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# 6.1: The Laws of Exponents

Created by: CK-12

## Multiplying and Dividing Exponential Terms

Objectives

The lesson objectives for the Laws of Exponents are:

• Multiplying Exponential Terms
• Dividing Exponential Terms

Introduction

If $a \varepsilon R$ and $n \varepsilon N$, then the $n$ th power of ‘$a$’ is written as:

$& a^n = \underleftrightarrow{a \times a \times a \times \ldots \times a}\\& \qquad \qquad \qquad {\color{red}\downarrow}\\& \qquad \qquad \ {\color{red}n \ \text{factors}}$

where $n$ is called the exponent and $a$ is called the base. The term $a^n$ is known as a power. In other words $\boxed{ 4^3 = 4 \times 4 \times 4 = 64}$ and $\boxed{2^6=2 \times 2 \times 2 \times 2 \times 2 \times 2=64}$.

Based on this concept, there are five laws of exponents, which will be illustrated in the following lesson.

Watch This

Guidance

1. To multiply two powers with the same base, add the exponents.

$& a^m \times a^n = \underleftrightarrow{(a \times a \times \ldots \times a)} \ \underleftrightarrow{(a \times a \times \ldots \times a)}\\& \qquad \qquad \qquad \qquad \ {\color{red}\downarrow} \qquad \qquad \qquad \quad \ {\color{red}\downarrow}\\& \qquad \qquad \qquad {\color{red} m \ \text{factors}} \qquad \qquad {\color{red} n \ \text{factors}}\\& a^m \times a^n = \underleftrightarrow{(a \times a \times a \ldots \times a)}\\& \qquad \qquad \qquad \qquad \ {\color{red}\downarrow}\\& \qquad \qquad \qquad {\color{red} m+n \ \text{factors}}\\& a^m \times a^n=a^{{\color{red}m+n}}$

2. To divide two powers with the same base, subtract the exponents.

$& \qquad \qquad \ {\color{red} m \ \text{factors}}\\& \qquad \qquad \qquad {\color{red}\uparrow}\\& \frac{a^m}{a^n}=\frac{\overleftrightarrow{(a \times a \times \ldots \times a)}}{\underleftrightarrow{(a \times a \times \ldots \times a)}} \ m>n;a \neq 0\\& \qquad \qquad \qquad {\color{red}\downarrow}\\& \qquad \qquad \ {\color{red} n \ \text{factors}}\\& \frac{a^m}{a^n}=\underleftrightarrow{(a \times a \times \ldots \times a)}\\& \qquad \qquad \qquad {\color{red}\downarrow}\\& \qquad \qquad \ {\color{red} m-n \ \text{factors}}\\& \frac{a^m}{a^n}=a^{\color{red}m-n}$

3. To raise a power to a new power, multiply the exponents.

$& (a^m)^n = \underleftrightarrow{(a \times a \times \ldots \times a)^n}\\& \qquad \qquad \qquad \quad {\color{red}\downarrow}\\& \qquad \qquad \quad {\color{red}m} \ \text{{\color{red} factors}}\\& (a^m)^n=\underleftrightarrow{(a \times a \times \ldots \times a)} \times \underleftrightarrow{(a \times a \times \ldots \times a)} \ \underleftrightarrow{(a \times a \times \ldots \times a)}\\& \qquad \qquad \qquad \quad \ {\color{red}\downarrow} \qquad \qquad \qquad \qquad {\color{red}\downarrow} \qquad \qquad \qquad \quad {\color{red}\downarrow}\\& \qquad \quad \quad \underleftrightarrow{\quad {\color{red}m} \ \text{{\color{red}factors}} \qquad \qquad \ \ {\color{red}m} \ \text{{\color{red}factors}} \qquad \quad \ {\color{red}m} \ \text{{\color{red}factors}} \ \ }\\& \qquad \qquad \qquad \qquad \qquad \qquad \qquad {\color{red}n \ times}\\& (a^m)^n=\underleftrightarrow{a \times a \times a \ldots \times a}\\& \qquad \qquad {\color{red}mn \ \text{factors}}\\& (a^m)^n = a^{\color{red}mn}$

