this question yourself.

$& \mathbf{Numerator} && \mathbf{Denominator}\\& 64^{\frac{2}{3}} && 216^{-\frac{1}{3}}\\& 64^{\frac{2}{3}}=\left(\sqrt[3]{64}\right)^2 && 216^{-\frac{1}{3}}=\frac{1}{216^{\frac{1}{3}}}\\& \left(\sqrt[3]{64}\right)^2=(4)^2 && 216^{-\frac{1}{3}}=\frac{1}{\sqrt[3]{216}}\\& (4)^2=16 && \frac{1}{\sqrt[3]{216}}=\frac{1}{6}$

Numerator divided by denominator:

$& 16 \div \frac{1}{6}\\& 16 \times \frac{6}{1}=96\\& \boxed{\frac{64^{\frac{2}{3}}}{216^{-\frac{1}{3}}}=96}$

Summary

In this lesson you have learned more laws of exponents. You have learned the laws of exponents that involve exponents that belong to the rational number system. The new law that was presented was the law for rational exponents: $\boxed{a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m \ m,n \varepsilon N}$ This law was used with the laws that you had previously learned to evaluate exponential expressions, to simplify exponential expressions and to perform indicated operations with exponential expressions.

Problem Set

Express each of the following as a radical and if possible, simplify.

1. $x^{\frac{1}{2}}$
2. $5^{\frac{3}{4}}$
3. $2^{\frac{3}{2}}$
4. $2^{-\frac{1}{2}}$
5. $9^{-\frac{1}{5}}$

Express each of the following using exponents:

1. $\sqrt{26}$
2. $\sqrt[3]{5^2}$
3. $\left(\sqrt[6]{a}\right)^5$
4. $\sqrt[4]{m}$
5. $\left(\sqrt[3]{7}\right)^2$

Evaluate each of the following using the laws of exponents:

1. $3^{\frac{2}{5}} \times 3^{\frac{3}{5}}$
2. $(6^{0.4})^5$
3. $2^{\frac{1}{7}} \times 4^{\frac{3}{7}}$
4. $\left(\frac{64}{125}\right)^{-\frac{1}{2}}$
5. $(81^{-1})^{-\frac{1}{4}}$

Simplify each of the following using the laws of exponents:

1. $5^{\frac{1}{2}} \times 5^{\frac{1}{3}}$
2. $(d^4 e^8 f^{12})^{\frac{1}{4}}$
3. $\sqrt[4]{\frac{y^{\frac{1}{2}} \sqrt{xy}}{x^{\frac{2}{3}}}}$
4. $(32a^{20}b^{-15})^{\frac{1}{5}}$
5. $(729x^{12}y^{-6})^{\frac{2}{3}}$

Express each of the following...

1. $x^{\frac{1}{2}}$

$& \boxed{x^{\frac{1}{2}}=\sqrt{x}}$

1. $2^{\frac{3}{2}}$

$& 2^{\frac{3}{2}}=\sqrt{2^3}\\& 2^{\frac{3}{2}}=\sqrt{2^3}\\& 2^{\frac{3}{2}}=\sqrt{2 \cdot 2 \cdot 2}\\& \boxed{2^{\frac{3}{2}}=2 \sqrt{2}}$

1. $9^{-\frac{1}{5}}$

$& 9^{-\frac{1}{5}}=\left(\frac{1}{9}\right)^{\frac{1}{5}}\\& \boxed{\left(\frac{1}{9}\right)^{\frac{1}{5}}=\sqrt[5]{\frac{1}{9}}}$

Express each of the following...

1. $\sqrt{26}$

$& \boxed{\sqrt{26}=26^{\frac{1}{2}}}$

1. $\left(\sqrt[6]{a}\right)^5$

$& \boxed{\left(\sqrt[6]{a}\right)^5=a^{\frac{5}{6}}}$

1. $\left(\sqrt[3]{7}\right)^2$

$& \boxed{\left(\sqrt[3]{7}\right)^2=7^{\frac{2}{3}}}$

Evaluate each of the following...

