UK Secondary (7-11)
Simplify expressions using multiple index laws with variable bases
Lesson

We have come across a number of index laws up to this point: the multiplication law, the division law, the power of a power law and the zero index law.

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.

Summary of laws
The multiplication law: $a^m\times a^n=a^{m+n}$am×an=am+n
The division law: $a^m\div a^n=\frac{a^m}{a^n}$am÷​an=aman$=$=$a^{m-n}$amn, $a^n\ne0$an0
The power of a power law: $\left(a^m\right)^n=a^{m\times n}$(am)n=am×n
The zero power law: $a^0=1$a0=1, $a\ne0$a0

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.

#### Worked example

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) $=$= $\frac{24p^7}{6p^4}$24p76p4​ (Rewrite as a fraction) $=$= $\frac{4p^7}{p^4}$4p7p4​ (Perform numeric division first) $=$= $4p^{7-4}$4p7−4 (By using division law) $=$= $4p^3$4p3

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:

$p^5\times p^2\div p^4=p^{5+2-4}$p5×p2÷​p4=p5+24

And so the final simplified form is $4p^{5+2-4}=4p^3$4p5+24=4p3, as expected.

#### Worked example

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 $=$= $26$26

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.

#### Practice questions

##### Question 1

Simplify the following, giving your answer with a positive index: $m^9\div m^5\times m^4$m9÷​m5×m4

##### Question 2

Simplify the following expression: $\left(4y^3\right)^2\times\left(4y^4\right)^3$(4y3)2×(4y4)3.

##### Question 3

Simplify the following expression: $\left(x^5y^4\right)^4$(x5y4)4.