4. To raise a product to a power, raise each of the factors to the power.

$&(ab)^n=\underleftrightarrow{(ab) \times (ab) \times \ldots \times (ab)}\\& \qquad \qquad \qquad \qquad {\color{red}\downarrow}\\& \qquad \qquad \qquad {\color{red}n} \ {\color{red}\text{factors}}\\& (ab)^n=\underleftrightarrow{(a \times a \times \ldots \times a)} \times \underleftrightarrow{(b \times b \times \ldots \times b)}\\& \qquad \qquad \qquad \quad {\color{red}\downarrow} \qquad \qquad \qquad \qquad {\color{red}\downarrow}\\& \qquad \qquad \quad \ {\color{red}n} \ {\color{red}\text{factors}} \qquad \qquad \ {\color{red}n} \ {\color{red}\text{factors}}\\& (ab)^n=a^{{\color{red}n}} b^{{\color{red}n}}$

5. To raise a quotient to a power, raise both the numerator and the denominator to the power.

$& \left(\frac{a}{b} \right)^n= \underleftrightarrow{\frac{a}{b} \times \frac{a}{b} \times \ldots \times \frac{a}{b}}\\& \qquad \qquad \qquad \quad \ {\color{red}\downarrow}\\& \qquad \qquad \quad \quad {\color{red}n} \ {\color{red}\text{factors}}\\& \qquad \qquad \quad \quad {\color{red}n} \ {\color{red}\text{factors}}\\& \qquad \qquad \qquad \quad \ {\color{red}\uparrow}\\& \left(\frac{a}{b}\right)^n=\frac{\overleftrightarrow{(a \times a \times \ldots \times a)}}{\underleftrightarrow{(b \times b \times \ldots \times b)}}\\& \qquad \qquad \qquad \quad \ {\color{red}\downarrow}\\& \qquad \qquad \quad \quad {\color{red}n} \ {\color{red}\text{factors}}\\& \left(\frac{a}{b} \right)^n=\frac{a^{\color{red}n}}{b^{\color{red}n}} \ (b \neq 0)$

The following table will summarize the above five laws of exponents:

Laws of Exponents

If $a \varepsilon R$ and $m, n \varepsilon N$, then

1. $a^m \times a^n = a^{m+n}$
2. $\frac{a^m}{a^n}=a^{m-n}$ (if $m>n, a \neq 0$)
3. $(a^m)^n=a^{mn}$
4. $(ab)^n=a^n b^n$
5. $\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n} \ (b \neq 0)$

Example A

$\boxed{a^m \times a^n=a^{m+n}}$

Use this law of exponents to evaluate each of the following expressions:

a) $3^2 \times 3^3$

b) $(x^3) (x^6)$

c) $y^5 \cdot y^2$

d) $5x^2 y^3 \cdot 3xy^2$

a) $& 3^2 \times 3^3 && \text{The base is} \ 3.\\& 3^{2+3} && \text{Keep the base of} \ 3 \ \text{and add the exponents.}\\& 3^{\color{red}5} && \text{This answer is in exponential form.}$

The answer can be taken one step further. The base is numerical so the term can be evaluated.

$& 3^5=3 \times 3 \times 3 \times 3 \times 3\\& {\color{red}3^5}={\color{red}243}\\& \boxed{3^2 \times 3^3 = 3^5=243}$

b) $& (x^3)(x^6) && \text{The base is} \ x.\\& x^{3+6} && \text{Keep the base of} \ x \ \text{and add the exponents.}\\& x^{\color{red}9} && \text{The answer is in exponential form.}\\& \boxed{(x^3)(x^6)=x^9}$

c) $& y^5 \cdot y^2 && \text{The base is} \ y.\\& y^{5+2} && \text{Keep the base of} \ y \ \text{and add the exponents.}\\& y^{\color{red}7} && \text{The answer is in exponential form.}\\& \boxed{y^5 \cdot y^2=y^7}$

d) $& 5x^2 y^3 \cdot 3xy^2 && \text{The bases are} \ x \ \text{and} \ y.\\& 15(x^2 y^3)(xy^2) && \text{Multiply the coefficients -} \ 5 \times 3=15. \ \text{Keep the bases of} \ x \ \text{and} \ y \ \text{and add}\\& && \text{the exponents of the same base. If a base does not have a written}\\& && \text{exponent, it is understood as} \ 1.\\& 15x^{2+1} y^{3+2}\\& 15x^{\color{red}3} y^{\color{red}5} && \text{The answer is in exponential form.}\\& \boxed{5x^2 y^3 \cdot 3xy^2=15x^3y^5}$

