Laws of indices
What are indices?
Indices (sometimes known as powers) are used to display how many times a number has been multiplied by itself. They can also be used to represent roots and fractions, such as the square root. The laws of indices make it possible to alter expressions involving powers more quickly than if they were written out whole.
This guide will introduce six properties of indices: multiplication, division, brackets, powers of zero, negative indices and fractional indices.
Indices are used in many areas of science as a way of representing growth. Economics, biology, demographics and many more fields rely on this sort of growth; chemicals, radioactive waste, and sound are all examples of decay.
The index of a variable (or a constant) is a value that represents how many times a variable is multiplied by itself. Indices are also known as powers or exponents. It is represented in the form
\[ a^r =\underbrace{a\cdot a \cdot a \cdot \ldots \cdot a}_{r\textsf{ times}}\]
Here, \(a\) is the base and \(r\) is the index. Indices are a shorthand way of writing multiplications of the same number.
As \(a\) multiplied by itself once is \(a\), it follows that anything to the power of one is equal to itself \(a^1 = a\) and \(1\) to the power of anything is \(1\); so \(1^r=1\) for all real numbers \(r\).
Suppose you have \[2·2·2·2·2\] you can write this as ‘\(2\) to the power of \(5\)’ \[2^5\] So \[2^5 = 2·2·2·2·2 = 32\] In this example, \(2\) is the base and \(5\) is the index.
Laws of indices
In mathematics, you are not restricted to using positive whole numbers for indices. For example, what do these expressions mean: \(a^{-1}\) or \(a^0\) or \(a^{\tfrac{1}{2}}\)?
To understand these expressions, you can learn rules to help you deal with these strange looking powers.
If the two terms have the same base (in this case \(a\)) and are multiplied together, then their indices \(r\) and \(s\) are added (in this case \(r+s\))
\[a^r·a^s = a^{r+s}\]
For example, suppose you have \(3^2\) and you want to multiply it by \(3^3\). That is \[3^2·3^3 = (3·3) · (3·3·3)\] which is equal to \(3^5\).
Therefore, using Law 1 you can get \[3^2·3^3 = 9·27 = 243 = 3^5 = 3^{2+3}\]
Simplify the following expression: \(x^2y^3· x^3y\).
You can treat \(x\) and \(y\) as two separate terms and you are now simplifying \(x^2·x^3\) and \(y^3·y\). Using Law 1, you can add the indices to get: \(x^{2+3}\) and \(y^{3+1}\). Remembering to multiply the \(x\) and \(y\) terms, you then get \[x^2y^3· x^3y = x^{2+3}y^{3+1} = x^5y^4\]
If the two terms have the same base (in this case \(a\)) and are to be divided, then their indices are subtracted (in this case \(r-s\))
\[\dfrac{a^r}{a^s} = a^{r-s}\] For example suppose you want to divide \(2^5\) by \(2^3\). \[\dfrac{2^5}{2^3} = \dfrac{2·2·2·2·2}{2·2·2}\] You can now cancel the common factors of \(2\). Three of the \(2\)’s at the bottom can be cancelled out, so you are left with \[\dfrac{2·2}{1} = 2^2 = 4\] Therefore, you can get \[\dfrac{2^5}{2^3} = 4 = 2^2 = 2^{5-3}\]
Simplify the following expression: \(\dfrac{x^2y^3}{x^5y}\)
You can treat \(\frac{x^2y^3}{x^5y}\) as the product of two fractions: \(\frac{x^2}{x^5}\) and \(\frac{y^3}{y}\). Using Law 2, you can simplify the fractions to get: \[\dfrac{x^2}{x^5} = x^{2-5} = x^{-3}\] and \[\dfrac{y^3}{y} = y^{3-1} = y^2\] Remembering to multiply the fractions, you get \[\dfrac{x^2y^3}{x^5y} = x^{2-5}y^{3-1} = x^{-3}y^2\] You can notice you have a negative index here; more on those later!
To raise a value or variable (in this case \(a\)) presented in index form to another index, multiply the indices together (\(r\) and \(s\)) \[(a^r)^s = a^{r·s}\] For example suppose you had \(3^2\) and you want to raise it all to the power of \(3\); that is \((3^2)^3\), which means \[3^2·3^2·3^2.\] Law 1 tells you to add the indices together. So you can get \[3^6\] But note that \(6\) is \(2·3\). This suggests that Law 3 also works. Therefore, using Law 3 you can get \[(3^2)^3 = 729 = 3^6 = 3^{2·3}\]
Simplify the following expression: \((x^2y^3)^4\)
Using Law 3, you get \[(x^2y^3)^4 = (x^2)^4·(y^3)^4 = x^{2·4}y^{3·4} = x^8y^{12}\]
Any non-zero real number \(a\) to the power of zero is equal to one: \[a^0 = 1\]
\(0^0\) is indeterminate, anything to the power of \(0\) is \(1\), but \(0\) to the power of anything is \(0\), so it is not defined.
