When multiplying a number by itself repeatedly, we are able to use index notation to write the expression more simply. Here we are going to look at a rule that allows us simplify products that involve the multiplication of index terms.
Consider the expression $a^5\times a^3$a5×a3. 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 $a$as being multiplied together, and notice that $8$8 is the sum of the powers in the original expression.
So, in our example above,
$a^5\times a^3$a5×a3  $=$=  $a^{5+3}$a5+3 
$=$=  $a^8$a8 
We can avoid having to write each expression in expanded form by using the multiplication law.
For any base number $a$a, and any numbers $m$m and $n$n as powers,
$a^m\times a^n=a^{m+n}$am×an=am+n
That is, when multiplying terms with a common base:
Simplify the following, giving your answer in index form:
$x^6\times8x^3$x6×8x3
The method to divide power terms is similar to the multiplication law, however in this case we subtract the powers from one another, rather than add them. Let's look at an expanded example to see why this is the case.
If we wanted to simplify the expression $a^6\div a^2$a6÷a2, we could write it as:
We can see that there are six $a$as being divided by two $a$as to give a result of four $a$as, and notice that $4$4 is the difference of the powers in the original expression.
So, in our example above,
$a^6\div a^2$a6÷a2  $=$=  $a^{62}$a6−2 
$=$=  $a^4$a4 
We can avoid having to write each expression in expanded form by using the division law (which is also known as the quotient law).
$\frac{a^m}{a^n}=a^{mn}$aman=am−n, where $a$a is any number,
That is, when dividing terms with a common base:
We can also write the division law in the form:
$a^m\div a^n=a^{mn}$am÷an=am−n.
Simplify the following, giving your answer in index form:
$\frac{x^{11}}{5x^8}$x115x8
For any base $a$a, and any numbers $m$m and $n$n as powers,
$\left(a^m\right)^n=a^{m\times n}$(am)n=am×n
That is, when simplifying a term with a power that itself has a power:
Since "powers of powers" involve expressions with brackets, it's important to remember that everything inside the brackets is raised to the outside power.
Let's say we want to simplify the expression $\left(2x^2\right)^3$(2x2)3:
A common mistake is to only apply the outside power to the algebraic term. If we did this, we would get an answer of $2x^{2\times3}=2x^6$2x2×3=2x6, which is not correct.
Consider the expression in expanded form: $\left(2x^2\right)^3=2x^2\times2x^2\times2x^2$(2x2)3=2x2×2x2×2x2
$\left(2x^2\right)^3$(2x2)3  $=$=  $2x^2\times2x^2\times2x^2$2x2×2x2×2x2 
$=$=  $\left(2\times2\times2\right)\times\left(x^2\times x^2\times x^2\right)$(2×2×2)×(x2×x2×x2)  
$=$=  $2^3\times\left(x^2\right)^3$23×(x2)3  
$=$=  $8x^6$8x6 
You can see that not only is $x^2$x2 multiplied $3$3 times, $\left(x^2\right)^3$(x2)3, but $2$2 is also multiplied $3$3 times, $2^3$23.
So we need to raise $2$2 to the power of $3$3 as well as $x^2$x2 to the power of $3$3.
$\left(2x^2\right)^3=8x^6$(2x2)3=8x6
$\left(2x^2\right)^3=8x^6$(−2x2)3=−8x6
Express the following in simplified index form:
$\left(j^3\right)^6$(j3)6
What happens if we want to divide one term by another and when we perform the subtraction and we are left with a power of $0$0? For example,
$x^5\div x^5$x5÷x5  $=$=  $x^{55}$x5−5 
$=$=  $x^0$x0 
To think about what value we can assign to the term $x^0$x0, let's write this division problem as the fraction $\frac{x^5}{x^5}$x5x5. Since the numerator and denominator are the same, the fraction simplifies to $1$1. Notice that this will also be the case with $\frac{k^{20}}{k^{20}}$k20k20 or any expression where we are dividing like bases whose powers are the same.
So the result we arrive at by using index laws is $x^0$x0, and the result we arrive at by simplifying fractions is $1$1. This must mean that $x^0=1$x0=1.
There is nothing special about $x$x, so we can extend this observation to any base. This result is summarised by the zero power law.
For any base $a$a,
$a^0=1$a0=1
This says that taking the zeroth power of any number will always result in $1$1.
Simplify the following expression:
$18a^0$18a0.
Now we are going to look at questions that can be solved by using a combination of these rules. It's important to remember the order of operations when solving such questions.
We may also come across expressions of the form $\left(a^m\times b^n\right)^p$(am×bn)p, and we can use a combination of the multiplication law and the power of a power law to see that
$\left(a^m\times b^n\right)^p=a^{m\times p}\times b^{n\times p}$(am×bn)p=am×p×bn×p.
Simplify $3p^5\times8p^2\div\left(6p^4\right)$3p5×8p2÷(6p4).
Think: Each important term in the expression has a coefficient, a base of $p$p, and some power. We also have a multiplication operation and a division operation. Considering the order of operations, we can simplify this expression by moving from left to right, using the multiplication law then the division law. Notice that $a\div\left(bc\right)$a÷(bc) is to be interpreted as $a\div\left(b\times c\right)$a÷(b×c), and not as $a\div b\times c=a\times c\div b$a÷b×c=a×c÷b.
Do:
$3p^5\times8p^2\div\left(6p^4\right)$3p5×8p2÷(6p4)  $=$=  $3\times8p^5\times p^2\div\left(6p^4\right)$3×8p5×p2÷(6p4)  
$=$=  $24p^5\times p^2\div\left(6p^4\right)$24p5×p2÷(6p4) 
Perform numeric multiplication first 