Example B

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

Use this law of exponents to evaluate each of the following expressions:

a) $2^7 \div 2^3$

b) $\frac{x^8}{x^2}$

c) $\frac{y^3}{y^{-5}}$

d) $\frac{16x^2 y^5}{4x^5 y^3}$

a) $& 2^7 \div 2^3 && \text{The base is} \ 2.\\& 2^{7-3} && \text{Keep the base of} \ 2 \ \text{and subtract the exponents}.\\& 2^{\color{red}4} && \text{The answer is in exponential form}.$

The answer can be taken one step further. The base is numerical so the term can be evaluated.

$& 2^4 = 2 \times 2 \times 2 \times 2\\&{\color{red}2^4} = {\color{red}16}\\& \boxed{2^7 \div 2^3 =2^4=16}$

b) $& \frac{x^8}{x^2} && \text{The base is} \ x.\\& x^{8-2} && \text{Keep the base of} \ x \ \text{and subtract the exponents.}\\& x^{\color{red}6} && \text{The answer is in exponential form.}\\& \boxed{\frac{x^8}{x^2}=x^6}$

c) $& \frac{y^3}{y^{-5}} && \text{The base is} \ y.\\& y^{3 \overset{{\color{red}+}}{--} 5} && \text{Keep the base of} \ y \ \text{and subtract the exponents.}\\& y^{\color{red}8} && \text{The answer is in exponential form.}\\& \boxed{\frac{y^3}{y^{-5}} = y^8}$

d) $& \frac{16x^2 y^5}{4x^5 y^3} && \text{The bases are} \ x \ \text{and} \ y.\\& 4 \left( \frac{x^2 y^5}{x^5 y^3} \right) && \text{Divide the coefficients -} \ 16 \div 4=4. \ \text{Keep the bases of} \ x \ \text{and} \ y \ \text{and}\\& && \text{subtract the exponents of the same base.}\\& 4x^{2-5}y^{5-3}\\& 4x^{{\color{red}-3}} y^{\color{red}2} && \text{All answers must be written with positive exponents. If the base is in the}\\& && \text{numerator with a negative exponent, write it in the denominator with a}\\ & && \text{positive exponent.}\\& \frac{4y^{\color{red}2}}{x^{\color{red}3}} && \text{The answer is in exponential form.}\\& \boxed{\frac{16x^2 y^5}{4x^5 y^3}=\frac{4y^2}{x^3}}$

Example C

$\boxed{(a^m)^n=a^{mn}}$

Use this law of exponents to evaluate each of the following expressions:

a) $(2^3)^2$

b) $(x^7)^4$

c) $(3^2)^3$

d) $(y^4)^2$

a) $& (2^3)^2 && \text{The base is} \ 2^3.\\& 2^{3 \times 2} && \text{Keep the base of} \ 2^3 \ \text{and multiply the exponents}.\\& 2^{\color{red}6} && \text{The answer is in exponential form.}$

The answer can be taken one step further. The base is numerical so the term can be evaluated.

$& 2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\\& {\color{red}2^6}={\color{red}64}\\& \boxed{(2^3)^2=2^6=64}$

b) $& (x^7)^4 && \text{The base is} \ x^7.\\& x^{7 \times 4} && \text{Keep the base of} \ x^7 \ \text{and multiply the exponents}.\\& x^{\color{red}28} && \text{The answer is in exponential form.}\\& \boxed{(x^7)^4 = x^{28}}$

c) $& (3^2)^3 && \text{The base is} \ 3^2.\\& 3^{2 \times 3} && \text{Keep the base of} \ 3^2 \ \text{and multiply the exponents}.\\& 3^{\color{red}6} && \text{The answer is in exponential form}.$

The answer can be taken one step further. The base is numerical so the term can be evaluated.

