# Indices

Indices are also known as powers, exponents or sometimes even orders. Note that indices is plural and index is singular.

Note that in the following:

$x^a$

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.

It can be shown that indices abide by the rules that follow.

## The Laws of Indices

- $x^a\times x^b=x^{a+b}$ – 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.

## 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.

## Indices Examples

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

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

Solve the simultaneous equations:

$\begin{array}{c} 27^{2x}&=&9^{y-7}\\y&=&x+5\end{array}$

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.

## Questions by Topic

## Indices Videos

## What Next?

- Go back to PURE MATHS
- See QUESTIONS BY TOPIC
- Go to PAST and PRACTICE PAPERS