1. $3^{\frac{2}{5}} \times 3^{\frac{3}{5}}$

$& 3^{\frac{2}{5}} \times 3^{\frac{3}{5}}=3^{\frac{5}{5}}\\& 3^1=3$

1. $2^{\frac{1}{7}} \times 4^{\frac{3}{7}}$

$& 2^{\frac{1}{7}} \times 4^{\frac{3}{7}}=2^{\frac{1}{7}} \times (2^2)^{\frac{3}{7}}\\& 2^{\frac{1}{7}} \times (2^2)^{\frac{3}{7}}=2^{\frac{1}{7}} \times 2^{\frac{6}{7}}\\& 2^\frac{1}{7} \times 2^{\frac{6}{7}}=2^{\frac{7}{7}}\\& 2^{\frac{7}{7}}=2$

1. $(81^{-1})^{-\frac{1}{4}}$

$& (81^{-1})^{-\frac{1}{4}}=81^{\frac{1}{4}}\\& 81^{\frac{1}{4}}=(3^4)^{\frac{1}{4}}\\& (3^4)^{\frac{1}{4}}=3^{\frac{4}{4}}\\& 3^{\frac{4}{4}}=3$

Simplify each of the following...

1. $5^{\frac{1}{2}} \times 5^{\frac{1}{3}}$

$& 5^{\frac{1}{2}} \times 5^{\frac{1}{3}}=5^{\frac{3}{6}} \times 5^{\frac{2}{6}}\\& 5^{\frac{3}{6}} \times 5^{\frac{2}{6}}=5^{\frac{5}{6}}$

1. $\sqrt[4]{\frac{y^{\frac{1}{2}} \sqrt{xy}}{x^{\frac{2}{3}}}}$

$& \sqrt[4]{\frac{y^{\frac{1}{2}}\sqrt{xy}}{x^{\frac{2}{3}}}}=\sqrt[4]{\frac{y^{\frac{1}{2}}x^{\frac{1}{2}}y^{\frac{1}{2}}}{x^{\frac{2}{3}}}}\\& \sqrt[4]{\frac{y^{\frac{1}{2}}x^{\frac{1}{2}}y^{\frac{1}{2}}}{x^{\frac{2}{3}}}}=\sqrt[4]{\frac{x^{\frac{1}{2}}y}{x^{\frac{2}{3}}}}\\& \sqrt[4]{\frac{x^{\frac{1}{2}}y}{x^{\frac{2}{3}}}}=\sqrt[4]{\frac{x^{\frac{3}{6}}y}{x^{\frac{4}{6}}}}\\& \sqrt[4]{\frac{x^{\frac{3}{6}}y}{x^{\frac{4}{6}}}}=\sqrt[4]{x^{-\frac{1}{6}}y}\\& \sqrt[4]{x^{-\frac{1}{6}}y}=\left(x^{-\frac{1}{6}} y\right)^{\frac{1}{4}}\\& \left(x^{-\frac{1}{6}}y\right)^{\frac{1}{4}}=x^{-\frac{1}{24}}y^{\frac{1}{4}}\\& x^{-\frac{1}{24}}y^{\frac{1}{4}}=\frac{y^{\frac{1}{4}}}{x^{\frac{1}{24}}}$

1. $(729x^{12}y^{-6})^{\frac{2}{3}}$

$& (729x^{12}y^{-6})^{\frac{2}{3}}= \left [(3)^6x^{12}y^{-6} \right ]^{\frac{2}{3}}\\& \left [(3)^6x^{12}y^{-6} \right ]^{\frac{2}{3}}=3^4x^8y^{-4}\\& 3^4x^8y^{-4}=\frac{81x^8}{y^4}$

Jan 16, 2013

Jun 04, 2014