$& 3^6=3 \times 3 \times 3 \times 3 \times 3 \times 3\\& {\color{red}3^6} = {\color{red}729}\\& \boxed{(3^2)^3=3^6=729}$

d) $& (y^4)^2 && \text{The base is} \ y^4.\\& y^{4 \times 2} && \text{Keep the base of} \ y^4 \ \text{and multiply the exponents}.\\& y^{\color{red}8} && \text{The answer is in exponential form}.\\& \boxed{(y^4)^2=y^8}$

Example D

$\boxed{(ab)^n=a^nb^n}$

Use this law of exponents to evaluate each of the following expressions:

a) $(-3x)^2$

b) $(5x^2 y^4)^3$

c) $(2^3 \times 3^2)^2$

d) $(3x^{-2} y^5)^3$

a) $& (-3x)^2 && \text{The base is} \ -3x.\\& (-3)^{1 \times 2} \cdot (x)^{1 \times 2} && \text{Keep the base of} \ -3x \ \text{and raise each factor of}\\& && \text{the base to the power of} \ 2.\\& (-3)^{\color{red}2} \cdot (x)^{\color{red}2} && \text{Simplify. Apply the exponent to each factor of the base.}\\& {\color{red}9}x^{\color{red}2} && \text{The answer is in exponential form.}\\& \boxed{(-3x)^2=9x^2}$

b) $& (5x^2 y^4)^3 && \text{The base is} \ 5x^2 y^4.\\& (5)^{1 \times 3} \cdot (x)^{2 \times 3} \cdot (y)^{4 \times 3} && \text{Keep the base of} \ 5x^2 y^4 \ \text{and raise each}\\& && \text{factor of the base to the power of} \ 3.\\& (5)^{\color{red}3} \cdot (x)^{\color{red}6} \cdot (y)^{\color{red}12} && \text{Simplify. Apply the exponent to each factor of the base.}\\& {\color{red}125} x^{\color{red}6} y^{\color{red}12} && \text{The answer is in exponential form.}\\& \boxed{(5x^2 y^4)^3 = 125x^6 y^{12}}$

c) $& (2^3 \times 3^2)^2 && \text{The base is} \ 2^3 \times 3^2.\\& (2)^{3 \times 2} \cdot (3)^{2 \times 2} && \text{Keep the base of} \ 2^3 \times 3^2 \ \text{and raise each}\\& && \text{factor of the base to the power of} \ 2.\\& 2^{\color{red}6} \times 3^{\color{red}4} && \text{Simplify. Apply the exponent to each factor of the base.}\\& 2^{\color{red}6} \times 3^{\color{red}4} && \text{The answer is in exponential form.}$

The answer can be taken one step further. The base of each factor is numerical so each term can be evaluated. The final answer will be the product of the two answers.

$& 2^6=2 \times 2 \times 2 \times 2 \times 2 \times 2\\& 2^6 = 64\\& 3^4=3 \times 3 \times 3 \times 3\\& 3^4 = 81\\& {\color{red}64 \times 81}={\color{red}5184}\\& \boxed{(2^3 \times 3^2)^2=2^6 \times 3^6=5184}$

d) $& (3x^{-2} y^5)^3 && \text{The base is} \ 3x^{-2} y^5.\\& (3)^{1 \times 3} \cdot (x)^{-2 \times 3} \cdot (y)^{5 \times 3} && \text{Keep the base of} \ 3x^{-2} y^5 \ \text{and raise each}\\& && \text{factor of the base to the power of} \ 3.\\& 3^{\color{red}3} \cdot x^{{\color{red}-6}} \cdot y^{{\color{red}15}} && \text{Simplify. Apply the exponent to each factor of the base.}\\& {\color{red}27} x^{{\color{red}-6}} y^{\color{red}15} && \text{Write} \ x \ \text{in the denominator with a positive exponent.}\\& \frac{{\color{red}27} y^{{\color{red}15}}}{x^{\color{red}6}} && \text{The answer has been simplified and is in exponential form.}\\& \boxed{(3x^{-2} y^5)^3 = \frac{27 y^{15}}{x^6}}$

Example E

$\boxed{\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}}$

Use this law of exponents to evaluate each of the following expressions:

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

b) $\left(\frac{x}{6}\right)^3$

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

d) $\left(\frac{4a^5b^3}{6ab}\right)^3$

a) $& \left(\frac{2}{3}\right)^2 && \text{The base is} \ \frac{2}{3}.\\& \frac{2^{1 \times 2}}{3^{1 \times 2}} && \text{Keep the base of} \ \frac{2}{3} \ \text{and multiply the exponents of both the numerator and the}\\& && \text{by} \ 2.\\& \frac{2^{\color{red}2}}{3^{\color{red}2}} && \text{The answer is in exponential form}.$

The answer can be taken one step further. The base is numerical so each term can be evaluated.