Simplify the following expression: \((x^2 + y^5)^0\)
Using Law 4, you can get \[(x^2 + y^5)^0 = 1\]
If the index is a negative value, then it is reciprocal of the positive index raised to the same variable \[a^{-r} = \dfrac{1}{a^r}\]
For example, suppose you want to divide \(2^3\) by \(2^7\) \[\dfrac{2^3}{2^7}=\dfrac{2·2·2}{2·2·2·2·2·2·2}\] Again, you can find the common factors of \(2\). The three \(2\)’s at the top can be cancelled out, so you are left with \[\dfrac{2^3}{2^7}=\dfrac{1}{2·2·2·2}=\dfrac{1}{2^4}\]
Now consider Law 2, you can do the similar calculation by subtracting the indices \[\dfrac{2^3}{2^7}=2^{3-7}=2^{-4}\] You have done the calculation in two different ways. So you get \[\dfrac{2^3}{2^7}=2^{3-7}=2^{-4}=\dfrac{1}{2^4}\]
Simplify the following expression: \((3x+4y)^{-1}\)
Using Law 5, you can get \[(3x+4y)^{-1} = \dfrac{1}{3x+4y}\]
An index in the form \(1/n\) (where \(n\) is a positive whole number) is an \(n\)th root of \(a\). So \[a^{1/n} = \sqrt[n]{a}.\]
For example: suppose you found that \[2^r·2^r=2\] This means that multiplying \(2^r\) by \(2^r\) gives the result \(2\). You know that \(2=2^1\), so consider Law 1, \(2^r·2^r=2^{2r}\). If \(2^r·2^r=2\) then you can get \[2^{2r}=2^1\] from which \(r=1\); so \(r=\frac{1}{2}\).This shows that \[2^{\frac{1}{2}}=\sqrt[2]{2}=\sqrt{2}\]
What about an expression of the form \(a^{r/n}\), where \(r\) is a real number and \(n\) is a positive whole number? Combining Laws 3 and 6 yields the following: \[a^{r/n} = (a^{1/n})^r = (\sqrt[n]{a})^r\] or equivalently \(a^{r/n} = (a^r)^{1/n}\).
- Simplify \(10000^{(\frac{3}{4})}\).
Using Law 7, you can re-express \(10000^{\frac{3}{4}}\) as \((\sqrt[4]{10000})^3\). Since \(10^4=10000\), it follows that \(\sqrt[4]{10000} = 10\). Finally, you can say that \((\sqrt[4]{10000})^3 = 10^3 = 1000\).
- Simplify \(27^{-\frac{2}{3}}\).
Using Law 7, you can write this expression as \(27^{-\frac{2}{3}} = (\sqrt[3]{27})^{-2}\). Since the cube root of \(27\) is \(3\), it follows that \[ (\sqrt[3]{27})^{-2} = 3^{-2}\] which, using Law 5, becomes \(3^{-2} = \frac{1}{3^2} = \frac{1}{9}.\)
The first six laws above focused on the index and when the index changes; and in this case, the base remains constant. The final four laws focus on when the base changes and the index remains constant.
This next law follows from the fact that multiplication is commutative; so \(ab = ba\) for any two real numbers \(a,b\).
When multiplying variables with the same powers you can expand and rewrite them into a nicer format: \[a^r \cdot b^r = (ab)^r\]
Where the bases are multiplied together then raised to the \(r\)th power. For example \[2^3 \cdot 3^3 = \underbrace{(2\cdot 2\cdot 2)}_{3\textsf{ times}}\cdot \underbrace{(3\cdot 3\cdot 3)}_{3\textsf{ times}}= (2\cdot 3) \cdot (2 \cdot 3)\cdot (2 \cdot 3) = (2\cdot 3)^3.\]
Here, you can use Law 7 to say that \((3x)^3 = 3^3\cdot x^3 = 27x^3\). Notice that \((3x)^3\) is not necessarily equal to \(3x^3\)!
