From this definition, we can deduce some basic rules that exponentiation must follow angstrom well as some hand extra cases that follow from the rules. In the process, we ‘ll define exponentials $ x^a $ for exponents $ a $ that are n’t positive integers .

The rules and limited cases are summarized in the stick to board. Below, we give details for each one .

#### The rules

##### Product of exponentials with same base

If we take the merchandise of two exponentials with the same base, we just add the exponents : \begin { gather } x^ax^b = x^ { a+b }. \label { merchandise } \end { gather }

To see this rule, we merely expand out what the exponents mean. Let ‘s start out with a couple childlike examples. \begin { align * } 3^4 3^2 & = ( 3 \times 3 \times 3 \times 3 ) \times ( 3 \times 3 ) \\ & = 3 \times 3 \times 3 \times 3 \times 3 \times 3\\ & = 3^6 \end { align * } \begin { align * } y^2 y^3 & = ( yttrium \times y ) \times ( y \times y \times yttrium ) \\ & = yttrium \times y \times yttrium \times yttrium \times y\\ & = y^5 \end { align * }

The general font works the like way. We barely need to keep track of the number of factors we have. \begin { align * } x^ax^b & = \underbrace { adam \times \cdots \times x } _ { a \text { times } } \times \underbrace { ten \times \cdots \times x } _ { boron \text { times } } \\ [ 0.2cm ] & = \underbrace { adam \times \cdots \times x } _ { a+b \text { times } } \\ [ 0.2cm ] & =x^ { a+b } \end { align * }

##### Quotient of exponentials with same base

If we take the quotient of two exponentials with the lapp basis, we plainly subtract the exponents : \begin { gain } \frac { x^a } { x^b } = x^ { a-b } \label { quotient } \end { gather }

$ \cancel { } $

This principle results from canceling coarse factors in the numerator and denominator. For example : \begin { align * } \frac { y^5 } { y^3 } & = \frac { yttrium \times yttrium \times y \times yttrium \times yttrium } { y \times y \times y } \\ & = \frac { ( y \times y ) \times \cancel { ( y \times yttrium \times y ) } } { \cancel { yttrium \times y \times y } } \\ & = y \times y = y^2. \end { align * } To show this in general, we look at two different cases. If we imagine that $ a > b $, then this rule follows from canceling the park $ bel $ factors of $ x $ that occur in both the numerator and denominator. We are left with precisely $ b-a $ factors of $ ten $ in the numerator. \begin { align * } \frac { x^a } { x^b } & = \frac { \quad \overbrace { x \times \cdots \times x } ^ { a \text { times } } \quad } { \underbrace { adam \times \cdots \times x } _ { bacillus \text { times } } } \\ [ 0.2cm ] & = \frac { \quad \overbrace { ten \times \cdots \times x } ^ { a-b \text { times } } \times\overbrace { \cancel { adam \times \cdots \times x } } ^ { boron \text { times } } \quad } { \underbrace { \cancel { adam \times \cdots \times x } } _ { bacillus \text { times } } } \\ [ 0.2cm ] & = \underbrace { adam \times \cdots \times x } _ { a-b \text { times } } \\ [ 0.2cm ] & =x^ { a-b } \end { align * }