$& 2^2 = 2 \times 2 \qquad \ 3^2=3 \times 3\\& 2^2=4 \qquad \qquad 3^2=9\\& \frac{{\color{red}4}}{{\color{red}9}}\\& \boxed{\left(\frac{2}{3}\right)^2=\frac{2^2}{3^2}=\frac{4}{9}}$

b) $& \left(\frac{x}{6}\right)^3 && \text{The base is} \ \frac{x}{6}.\\& \frac{x^{1 \times 3}}{6^{1 \times 3}} && \text{Keep the base of} \ \frac{x}{6} \ \text{and multiply the exponents of both the numerator and the}\\& && \text{by} \ 3.\\& \frac{x^{\color{red}3}}{6^{\color{red}3}} && \text{The answer is in exponential form.}$

The answer can be taken one step further. The denominator is numerical so the term can be evaluated.

$& 6^3=6 \times 6 \times 6\\& 6^3=216\\& \frac{x^{\color{red}3}}{{\color{red}216}}\\& \boxed{\left(\frac{x}{6}\right)^3=\frac{x^3}{6^3}=\frac{x^3}{216}}$

c) $& \left(\frac{3x^2}{4y^3}\right)^2 && \text{The base is} \ \frac{3x^2}{4y^3}.\\& \frac{3^{1 \times 2} x^{2 \times 2}}{4^{1 \times 2} y^{3 \times 2}} && \text{Keep the base of} \ \frac{3x^2}{4y^3} \ \text{and multiply the exponents of both the}\\& && \text{numerator and the denominator by} \ 2.\\& \frac{3^{\color{red}2} x^{\color{red}4}}{4^{\color{red}2} y^{\color{red}6}} && \text{The answer is in exponential form.}$

The answer can be taken one step further. The denominator and the numerator both have numerical coefficients to be evaluated.

$& 3^2 = 3 \times 3 \qquad \ 4^2=4 \times 4\\& 3^2=9 \qquad \qquad 4^2=16\\& \frac{{\color{red}9}x^{\color{red}4}}{{\color{red}16}y^{\color{red}6}}\\& \boxed{\left(\frac{3x^2}{4y^3}\right)^2=\frac{3^2x^4}{4^2y^6}=\frac{9x^4}{16y^6}}$

d) $& \left(\frac{4a^5b^3}{6ab}\right)^3 && \text{The base is} \ \frac{4a^5b^3}{6ab}. \ \text{Begin by simplifying the base.}\\& \frac{2}{3}a^{{\color{red}5-1}}b^{{\color{red}3-1}}=\frac{2a^{\color{red}4}b^{\color{red}2}}{3} && \text{Write the problem with the simplified base.}\\& \left(\frac{2a^4b^2}{3}\right)^3\\& \frac{2^{1 \times 3} a^{4 \times 3} b^{2 \times 3}}{3^{1 \times 3}} && \text{Keep the base of} \ \frac{2a^4b^2}{3} \ \text{and multiply the exponents of}\\& && \text{both the numerator and the denominator by} \ 3.\\& \frac{2^{\color{red}3}a^{{\color{red}12}}b^{\color{red}6}}{3^{\color{red}3}} && \text{The answer is in exponential form.}$

The answer can be taken one step further. The denominator and the numerator both have numerical coefficients to be evaluated.

$& 2^3=2 \times 2 \times 2 \quad 3^3 = 3 \times 3 \times 3\\& 2^3=8 \qquad \qquad \ \ 3^3=27\\& \frac{{\color{red}8}a^{{\color{red}12}}b^{\color{red}6}}{{\color{red}27}}\\& \boxed{\left(\frac{4a^5b^3}{6ab}\right)^3=\left(\frac{2a^4b^2}{3}\right)^3=\frac{2^3a^{12}b^6}{3^3}=\frac{8a^{12}b^6}{27}}$

Example F

In many of the previous examples, powers with numerical bases were evaluated by expanding the power into its factors and determining the product of the factors.

