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, bÂ° = 1.

There are six rules that govern the behavior of exponents:

Rule 1:
xa â€¢ xb = xa + b

Example:
23 â€¢ 22 = 23+2 = 25 = 32.
Caution:
xa + xb â‰  x a+ b

Rule 2:
( xa )b = xab
Example:
(2 3)2= 23â€¢2 = 26 = 64

Rule 3:
(x y)a = x a â€¢ ya
Example:
(2y)3 = 23 â€¢ y3 = 8y3

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 ( x a )b = xab to the expression $\frac{x{\left({x}^{5}\right)}^{2}}{{x}^{4}}$ :
$\frac{xÂ·{x}^{5Â·2}}{{x}^{4}}$ = $\frac{xÂ·{x}^{10}}{{x}^{4}}$

Next, apply the rule xa â€¢ xb = xa +b :

$\frac{xÂ·{x}^{10}}{{x}^{4}}$ = $\frac{{x}^{11}}{{x}^{4}}$

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

$\frac{{x}^{11}}{{x}^{4}}$ = x11-4 = x7

Note: Typically, there are many ways of solving these types of problems. For this example, we could have begun with Rule 5, $\frac{{x}^{a}}{{x}^{b}}$ = $\frac{1}{{x}^{b-a}}$ :
$\frac{x{\left({x}^{5}\right)}^{2}}{{x}^{4}}$ = $\frac{x{\left({x}^{5}\right)}^{2}}{{x}^{4-1}}$ = $\frac{x{\left({x}^{5}\right)}^{2}}{{x}^{3}}$

Then apply Rule 2,(xa)b = xab:

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

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

$\frac{{x}^{10}}{{x}^{3}}$ = x7

Example 2:

 Column A Column B $\frac{3Â·3Â·3Â·3}{9Â·9Â·9Â·9}$ ${\left(\frac{1}{3}\right)}^{4}$
Canceling the common factor 3 in Column A yields $\frac{1Â·1Â·1Â·1}{3Â·3Â·3Â·3}$, 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
24â€¢32

First, factor Column A:
$\frac{{\left(2Â·3\right)}^{4}}{{3}^{2}}$

Next, apply the rule (x y)a = x a â€¢ ya :
$\frac{{2}^{4}Â·{3}^{4}}{{3}^{2}}$

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

24â€¢32

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