Indices

Indices, also referred to as powers, exponents, or orders, are a key mathematical tool to signify multiplication of a number or base by itself a certain number of times. The power indicates the number of duplications of the base. Indices follow certain laws that enable simplification and computation of complex algebraic and numerical expressions. From understanding exponential growth in finance and economics to representing data scales in the field of computer science, indices play a crucial role in various real-world applications..

Note that indices is plural and index is singular. Note that in x^a, a is the power/exponent and x is the base. In English, when a letter is smaller and on the upper right side of the bigger letter, we call it a superscript. In Maths, it is often a power. (What are subscripts and superscripts?).

Bear in mind that when you see an expression such as 2x^3, this is 2 lots of x cubed. This follows from BIDMAS where we apply powers before multiplication. This confuses many students – they often cube 2x which, of course, gives a different answer of 8x^3. Students may find it hard to perform tasks with indices at first, especially in an algebraic setting. If you find that you are struggling, take a step back, try doing the calculations with numbers first.

The Laws of Indices

It can be shown that indices abide by the following rules:

  • x^a\times x^b=x^{a+b} – think of multiplying x^2 by x^3. You can write it out in full as x\times x\times x\times x\times x. Hence, add the powers. Note that this is only true if the base is the same and should not be applied to x^2 by y^3, for instance.
  • x^a\div x^b=x^{a-b} – similar to the previous example, however, when you are dividing algebraic terms you should subtract the powers.
  • x^0=1 – anything to the power of zero is 1. You can see this from the previous bullet point by choosing a and b to be the same number.
  • \left(x^a\right)^b=x^{ab} – consider taking x^2 to the power of 3, i.e. multiplying by itself 3 times. We have \left(x^2\right)^3=x^2\times x^2\times x^2=x^6. It follows that we multiply the powers.
  • x^{-n}=\frac{1}{x^n} – this is easy to see if you consider x^3\div x^5=x^{-2} and subtracting the powers, then writing it as a fraction: \frac{x^3}{x^5}=\frac{1}{x^2}.
  • x^{\frac{1}{n}}=\sqrt[n]{x} – can be seen if you consider x^{\frac{1}{2}}\times x^{\frac{1}{2}}=x and so x^{\frac{1}{2}} must be the square root of x. This is because x was the result when multiplying something by itself. Multiplying x^{\frac{1}{3}} by itself 3 times shows that x^{\frac{1}{3}}=\sqrt[3]{x} and the same applies for other fractions.  It follows from this rule that x^{\frac{m}{n}}=\left(\sqrt[n]{x}\right)^m.

See Examples 1 and 2.

Changing the Base

Some exam questions may require you to change the base of an expression. By making the bases consistent in an equation, it makes it possible to compare the powers. For example, suppose you are given the equation:

4^{x+1}=2^{3x}

We can solve this by writing 4=2^2 and so 4^{x+1}=\left(2^2\right)^{x+1}=2^{2(x+1)}. It follows that

2^{2x+2}=2^{3x}

and we can compare the powers: 2x+2=3x and so the solution is x=2. Other questions may require you to change both bases or even include one equation in a pair of simultaneous equations. See Example 3.

Examples

Simplify the following:
2^7\times 2^9
\frac{4x^7}{2x^3}
\left(3p^2\right)^4=3^4(p^2)^4
16\times 2^{-3}

Solution:

2^7\times 2^9=2^{16}
\frac{4x^7}{2x^3}=2x^4
\left(3p^2\right)^4=3^4(p^2)^4=81p^8
16\times 2^{-3}=16\times \frac{1}{2^3}=16\times \frac{1}{8}=2

Evaluate \left(\frac{27}{8}\right)^{-\frac{3}{2}}.

Solution:

\left(\frac{27}{8}\right)^{-\frac{3}{2}}=\left(\frac{8}{27}\right)^{\frac{2}{3}}=\left(\left(\frac{8}{27}\right)^{\frac{1}{3}}\right)^2=\left(\sqrt[3]{\frac{8}{27}}\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}

Solve the simultaneous equations: \begin{array}{c} 27^{2x}=9^{y-7}\\y=x+5\end{array}

Solution:

We must change the bases of the first equation and obtain a linear equation for x and y. Note that 27=3^3 and 9=3^2 so we can write the first equation as:

\left(3^3\right)^{2x}=\left(3^2\right)^{y-7} or 3^{6x}=3^{2y-14}

It follows that 6x=2y-14. We can write the second equation in the pair as x=y-5. Substituting into 6x=2y-14 gives 6(y-5)=2y-14 or 6y-30=2y-14. Solving this gives y=4 and so x=-1. See more on simultaneous equations. 

Videos

https://youtu.be/in-hECgjJWQ

Using indices rules to terms in the form of others and using them to solve hidden quadratics.

Extra Resources

Questions by Topic

IndicesExamQuestions

Open Indices Questions by Topic in New Window

Board Example

AS Maths Algebra & Functions

A2 Maths Algebra & Functions