7.2 Exponent power rules

Algebra i for the community Page 1 / 1

This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The symbols, notations, and properties of numbers that form the basis of algebra, as well as exponents and the rules of exponents, are introduced in this chapter. Each property of real numbers and the rules of exponents are expressed both symbolically and literally. Literal explanations are included because symbolic explanations alone may be difficult for a student to interpret.Objectives of this module: understand the power rules for powers, products, and quotients.

Overview

The Power Rule for Powers
The Power Rule for Products
The Power Rule for quotients

The power rule for powers

The following examples suggest a rule for raising a power to a power:

${(a^{2})}^{3} = a^{2} \cdot a^{2} \cdot a^{2}$

Using the product rule we get

$\begin{array}{l} {(a^{2})}^{3} = a^{2 + 2 + 2} \\ {(a^{2})}^{3} = a^{3 \cdot 2} \\ {(a^{2})}^{3} = a^{6} \end{array}$

$\begin{array}{l} {(x^{9})}^{4} & = & x^{9} \cdot x^{9} \cdot x^{9} \cdot x^{9} \\ {(x^{9})}^{4} & = & x^{9 + 9 + 9 + 9} \\ {(x^{9})}^{4} & = & x^{4 \cdot 9} \\ {(x^{9})}^{4} & = & x^{36} \end{array}$

Power rule for powers

x

is a real number and

n

and

m

are natural numbers,

{(x^{n})}^{m} = x^{n \cdot m}

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

Sample set a

Simplify each expression using the power rule for powers. All exponents are natural numbers.

$\begin{matrix} {(x^{3})}^{4} = \begin{matrix} x^{3 \cdot 4} \end{matrix} x^{12} & The box represents a step done mentally. \end{matrix}$

${(y^{5})}^{3} = \begin{matrix} y^{5 \cdot 3} \end{matrix} = y^{15}$

${(d^{20})}^{6} = \begin{matrix} d^{20 \cdot 6} \end{matrix} = d^{120}$

${(x^{□})}^{△} = x^{□ △}$

Although we don’t know exactly what number $□ △$ is, the notation $□ △$ indicates the multiplication.

Practice set a

Simplify each expression using the power rule for powers.

${(x^{5})}^{4}$

$x^{20}$

${(y^{7})}^{7}$

$y^{49}$

The power rule for products

The following examples suggest a rule for raising a product to a power:

$\begin{array}{l} {(a b)}^{3} & = & a b \cdot a b \cdot a b & Use the commutative property of multiplication . \\ = & a a a b b b \\ = & a^{3} b^{3} \end{array}$

$\begin{array}{l} {(x y)}^{5} & = & x y \cdot x y \cdot x y \cdot x y \cdot x y \\ = & x x x x x \cdot y y y y y \\ = & x^{5} y^{5} \end{array}$

$\begin{array}{l} {(4 x y z)}^{2} & = & 4 x y z \cdot 4 x y z \\ = & 4 \cdot 4 \cdot x x \cdot y y \cdot z z \\ = & 16 x^{2} y^{2} z^{2} \end{array}$

Power rule for products

x

and

y

are real numbers and

n

is a natural number,

{(x y)}^{n} = x^{n} y^{n}

To raise a product to a power, apply the exponent to each and every factor.

Sample set b

Make use of either or both the power rule for products and power rule for powers to simplify each expression.

${(a b)}^{7} = a^{7} b^{7}$

${(a x y)}^{4} = a^{4} x^{4} y^{4}$

$\begin{matrix} {(3 a b)}^{2} = 3^{2} a^{2} b^{2} = 9 a^{2} b^{2} & Don't forget to apply the exponent to the 3! \end{matrix}$

$\begin{matrix} {(2 s t)}^{5} = 2^{5} s^{5} t^{5} = 32 s^{5} t^{5} \end{matrix}$

$\begin{array}{l} {(a b^{3})}^{2} = a^{2} {(b^{3})}^{2} = a^{2} b^{6} & We used two rules here . First, the power rule for \\ products . Second, the power rule for powers. \end{array}$