$2^4$ was expanded to $2 \times 2 \times 2 \times 2$

The product was determined: $2 \times 2={\color{red}4} \times 2={\color{red}8} \times 2={\color{red}16}$

Therefore $\boxed{2^4=16}$

This concept can also be reversed. Write 32 as a power of 2.

$32={\color{red}2} \times {\color{red}2}=4 \times {\color{red}2}=8 \times {\color{red}2}=16 \times {\color{red}2}=32$

There are 5 twos. Therefore $\boxed{32=2^{\color{red}5}}$

Use the above concept to answer the following:

a) Write 81 as a power of 3.

b) Write $(9)^3$ as a power of 3.

c) Write $(4^3)^2$ as a power of 2.

a) $81={\color{red}3} \times {\color{red}3}=9 \times {\color{red}3}=27 \times {\color{red}3}=81$

There are 4 threes. Therefore $\boxed{81=3^{\color{red}4}}$

b) $(9)^3$

$9={\color{red}3} \times {\color{red}3}=9$

There are 2 threes. Therefore $\boxed{9=3^{\color{red}2}}$

$(3^2)^3$ Apply the law of exponents for power to a power--multiply the exponents.

$3^{2 \times 3}=3^{\color{red}6}$

Therefore $\boxed{(9)^3=3^{\color{red}6}}$

c) $(4^3)^2$

$4={\color{red}2} \times {\color{red}2}=4$

There are 2 twos. Therefore $\boxed{4=2^{\color{red}2}}$

$\left((2^2)^3\right)^2$ Apply the law of exponents for power to a power--multiply the exponents.

$\boxed{2^{2 \times 3}=2^{\color{red}6}}$

$(2^6)^2$ Apply the law of exponents for power to a power--multiply the exponents.

$\boxed{2^{6 \times 2}=2^{\color{red}12}}$

Therefore $\boxed{(4^3)^2=2^{\color{red}12}}$

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.
Power
A power is simply the name given to an algebraic expression that is raised to an exponent. $2^5$ and $(-3y)^4$ are both examples of a power.

Guided Practice

1. Perform the following operations:

i) $(x^{10}) (x^{10})$

ii) $(7x^3)(3x^7)$

iii) $(x^3 y^2) (xy^3) (x^5 y)$

2. Perform the following operations:

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

ii) $\frac{x^2}{x^{-3}}$

iii) $\frac{x^6 y^8}{x^4 y^{-2}}$

3. Perform the following operations:

i) $(2x^{12})^3$

ii) $(x^5 y^{10})^7$

iii) $\left(\frac{2x^{10}}{3y^{20}}\right)^3$

1. $(x^{10})(x^{10})$ Keep the base of ‘$x$’ and add the exponents.

i) $\boxed{(x^{10})(x^{10})=x^{{\color{red}10+10}}=x^{{\color{red}20}}}$

$(7x^3)(3x^7)$ Multiply the coefficients $\boxed{7 \times 3=21}$, keep the base of ‘$x$’ and add the exponents.

ii) $\boxed{(7x^3)(3x^7)={\color{red}21}x^{{\color{red}3+7}}}=21x^{{\color{red}10}}$

$(x^3 y^2) (xy^3) (x^5 y)$

iii) $\boxed{(x^3 y^2) (xy^3) (x^5 y)=x^{{\color{red}3+1+5}} y^{{\color{red}2+3+1}}=x^{\color{red}9} y^{\color{red}6}}$

2. $\frac{(x^3)(x^2)}{(x^4)}$ In the numerator, keep the base of ‘$x$’ and add the exponents.

i) $\frac{(x^3)(x^2)}{(x^4)}=\frac{(x^{{\color{red}3+2}})}{(x^4)}=\frac{x^{\color{red}5}}{x^4}$

$\frac{x^5}{x^4}$ Keep the base of ‘$x$’ and subtract the exponents.

$\frac{x^5}{x^4}=x^{{\color{red}5-4}}=x^{\color{red}1}$ or $x$

$\boxed{\frac{(x^3)(x^2)}{(x^4)}=x^1}$ or $x$

ii) $\frac{x^2}{x^{-3}}$ Keep the base of ‘$x$’ and subtract the exponents.

$\frac{x^2}{x^{-3}}=x^{{\color{red}2 \overset{\overset{+}{\leftrightarrow}}{--}3}}=x^{{\color{red}2+3}}=x^{\color{red}5}$

$\boxed{\frac{x^2}{x^{-3}}=x^5}$

iii) $\frac{x^6 y^8}{x^4 y^{-2}}$ Keep the bases of $x$ and $y$ and subtract the exponents.