If $ a < b $, then what happens ? We cancel all the $ x $ 's from the numerator and are left with $ b-a $ of them in the denominator. \begin { align * } \frac { x^a } { x^b } & = \frac { \quad \overbrace { adam \times \cdots \times x } ^ { a \text { times } } \quad } { \underbrace { adam \times \cdots \times x } _ { bel \text { times } } } \\ [ 0.2cm ] & = \frac { \quad \overbrace { \cancel { ten \times \cdots \times x } } ^ { a \text { times } } \quad } { \underbrace { x \times \cdots \times x } _ { b-a \text { times } } \times \underbrace { \cancel { x \times \cdots \times x } } _ { a \text { times } } } \\ [ 0.2cm ] & = \frac { 1 } { \underbrace { ten \times \cdots \times x } _ { b-a \text { times } } } \\ [ 0.2cm ] \end { align * } To make the above principle work for this character, we must define a negative exponent to mean a power in the denominator. If $ n $ is a convinced integer, we define \begin { gather } x^ { -n } = \frac { 1 } { \underbrace { adam \times adam \times \cdots \times x } _ { north \text { times } } }. \end { gather } then the govern for the quotient of exponentials works even if $ a < b $ : \begin { align * } \frac { x^a } { x^b } & = \frac { \quad \underbrace { adam \times \cdots \times x } _ { a \text { times } } \quad } { \underbrace { adam \times \cdots \times x } _ { b \text { times } } } \\ [ 0.2cm ] & = \frac { 1 } { \underbrace { ten \times \cdots \times x } _ { b-a \text { times } } } \\ [ 0.2cm ] & =x^ { a-b }. \end { align * } When $ b > a $, the exponent $ a-b $ is a minus number. Since convention \eqref { quotient } is the same no matter the relationship between $ a $ and $ b-complex vitamin $, we do n’t need to worry about it and can just subtract the exponents .

##### Power of a power

We can raise exponential to another ability, or take a power of a world power. The leave is a single exponential where the office is the product of the original exponents : \begin { gain } ( x^a ) ^b = x^ { ab }. \label { power_power } \end { gather }

We can see this result by writing it as a product where the $ x^a $ is repeated $ barn $ times : \begin { gather * } ( x^a ) ^b = \underbrace { x^a \times x^a \times \cdots \times x^a } _ { b\text { times } }. \end { meet * } Next we apply rule \eqref { product } for the product of exponentials with the lapp base. We use this rule $ bel $ times to conclude that \begin { align * } ( x^a ) ^b & = \underbrace { x^a \times x^a \times \cdots \times x^a } _ { b\text { times } } \\ [ 0.2cm ] & = x^ { \overbrace { a + a + \cdots + a } ^ { b\text { times } } } \\ [ 0.2cm ] & = x^ { bachelor of arts }. \end { align * } In the end footfall, we had to remember that multiplication can be defined as repeat accession .

##### Power of a product

If we take the world power of a product, we can distribute the exponent over the different factors : \begin { gather } ( xy ) ^a = x^ay^a. \label { power_product } \end { gather }

We can show this rule in the lapp way as we show that you can distribute multiplication over addition. One way to show this distributive law for multiplication is is to remember that multiplication is defined as perennial addition : \begin { align * } ( x+y ) a & = \underbrace { ( x + y ) + ( x+y ) + \cdots + ( x+y ) } _ { a\text { times } } \\ [ 0.2cm ] & = \underbrace { x + x + \cdots + x } _ { a\text { times } } +\underbrace { y+ y + \cdots + y } _ { a\text { times } } \\ [ 0.2cm ] \\ & = xa +ya. \end { align * } In the like room, we can show the distributive law for exponentiation : \begin { align * } ( xy ) ^a & = \underbrace { ( xy ) \times ( xy ) \times \cdots \times ( xy ) } _ { a\text { times } } \\ [ 0.2cm ] & = \underbrace { x \times ten \times \cdots \times x } _ { a\text { times } } \times\underbrace { yttrium \times yttrium \times \cdots \times y } _ { a\text { times } } \\ [ 0.2cm ] \\ & = x^a y^a. \end { align * }

Read more : How To Do a Smokey Eye

This principle besides works for quotients \begin { meet * } \left ( \frac { x } { yttrium } \right ) ^a = \frac { x^a } { y^a }, \end { gather * } but it does NOT work for sums. For model, \begin { align * } ( 3+5 ) ^2 = 8^2 = 64, \end { align * } but this is NOT equal to \begin { align * } 3^2+5^2 = 9 + 25 =34. \end { align * }

#### Special cases

The following are special cases that follow from the rules .