$=$=  $24p^{5+2}\div\left(6p^4\right)$24p5+2÷(6p4) 
By using the multiplication law 

$=$=  $24p^7\div\left(6p^4\right)$24p7÷(6p4) 
Evaluate the addition in the power 

$=$=  $\frac{24p^7}{6p^4}$24p76p4 
Rewrite as a fraction 

$=$=  $\frac{4p^7}{p^4}$4p7p4 
Perform numeric division first 

$=$=  $4p^{74}$4p7−4 
By using division law 

$=$=  $4p^3$4p3 
Simplify the difference in the power 
Reflect: In this example we were able to rearrange some of the terms in the expression because multiplication is commutative (the order doesn't matter). Depending on the values of the coefficients and the powers, some rearrangements might be more convenient for simplification than others. We can be creative with the index laws to see how many different ways we can simplify the same expression.
Another approach we could have used to solve the above expression is to first group the numeric factors and the variable factors separately, like so:
$3p^5\times8p^2\div6p^4=\left(3\times8\div6\right)\times\left(p^5\times p^2\div p^4\right)$3p5×8p2÷6p4=(3×8÷6)×(p5×p2÷p4)
Now the numeric part of the expression simplifies to $3\times8\div6=4$3×8÷6=4, and the variable part can be simplified by combining the multiplication law and the division law:
$p^5\times p^2\div p^4=p^{5+24}$p5×p2÷p4=p5+2−4
And so the final simplified form is $4p^{5+24}=4p^3$4p5+2−4=4p3, as expected.
Evaluate the expression $\left(13m^4\right)^0+25\left(n^0\right)^8$(13m4)0+25(n0)8.
Think: This expression features powers of powers and zero powers. For each term in the sum, what is the base and what is the power?
Do:
$\left(13m^4\right)^0+25\left(n^0\right)^8$(13m4)0+25(n0)8  $=$=  $1+25\times1^8$1+25×18 
By using the zero power law 
$=$=  $1+25\times1$1+25×1 
$1$1 raised to any power is $1$1 

$=$=  $1+25$1+25 
Evaluate the product 

$=$=  $26$26 
Evaluate the addition 
Reflect: The term $\left(13m^4\right)^0$(13m4)0 has a base of $13m^4$13m4 and a power of $0$0. Regardless how complex the base may be, if the power is $0$0 then we know the term will evaluate to $1$1.
Simplify the following expression:
$\left(x^6y^3\right)^4$(x6y3)4.
Simplify the following expression:
$\frac{\left(2x^2y^0\right)^4}{x^5}$(2x2y0)4x5.