$\frac{x^6 y^8}{x^4 y^{-2}}=x^{{\color{red}6-4}} y^{{\color{red}8 \overset{\overset{+}{\leftrightarrow}}{--}2}}=x^{\color{red}2}y^{{\color{red}10}}$

$\boxed{\frac{x^6 y^8}{x^4 y^{-2}}=x^2y^{10}}$

3) $(2x^{12})^3$ Keep the base of ‘$2x^{12}$’ and multiply the exponents of each factor by 3.

i) $(2x^{12})^3=2^{{\color{red}1 \times 3}} x^{{\color{red}12 \times 3}}=2^{\color{red}3} x^{\color{red}36}$ Evaluate $2^{\color{red}3}$.

$2^{\color{red}3}=2 \times 2 \times 2= {\color{red}8}$

$\boxed{2^3 x^{36}=8x^{36}}$

ii) $(x^5 y^{10})^7$ Keep the bases of $x$ and $y$ and raise each factor to the power of 7.

$(x^5 y^{10})^7=x^{{\color{red}5 \times 7}} y^{{\color{red}10 \times 7}}=x^{{\color{red}35}} y^{{\color{red}70}}$

$\boxed{(x^5y^{10})^7=x^{35}y^{70}}$

iii) $\left(\frac{2x^{10}}{3y^{20}}\right)^3$ Keep the base of ‘$\frac{2x^{10}}{3y^{20}}$’ and multiply the exponents of each factor in the numerator by 3 and the exponents of each factor in the denominator by 3.

$\left(\frac{2x^{10}}{3y^{20}}\right)^3=\frac{2^{{\color{red}1 \times 3}}x^{{\color{red}10 \times 3}}}{3^{{\color{red}1 \times 3}}y^{{\color{red}20 \times 3}}}=\frac{2^{\color{red}3}x^{{\color{red}30}}}{3^{\color{red}3}y^{{\color{red}60}}}$ Evaluate $3^{\color{red}3}$ and $2^{\color{red}3}$

$2^{\color{red}3}=2 \times 2 \times 2 = {\color{red}8} \qquad 3^{\color{red}3}=3 \times 3 \times 3={\color{red}27}$

$\boxed{\left(\frac{2x^{10}}{3y^{20}}\right)^3=\frac{8x^{30}}{27y^{60}}}$

Summary

In this lesson you have learned to apply five laws of exponents. These laws were:

• $\boxed{\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}}$
• $\boxed{(ab)^n=a^nb^n}$
• $\boxed{(a^m)^n=a^{mn}}$
• $\boxed{\frac{a^m}{a^n}=a^{m-n}}$
• $\boxed{a^m \times a^n=a^{m+n}}$

In addition to learning these laws of exponents you also learned to write numbers as a power of another number. For example, you learned to write 8 as a power of 2. $8=2^3$. This concept is very important in solving exponential equations. You will learn this is a future lesson in this chapter.

Problem Set

Express each of the following as a power of 3. Do not evaluate.

1. $(3^3)^5$
2. $(3^9)(3^3)$
3. $(9)(3^7)$
4. $9^4$
5. $(9)(27^2)$

Apply the laws of exponents to evaluate each of the following without using a calculator. (Show Your work.)

1. $(2^3)(2^2)$
2. $6^6 \div 6^5$
3. $-(3^2)^3$
4. $(1^2)^3+(1^3)^2$
5. $\left(\frac{1}{3}\right)^6 \div \left(\frac{1}{3}\right)^8$

Use the laws of exponents to simplify each of the following.

1. $(4x)^2$
2. $(-3x)^3$
3. $(x^3)^4$
4. $(3x)(x^7)$
5. $(5x)(4x^4)$
6. $(-3x^2)(-6x^3)$
7. $(10x^8) \div (2x^4)$
8. $(-2x)^5 (2x^2)$
9. $(16x^{10}) \left(\frac{3}{4}x^5\right)$
10. $\frac{(x^{15})(x^{24})(x^{25})}{(x^7)^8}$

## Date Created:

Jan 16, 2013

Jan 14, 2015
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