You can use Law 7 in conjunction with other laws. For instance, how would you simplify \((x^3y^2z)^{-2}\)? Using Law 7 twice, you can write \[(x^3y^2z)^{-2} = (x^3)^{-2}\cdot(y^2)^{-2}\cdot z^{-2}\] Now, using Law 3, you can write \[(x^3)^{-2}\cdot(y^2)^{-2}\cdot z^{-2} = x^{-6}\cdot y^{-4} \cdot z^{-2}\] Finally, using Law 5 you can say that \[(x^3y^2z)^{-2} = x^{-6}\cdot y^{-4} \cdot z^{-2} = \frac{1}{x^6}\cdot\frac{1}{y^4}\cdot\frac{1}{z^2} = \frac{1}{x^6y^4z^2}.\]
When dividing variables with the same powers you can expand and rewrite them into a nicer format:
\[\dfrac{a^r}{b^r} = \left(\dfrac{a}{b}\right)^r\]
Where the numerator base \(a\) is divided by the denominator base \(b\) then raised to the \(r\)th power. For example:
\[\dfrac{4^4}{5^4} = \dfrac{\overbrace{4\cdot 4\cdot 4\cdot 4}^{4\textsf{ times}}}{\underbrace{5\cdot 5\cdot 5\cdot 5}_{4\textsf{ times}}} = \dfrac{4}{5}\cdot \dfrac{4}{5}\cdot \dfrac{4}{5}\cdot \dfrac{4}{5} = \left(\dfrac{4}{5}\right)^4\]
Here, you can use Law 8 (and Laws 3 and 7) to simplify \((2x/y^2)^3\) in the following way: \[\left(\frac{2x}{y^2}\right)^3 = \frac{(2x)^3}{(y^2)^3} = \frac{8x^3}{y^6}.\]
When multiplying by bases with fractional powers (some sort of nth root) :
\[a^{1/n} \cdot b=\sqrt[n]{a \cdot b^n}\]
Where the aim of the manipulation is to put everything inside the \(n\)th root sign, then to continue with the calculation.
The main purpose of this law is to combine bases if you can; as evidenced in the next example.
Simplify \(2^{1/3}\cdot 4\). Here, you can use Law 9 to write \[2^{1/3}\cdot 4 = \sqrt[3]{2\cdot 4^3}.\] You know that \(4 = 2^2\), and so using Law 3 and then Law 1 gives \[\sqrt[3]{2\cdot 4^3} = \sqrt[3]{2\cdot (2^2)^3} = \sqrt[3]{2\cdot 2^6} = \sqrt[3]{2^7}\] Finally, using Law 6 gives \[2^{1/3}\cdot 4 = \sqrt[3]{2^7} = 2^{7/3}.\]
When you have a term under a \(n\)th root sign which you can factorise (split up into different factors) this can be done and the \(n\)th root signs can be dealt with separately.
\[\begin{align*} \sqrt[n]{a\cdot b}=\sqrt[n]{a}\cdot\sqrt[n]{b}\end{align*}\]
Where you are using the commutativity of multiplication to split up the \(n\)th root and deal with terms separately.
This law is particularly useful for writing square roots more concisely. For example \[\sqrt{12} = \sqrt{4\cdot 3} = \sqrt{4}\cdot\sqrt{3} = 2\sqrt{3}\]
Simplify \(\sqrt[3]{27x^9}\). First of all, you can use Law 10 to say that \[\sqrt[3]{27x^9} = \sqrt[3]{27}\cdot\sqrt[3]{x^9}.\] You know that \(\sqrt[3]{27} = 3\); using Law 6 and Law 3 gives \[\sqrt[3]{27x^9} = \sqrt[3]{27}\cdot\sqrt[3]{x^9} = 3(x^9)^{1/3} = 3x^{9/3} = 3x^3.\]
Quick check problems
Determine whether the following equations are correct or not:
\(\dfrac{17x^{17}y^{14}}{15x^{18}y^{18}} = \dfrac{17}{15xy}\)
\((x^2·x^5·x^4)^2 = x^{22}\)
\(\dfrac{12x^{14}y^{-4}}{12x^7y^{11}} = \dfrac{x^6}{y^7}\)
\(\dfrac{3x^a}{y^{8b}}\cdot\dfrac{4z^a}{y^{d}}= \dfrac{12(xz)^a}{y^{d+8b}}\)
\(\dfrac{4^2}{2^4}=2\)
Further reading
For more questions on the subject, please go to Questions: Laws of indices
Version history and licensing
v1.0: initial version created 08/23 by Zheng Chen and Zoë Gemmell as part of a University of St Andrews STEP project.
- v1.1: edited 04/24 by tdhc.