$\begin{array}{l} {(7 a^{4} b^{2} c^{8})}^{2} & = 7^{2} {(a^{4})}^{2} {(b^{2})}^{2} {(c^{8})}^{2} \\ = 49 a^{8} b^{4} c^{16} \end{array}$

$\begin{matrix} If 6 a^{3} c^{7} \neq 0, then {(6 a^{3} c^{7})}^{0} = 1 & Recall that x^{0} = 1 for x \neq 0. \end{matrix}$

$\begin{array}{l} {[2 {(x + 1)}^{4}]}^{6} & = 2^{6} {(x + 1)}^{24} \\ = 64 {(x + 1)}^{24} \end{array}$

Practice set b

Make use of either or both the power rule for products and the power rule for powers to simplify each expression.

${(a x)}^{4}$

$a^{4} x^{4}$

${(3 b x y)}^{2}$

$9 b^{2} x^{2} y^{2}$

${[4 t (s - 5)]}^{3}$

$64 t^{3} {(s - 5)}^{3}$

${(9 x^{3} y^{5})}^{2}$

$81 x^{6} y^{10}$

${(1 a^{5} b^{8} c^{3} d)}^{6}$

$a^{30} b^{48} c^{18} d^{6}$

${[(a + 8) (a + 5)]}^{4}$

${(a + 8)}^{4} {(a + 5)}^{4}$

${[(12 c^{4} u^{3} {(w - 3)}^{2}]}^{5}$

$12^{5} c^{20} u^{15} {(w - 3)}^{10}$

${[10 t^{4} y^{7} j^{3} d^{2} v^{6} n^{4} g^{8} {(2 - k)}^{17}]}^{4}$

$10^{4} t^{16} y^{28} j^{12} d^{8} v^{24} n^{16} g^{32} {(2 - k)}^{68}$

${(x^{3} x^{5} y^{2} y^{6})}^{9}$

${(x^{8} y^{8})}^{9} = x^{72} y^{72}$

${(10^{6} \cdot 10^{12} \cdot 10^{5})}^{10}$

$10^{230}$

The power rule for quotients

The following example suggests a rule for raising a quotient to a power.

${(\frac{a}{b})}^{3} = \frac{a}{b} \cdot \frac{a}{b} \cdot \frac{a}{b} = \frac{a \cdot a \cdot a}{b \cdot b \cdot b} = \frac{a^{3}}{b^{3}}$

Power rule for quotients

x

and

y

are real numbers and

n

is a natural number,

{(\frac{x}{y})}^{n} = \frac{x^{n}}{y^{n}}, y \neq 0

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

Sample set c

Make use of the power rule for quotients, the power rule for products, the power rule for powers, or a combination of these rules to simplify each expression. All exponents are natural numbers.

${(\frac{x}{y})}^{6} = \frac{x^{6}}{y^{6}}$

${(\frac{a}{c})}^{2} = \frac{a^{2}}{c^{2}}$

${(\frac{2 x}{b})}^{4} = \frac{{(2 x)}^{4}}{b^{4}} = \frac{2^{4} x^{4}}{b^{4}} = \frac{16 x^{4}}{b^{4}}$

${(\frac{a^{3}}{b^{5}})}^{7} = \frac{{(a^{3})}^{7}}{{(b^{5})}^{7}} = \frac{a^{21}}{b^{35}}$

$\begin{matrix} {(\frac{3 c^{4} r^{2}}{2^{3} g^{5}})}^{3} = \frac{3^{3} c^{12} r^{6}}{2^{9} g^{15}} = \frac{27 c^{12} r^{6}}{2^{9} g^{15}} & or & \frac{27 c^{12} r^{6}}{512 g^{15}} \end{matrix}$

${[\frac{(a - 2)}{(a + 7)}]}^{4} = \frac{{(a - 2)}^{4}}{{(a + 7)}^{4}}$

${[\frac{6 x {(4 - x)}^{4}}{2 a {(y - 4)}^{6}}]}^{2} = \frac{6^{2} x^{2} {(4 - x)}^{8}}{2^{2} a^{2} {(y - 4)}^{12}} = \frac{36 x^{2} {(4 - x)}^{8}}{4 a^{2} {(y - 4)}^{12}} = \frac{9 x^{2} {(4 - x)}^{8}}{a^{2} {(y - 4)}^{12}}$

