**Exponents**

Exponents afford a convenient way of expressing long products of the same number. The expression ${b}^{n}$ is called a power and it stands for *b Ã— b Ã— b* Ã— â€¢ â€¢ â€¢ Ã— *b*, where there are *n* factors of *b. b* is called the base, and *n* is called the exponent. By definition, *b*Â° = 1.

There are **six ****rules** that govern the behavior of exponents:

**Rule 1:**

*x ^{a}* â€¢

*x*=

^{b}*x*

^{a + b}

**Example:**

2

^{3}â€¢ 2

^{2}= 2

^{3+2}= 2

^{5}= 32.

**Caution:**

*x*+

^{a}*x*â‰

^{b}*x*

^{a+ b}**Rule 2:**

(

*x*)

^{a}*=*

^{b}*x*

^{ab}**Example:**

(2

^{3})

^{2}= 2

^{3â€¢2}= 2

^{6}= 64

**Rule 3:**

(

*x y*)

*=*

^{a}*x*

^{a}â€¢ y^{a}**Example:**

(2

*y*)

^{3}= 2

^{3}â€¢

*y*

^{3}= 8

*y*

^{3}

**Rule 4:**

x y a = x a y a

**Example:**

${\left(\frac{x}{3}\right)}^{2}$ = $\frac{{x}^{2}}{{3}^{2}}$ = $\frac{{x}^{2}}{9}$

**Rule 5:**

x a x b = x a - b , if

*a*>

*b*

**Example:**

$\frac{{2}^{6}}{{2}^{3}}$ = ${2}^{6-3}$ = ${2}^{3}$ = 8

$\frac{{x}^{a}}{{x}^{b}}$ = $\frac{1}{{x}^{b-a}}$, if b>a.

**Example:**

$\frac{{2}^{3}}{{2}^{6}}$ = $\frac{1}{{2}^{6-3}}$ = $\frac{1}{{2}^{3}}$ = $\frac{1}{8}$

**Rule 6:**

x - a = 1 x a

**Example:**

*z*

^{-3}= $\frac{1}{{z}^{3}}$ Caution, a negative exponent does not make the number negative; it merely indicates that the base should be reciprocated. For example, 3

^{-2}â‰ - $\frac{1}{{3}^{2}}$

*or*- $\frac{1}{9}$.

Problems involving these six rules are common on the GRE, and they are often listed as hard problems. However, the process of solving these problems is quite mechanical: simply apply the six rules until they can no longer be applied.

**Example 1:**

If x â‰ 0, $\frac{x{\left({x}^{5}\right)}^{2}}{{x}^{4}}$ = ?

(A)* x *^{5}

(B) *x*^{ 6}

(C) *x*^{ 7}

(D) *x *^{8}

(E) *x*^{ 9}

First, apply the rule (

$\frac{x\xc2\xb7{x}^{5\xc2\xb72}}{{x}^{4}}$ = $\frac{x\xc2\xb7{x}^{10}}{{x}^{4}}$

Next, apply the rule

$\frac{x\xc2\xb7{x}^{10}}{{x}^{4}}$ = $\frac{{x}^{11}}{{x}^{4}}$

Finally, apply the rule $\frac{{x}^{a}}{{x}^{b}}$ =

$\frac{{x}^{11}}{{x}^{4}}$ =

The answer is (C).

*x**)*^{a}*=*^{b}*x*to the expression $\frac{x{\left({x}^{5}\right)}^{2}}{{x}^{4}}$ :^{ab}$\frac{x\xc2\xb7{x}^{5\xc2\xb72}}{{x}^{4}}$ = $\frac{x\xc2\xb7{x}^{10}}{{x}^{4}}$

Next, apply the rule

*x*â€¢^{a}*x*=^{b}*x*:^{a +b}$\frac{x\xc2\xb7{x}^{10}}{{x}^{4}}$ = $\frac{{x}^{11}}{{x}^{4}}$

Finally, apply the rule $\frac{{x}^{a}}{{x}^{b}}$ =

*x*:^{a-b}$\frac{{x}^{11}}{{x}^{4}}$ =

*x*^{11-4}=*x*^{7}The answer is (C).

**Note:**Typically, there are many ways of solving these types of problems. For this example, we could have begun with Rule 5, $\frac{{x}^{a}}{{x}^{b}}$ = $\frac{1}{{x}^{b-a}}$ :

$\frac{x{\left({x}^{5}\right)}^{2}}{{x}^{4}}$ = $\frac{x{\left({x}^{5}\right)}^{2}}{{x}^{\mathrm{4-1}}}$ = $\frac{x{\left({x}^{5}\right)}^{2}}{{x}^{3}}$

Then apply Rule 2,(x

^{a})

^{b}= x

^{ab}:

$\frac{{\left({x}^{5}\right)}^{2}}{{x}^{3}}$ = $\frac{{x}^{10}}{{x}^{3}}$

Finally, apply the other version of Rule 5, $\frac{{x}^{a}}{{x}^{b}}$ =

*x*

^{a - b}:

$\frac{{x}^{10}}{{x}^{3}}$ =

*x*

^{7}

**Example 2:**

Column A |
Column B |

$\frac{3\xc2\xb73\xc2\xb73\xc2\xb73}{9\xc2\xb79\xc2\xb79\xc2\xb79}$ | ${\left(\frac{1}{3}\right)}^{4}$ |

Canceling the common factor 3 in Column A yields $\frac{1\xc2\xb71\xc2\xb71\xc2\xb71}{3\xc2\xb73\xc2\xb73\xc2\xb73}$, or $\frac{1}{3}$â€¢$\frac{1}{3}$â€¢$\frac{1}{3}$â€¢$\frac{1}{3}$ . Now, by the definition of a power, $\frac{1}{3}$â€¢$\frac{1}{3}$â€¢$\frac{1}{3}$â€¢$\frac{1}{3}$ = ${\left(\frac{1}{3}\right)}^{4}$ Hence, the columns are equal and the answer is (C).

**Example 3:**

**Column A**

$\frac{{6}^{4}}{{3}^{2}}$

**Column B**

2^{4}â€¢3^{2}

First, factor Column A:

$\frac{{\left(2\xc2\xb73\right)}^{4}}{{3}^{2}}$

Next, apply the rule (x y)

$\frac{{2}^{4}\xc2\xb7{3}^{4}}{{3}^{2}}$

Finally, apply the rule $\frac{{x}^{a}}{{x}^{b}}$ =

2

Hence, the columns are equal and the answer is (C).

$\frac{{\left(2\xc2\xb73\right)}^{4}}{{3}^{2}}$

Next, apply the rule (x y)

^{a}= x^{a}â€¢ y^{a}:$\frac{{2}^{4}\xc2\xb7{3}^{4}}{{3}^{2}}$

Finally, apply the rule $\frac{{x}^{a}}{{x}^{b}}$ =

*x*^{a - b}:2

^{4}â€¢3^{2}Hence, the columns are equal and the answer is (C).