3.09 Variables and exponents

Multiplying terms with exponents

When multiplying a number by itself repeatedly, we are able to use exponent 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 exponent 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,

The exponent law of multiplication

We can avoid having to write each expression in expanded form by using the multiplication law.

Practice question

Question 1

Dividing terms with exponents

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,

The exponent law of division

We can avoid having to write each expression in expanded form by using the division law (which is also known as the quotient law).

Practice question

Question 2

The power of a power law

Since "powers of powers" involve expressions with brackets, it's important to remember that everything inside the brackets is raised to the outside power.

Exploration

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

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

Practice question

Question 3

The zero exponent

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,

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 exponent 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.

Practice question

Question 4

Mixed operations

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.

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:

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 exponent 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+2−4

And so the final simplified form is $4p^{5+2-4}=4p^3$4p5+2−4=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:

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 5

Question 6