$\begin{array}{l} {(\frac{a^{3} b^{5}}{a^{2} b})}^{3} & = & {(a^{3 - 2} b^{5 - 1})}^{3} & We can simplify within the parentheses . We \\ have a rule that tells us to proceed this way . \\ = & {(a b^{4})}^{3} \\ = & a^{3} b^{12} \\ {(\frac{a^{3} b^{5}}{a^{2} b})}^{3} & = & \frac{a^{9} b^{15}}{a^{6} b^{3}} = a^{9 - 6} b^{15 - 3} = a^{3} b^{12} & We could have actually used the power rule for \\ quotients first . Distribute the exponent, then \\ simplify using the other rules . \\ It is probably better, for the sake of consistency, \\ to work inside the parentheses first. \end{array}$

${(\frac{a^{r} b^{s}}{c^{t}})}^{w} = \frac{a^{r w} b^{s w}}{c^{t w}}$

Practice set c

Make use of the power rule for quotients, the power rule for products, the power rule for powers, or a combination of these rules to simplify each expression.

${(\frac{a}{c})}^{5}$

$\frac{a^{5}}{c^{5}}$

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

$\frac{8 x^{3}}{27 y^{3}}$

${(\frac{x^{2} y^{4} z^{7}}{a^{5} b})}^{9}$

$\frac{x^{18} y^{36} z^{63}}{a^{45} b^{9}}$

${[\frac{2 a^{4} (b - 1)}{3 b^{3} (c + 6)}]}^{4}$

$\frac{16 a^{16} {(b - 1)}^{4}}{81 b^{12} {(c + 6)}^{4}}$

${(\frac{8 a^{3} b^{2} c^{6}}{4 a^{2} b})}^{3}$

$8 a^{3} b^{3} c^{18}$

${[\frac{{(9 + w)}^{2}}{{(3 + w)}^{5}}]}^{10}$

$\frac{{(9 + w)}^{20}}{{(3 + w)}^{50}}$

${[\frac{5 x^{4} (y + 1)}{5 x^{4} (y + 1)}]}^{6}$

$1, if x^{4} (y + 1) \neq 0$

${(\frac{16 x^{3} v^{4} c^{7}}{12 x^{2} v c^{6}})}^{0}$

$1, if x^{2} v c^{6} \neq 0$

Exercises

Use the power rules for exponents to simplify the following problems. Assume that all bases are nonzero and that all variable exponents are natural numbers.

