An expression of the form $b^n$bn is called an exponential or power expression. This expression is read "$b$b raised to the $n^{th}$nth power". $b$b is the base and $n$n is the index or exponent.
$a^m\times a^n=a^{m+n}$am×an=am+n
$\frac{a^m}{a^n}=a^{mn}$aman=am−n
$\left(a^m\right)^n=a^{m\times n}$(am)n=am×n
$\left(ab\right)^m=a^mb^m$(ab)m=ambm
$\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}$(ab)m=ambm
$a^0=1,\text{for }a\ne0$a0=1,for a≠0
Let's look at some examples that illustrate how these identities are consistent and how to apply them.
Consider the expression $a^5a^3$a5a3. Notice that the terms share like bases. Let's think about what this would look like if we expanded the expression:
We can see that there are eight $a$a's being multiplied together, and notice that $8$8 is the sum of the powers in the original expression.
Simplify the expression $8^2\times8^9\div8^3$82×89÷83.
Think: This expression contains three terms and two operations. We can perform one operation on two of the terms, then performing the remaining operation on the resulting expression.
Do: Let's begin with the multiplication, then do the division.
$8^2\times8^9\div8^3$82×89÷83  $=$=  $8^{2+9}\div8^3$82+9÷83 
$=$=  $8^{11}\div8^3$811÷83  
$=$=  $8^{113}$811−3  
$=$=  $8^8$88 
Reflect: We combined the first two terms using the multiplication law. Once we had evaluated the new power we then used the division law to complete the simplification. Alternatively, we could have started by simplifying the last two terms using the division law, then completing the process with the multiplication law, as follows.
$8^2\times8^9\div8^3$82×89÷83  $=$=  $8^2\times8^{93}$82×89−3 
$=$=  $8^2\times8^6$82×86  
$=$=  $8^{2+6}$82+6  
$=$=  $8^8$88 
So by the two approaches we get the same answer. This can be useful when one approach clearly involves less work than the other.
Simplify the expression $\frac{\left(20^3\right)^5}{\left(20^0\right)^7}$(203)5(200)7 using exponent laws.
Think: Right away we can see that simplifying this expression will involve the power of a power law and the zero power law. Once we complete the simplification we need to make sure our answer has a positive power.
Do:
$\frac{\left(20^3\right)^5}{\left(20^0\right)^7}$(203)5(200)7  $=$=  $\frac{\left(20^3\right)^5}{1^7}$(203)517 
Using the zero power law 
$=$=  $\frac{\left(20^3\right)^5}{1}$(203)51 
Simplify $1^7$17 

$=$=  $\left(20^3\right)^5$(203)5 
Simplify the quotient 

$=$=  $20^{3\times5}$203×5 
Using the power of a power law 

$=$=  $20^{15}$2015 
Evaluate the product in the exponent 
Reflect: We could have used the power of a power law for both the numerator and the denominator from the beginning, which would give the step $\frac{20^{15}}{20^0}$2015200. And from there we could use the zero power law to simplify further. In this way we can be creative with the order that we use each law.
The same rules apply when we are dealing with negative bases, we just need to take care if we are asked to evaluate. We know that the product of two negative numbers is positive, and the product of a positive and a negative number is negative. This means we need to be extra careful when evaluating powers of negative bases.
A negative base raised to an even power will evaluate to a positive answer.
A negative base raised to a odd power will evaluate to a negative answer.
When raising a fractional base, we apply the power to both the numerator and the denominator.
Let's consider a simple example like $\left(\frac{1}{2}\right)^2$(12)2. This expands to $\frac{1}{2}\times\frac{1}{2}=\frac{1\times1}{2\times2}$12×12=1×12×2, which evaluates to $\frac{1}{4}$14 or $\frac{1^2}{2^2}$1222.
Similarly, a slightly harder expression like $\left(\frac{2}{3}\right)^3$(23)3 expands to $\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}$23×23×23 giving us $\frac{2\times2\times2}{3\times3\times3}$2×2×23×3×3. So we can see that $\left(\frac{2}{3}\right)^3=\frac{2^3}{3^3}$(23)3=2333.
This can be generalised to give us the following rule:
For any base number of the form $\frac{a}{b}$ab, and any number $n$n as a power,
$\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$(ab)n=anbn
Fill in the blank to make the equation true.
$11^{11}\times11^{\editable{}}=11^{19}$1111×11=1119
Simplify the following using exponent laws, giving your answer in simplest exponential form: $\left(\frac{29}{41}\right)^7$(2941)7
Using exponent laws, evaluate $\left(4\right)^{11}\div\left(4\right)^7$(−4)11÷(−4)7.
Simplify $\frac{\left(17^5\right)^8}{17^{32}}$(175)81732, giving your answer in exponential form.
Analyse, through the use of patterning, the relationships between the exponents of powers and the operations with powers, and use these relationships to simplify numeric and algebraic expressions.