##### The power of one

The simplest particular sheath is that raising any phone number to the world power of 1 does n’t do anything : \begin { assemble } x^1=x.\label { power_one } \end { assemble }

##### The power of zero

ampere long as $ x $ is n’t zero, raising it to the world power of zero must be 1 : $ $ x^0=1. $ $ We can see this, for example, from the quotient rule, as $ $ 1 = \frac { x^a } { x^a } = x^ { a-a } =x^0. $ $

The saying $ 0^0 $ is indeterminate. You can see that it must be indeterminate, because you can come up with full reasons for it to be two different values .

first gear, from above, if $ x \ne 0 $, then $ x^0=1 $, no matter how modest $ x $ is. If we just let $ x $ go all the manner to zero ( take the restrict as $ ten $ goes to zero ), then it seems that $ 0^0 $ should be 1 .

On the early hired hand, $ 0^a=0 $ a long as $ a \ne 0 $. Repeated multiplication of $ 0 $ hush gives zero, and we can use the above rules to show $ 0^a $ still is zero, no matter how minor $ a $ is, adenine long as it is nonzero. If fair let $ a $ go all the direction to zero ( take the limit as $ a $ goes to zero ), then it seems like $ 0^0 $ should be 0 .

In other words, if we start with $ x^a $ for non-zero $ adam $ and non-zero $ a $, we ‘ll get a unlike suffice for $ 0^0 $ depending on whether we let $ x $ go to zero first or $ a $ go to zero first. There actually is no direction for deciding on a value for $ 0^0 $, so we are forced to leave it indeterminate. You can check out this applet to visualize this argumentation .

##### The power of negative one

negative one is a special prize for an exponent, because taking a number to the world power of negative one gives its reciprocal : $ $ x^ { -1 } = \frac { 1 } { ten }. $ $

##### The changing sign of exponent

In a exchangeable vein, changing the sign of a exponent gives the reciprocal, so $ $ x^ { -a } = \frac { 1 } { x^a }. $ $

##### Fractional exponents

The office of power rule \eqref { power_power } allows us to define fractional exponents. For exercise, rule \eqref { power_power } tells us that \begin { gather * } 9^ { 1/2 } = ( 3^2 ) ^ { 1/2 } = 3^ { 2 \cdot 1/2 } = 3^1 = 3. \end { gather * } Taking a count to the office of $ \frac { 1 } { 2 } $ undo taking a number to the power of 2 ( or squaring it ). In other words, taking a numeral to the power of $ \frac { 1 } { 2 } $ is the like thing as taking a square beginning : \begin { gather * } x^ { 1/2 } = \sqrt { adam }. \end { gather * }

Since $ ( x^n ) ^ { 1/n } = x^1 = x $, we can generalize the result so that taking a number to the power of $ 1/n $ is the same thing as taking the $ north $ thursday root : \begin { gather } x^ { 1/n } = \sqrt [ nitrogen ] { ten }. \end { gather }

If $ a $ is any rational count, then it can be written as $ a=m/n $. We can define taking a number to the $ a $ thorium might as taking that numeral to the $ megabyte $ thorium power and the $ normality $ thursday settle. We ‘ll assume the base $ ten $ is non-negative so that we do n’t have to worry about doing things like taking the public square root of a minus numeral. then, the ordain does n’t matter and \begin { gather * } x^ { m/n } = \sqrt [ newton ] { x^m } = ( \sqrt [ n ] { x } ) ^m. \end { gather * } If $ a $ is an irrational number, like $ a=\pi $, then this work does n’t precisely work. But, since you can find a rational number equally close as you want to any irrational number, you can approximate $ x^a $ equally well as you like. ( To be accurate, you could define $ x^a $ in terms of a restrict of $ x^b $, where $ b $ are intellectual numbers approaching $ a $. )