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}. Notice that the terms share like bases.
So, in our example above, \begin{aligned}a^{5}\times a^{3} &= a^{5+3}\\&=a^{8}\end{aligned}
We can avoid having to write each expression in expanded form by using the multiplication law.
For any base number a, and any numbers m and n as powers, a^{m} \times a^{n}=a^{m+n}.
That is, when multiplying terms with a common base:
Keep the same base
Find the sum of the exponents
Simplify the following, giving your answer in index form: x^{4} \times 10x^{3}
For any base number a, and any numbers m and n as powers. a^{m} \times a^{n}=a^{m+n}
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}, we could write it as:
So, in our example above,
\begin{aligned} a^{6} \div a^{2} &= a^{6-2} \\ &= a^{4} \end{aligned}
We can avoid having to write each expression in expanded form by using the division law (which is also known as the quotient law).
\dfrac{a^{m}}{a^{n}}=a^{m-n}, where a is any base number,
That is, when dividing terms with a common base:
Keep the same base
Find the difference in the power.
We can also write the division law in the form: a^{m}\div a^{n} = a^{m-n}.
Simplify the following, giving your answer in index form: \dfrac{x^{11}}{5x^{8}}
The division law:
\dfrac{a^{m}}{a^{n}}=a^{m-n}\,\,\text{ or }\,\,a^{m}\div a^{n} = a^{m-n} where a is any base number.
For any base a, and any numbers m and n as powers, \left(a^{m}\right)^{n} = a^{m \times 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}:
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\times 3}=2x^{6}, which is not correct.
Consider the expression in expanded form:
\begin{aligned} \left(2x^{2}\right)^{3} &= 2x^{2} \times 2x^{2} \times 2x^{2} \\ &= \left(2\times2\times2\right)\times \left(x^{2} \times x^{2} \times x^{2}\right) \\ &= 2^{3} \times \left(x^{2}\right)^{3} \\ &=8x^{6}\end{aligned}
You can see that not only is x^{2} multiplied 3 times, \left(x^{2}\right)^{3}, but 2 is also multiplied 3 times, 2^{3}.
So we need to raise 2 to the power of 3 as well as x^{2} to the power of 3.\left(2x^{2}\right)^{3} = 8x^{6}
Beware of the signs when raising negative numbers to a power, as a negative number raised to an odd power has a negative result. For example: \left(-2x^{2}\right)^{3}=-8x^{6}
Express the following in simplified index form: \left(w^{7}\right)^{4}
For any base a, and any numbers m and n as powers, \left(a^{m}\right)^{n} = a^{m \times n}
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 latex 0? For example:\begin{aligned} x^{5} \div x^{5} &= x^{5-5} \\ &= x^{0} \end {aligned}
To think about what value we can assign to the term x^{0}, let's write this division problem as the fraction \dfrac{x^{5}}{x^{5}}. Since the numerator and denominator are the same, the fraction simplifies to 1. Notice that this will also be the case with \dfrac{k^{20}}{k^{20}} 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}, and the result we arrive at by simplifying fractions is 1. This must mean that x^{0}=1.
There is nothing special about x, so we can extend this observation to any base. This result is summarised by the zero power law.
For any base a, a^{0}=1.
This says that taking the zeroth power of any number will always result in 1.
Simplify the following expression: 18a^{0}
For any base a, a^{0}=1.
Taking the zeroth power of any number or term will always result in 1.
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.
The multiplication law: a^{m}\times a^{n}=a^{m+n}
The division law: a^{m}\div a^{n}=a^{m-n}, a^{n} \neq 0
The power of a power law: \left(a^{m}\right)^{n} = a^{m \times n}
The zero power law: a^{0}=1, a\neq0
We may also come across expressions of the form \left(a^{m}\times b^{n}\right)^{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}
Simplify the following expression: \left(x^{5}y^{4}\right)^{4}.
Simplify the following expression: \dfrac{\left(2x^{2}y^{0}\right)^{4}}{x^{5}}.
Some expressions require a combination of the power laws to simplify them. It's important to remember the order of operations when simiplifying these expressions.
The multiplication law: a^{m}\times a^{n}=a^{m+n}
The division law: a^{m}\div a^{n}=a^{m-n}, a^{n} \neq 0
The power of a power law: \left(a^{m}\right)^{n} = a^{m \times n}
The zero power law: a^{0}=1, a\neq0