${(a c)}^{5}$

$a^{5} c^{5}$

${(n m)}^{7}$

${(2 a)}^{3}$

$8 a^{3}$

${(2 a)}^{5}$

${(3 x y)}^{4}$

$81 x^{4} y^{4}$

${(2 x y)}^{5}$

${(3 a b)}^{4}$

$81 a^{4} b^{4}$

${(6 m n)}^{2}$

${(7 y^{3})}^{2}$

$49 y^{6}$

${(3 m^{3})}^{4}$

${(5 x^{6})}^{3}$

$125 x^{18}$

${(5 x^{2})}^{3}$

${(10 a^{2} b)}^{2}$

$100 a^{4} b^{2}$

${(8 x^{2} y^{3})}^{2}$

${(x^{2} y^{3} z^{5})}^{4}$

$x^{8} y^{12} z^{20}$

${(2 a^{5} b^{11})}^{0}$

${(x^{3} y^{2} z^{4})}^{5}$

$x^{15} y^{10} z^{20}$

${(m^{6} n^{2} p^{5})}^{5}$

${(a^{4} b^{7} c^{6} d^{8})}^{8}$

$a^{32} b^{56} c^{48} d^{64}$

${(x^{2} y^{3} z^{9} w^{7})}^{3}$

${(9 x y^{3})}^{0}$

${(\frac{1}{2} f^{2} r^{6} s^{5})}^{4}$

${(\frac{1}{8} c^{10} d^{8} e^{4} f^{9})}^{2}$

$\frac{1}{64} c^{20} d^{16} e^{8} f^{18}$

${(\frac{3}{5} a^{3} b^{5} c^{10})}^{3}$

${(x y)}^{4} (x^{2} y^{4})$

$x^{6} y^{8}$

${(2 a^{2})}^{4} {(3 a^{5})}^{2}$

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

$a^{18} b^{21}$

${(h^{3} k^{5})}^{2} {(h^{2} k^{4})}^{3}$

${(x^{4} y^{3} z)}^{4} {(x^{5} y z^{2})}^{2}$

$x^{26} y^{14} z^{8}$

${(a b^{3} c^{2})}^{5} {(a^{2} b^{2} c)}^{2}$

$\frac{{(6 a^{2} b^{8})}^{2}}{{(3 a b^{5})}^{2}}$

$4 a^{2} b^{6}$

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

$\frac{{(x^{6} y^{5})}^{3}}{{(x^{2} y^{3})}^{5}}$

$x^{8}$

$\frac{{(a^{8} b^{10})}^{3}}{{(a^{7} b^{5})}^{3}}$

$\frac{{(m^{5} n^{6} p^{4})}^{4}}{{(m^{4} n^{5} p)}^{4}}$

$m^{4} n^{4} p^{12}$

$\frac{{(x^{8} y^{3} z^{2})}^{5}}{{(x^{6} y z)}^{6}}$

$\frac{{(10 x^{4} y^{5} z^{11})}^{3}}{{(x y^{2})}^{4}}$

$1000 x^{8} y^{7} z^{33}$

$\frac{(9 a^{4} b^{5}) (2 b^{2} c)}{(3 a^{3} b) (6 b c)}$

$\frac{{(2 x^{3} y^{3})}^{4} {(5 x^{6} y^{8})}^{2}}{{(4 x^{5} y^{3})}^{2}}$

$25 x^{14} y^{22}$

${(\frac{3 x}{5 y})}^{2}$

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

$\frac{27 a^{3} b^{3}}{64 x^{3} y^{3}}$

${(\frac{x^{2} y^{2}}{2 z^{3}})}^{5}$

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

$\frac{27 a^{6} b^{9}}{c^{12}}$

${(\frac{4^{2} a^{3} b^{7}}{b^{5} c^{4}})}^{2}$

${[\frac{x^{2} {(y - 1)}^{3}}{(x + 6)}]}^{4}$

$\frac{x^{8} {(y - 1)}^{12}}{{(x + 6)}^{4}}$

${(x^{n} t^{2 m})}^{4}$

$\frac{{(x^{n + 2})}^{3}}{x^{2 n}}$

$x^{n + 6}$

${(x y)}^{△}$

$\frac{4^{3} a^{Δ} a^{□}}{4 a^{\nabla}}$

${(\frac{4 x^{Δ}}{2 y^{\nabla}})}^{□}$

'Two to the power square, x to the power the product of triangle and square' over 'y to the power the product of delta and square'.

'Sixteen a cube b to the power star' over 'five a to the power triangle b to the power delta'. The whole to the zeroth power.

Exercises for review

( [link] ) Is there a smallest integer? If so, what is it?

( [link] ) Use the distributive property to expand $5 a (2 x + 8)$ .

( [link] ) Find the value of $\frac{{(5 - 3)}^{2} + {(5 + 4)}^{3} + 2}{4^{2} - 2 \cdot 5 - 1}$ .

147

( [link] ) Assuming the bases are not zero, find the value of $(4 a^{2} b^{3}) (5 a b^{4})$ .

( [link] ) Assuming the bases are not zero, find the value of $\frac{36 x^{10} y^{8} z^{3} w^{0}}{9 x^{5} y^{2} z}$ .

$4 x^{5} y^{6} z^{2}$

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Source: OpenStax, Algebra i for the community college. OpenStax CNX. Dec 19, 2014 Download for free at http://legacy.cnx.org/content/col11598/1.3

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