This article is Part 2 of a two-part series.

Part 1 covers all the Foundation content – basics, laws of indices, negative indices – and you can find it here.

In this post we’re picking up where that one left off, and moving on to look at fractional indices (needed for Higher GCSE spec and for A-level).

The image below shows an example of the kind of expression using fractional indices that you can expect to encounter and should be able to handle.

The laws of indices

We covered these in the previous lesson, but here’s a reminder:

1st law: $x^a \times x^b = x^{a + b}$
2nd law: $x^a \div x^b = x^{a - b}$
3rd law: $(x^a)^b = x^{ab}$

How do fractional indices work?

Take a look at the expressions below, which involve simple fractional indices, and apply the laws of indices to simplify them as far as possible:

$x^\frac{1}{2} \times x^ \frac{1}{2}$
$(x^\frac{1}{2})^2$
$(x^2)^\frac{1}{2}$

In each case you should get the same answer: $x^1$ , which of course is simply $x$ .

What do you have to multiply by itself to get $x$ ? Compare your answer with example 1.
What do you have to square to get $x$ ? Compare your answer with example 2.
What do you have to do to get from $x^2$ to just $x$ ? Compare your answer with example 3.

In each case we need the square root…

… so $x^\frac{1}{2} = \sqrt{x}$

What if we’d started with $x^\frac{1}{3}$ ? How many of those would you need to multiply together to get $x$ ?

Answer: $(x^ \frac{1}{3})^3 = x$

… so $x^\frac{1}{3} = \sqrt[3]{x}$

And more generally,

$\boxed{ x^\frac{1}{n} = \sqrt[n]{x}}$

Note: As with surds, a fractional power indicates only the positive root; you only need to worry about the possibility of a second, negative, root if you’ve started off with the statement $x^2 = k$ or $x^4 = k$ , etc.

Your turn: Work out the values of:

$9^\frac{1}{2}$
$64^\frac{1}{3}$
$10000^\frac{1}{4}$
$(\frac{9}{16})^\frac{1}{2}$
$(\frac{3}{4})^\frac{1}{2}$

(Check answers)

What if the numerator is greater than 1?

If the numerator is not simply 1 then we need to apply the 3rd law, but having the power in fractional form makes it look a little different from what we’ve encountered before:

$\boxed{x^\frac{a}{b} = x^{a \times \frac{1}{b}} = (x^\frac{1}{b})^a = (x^a)^\frac{1}{b}}$

Of course, $(x^\frac{1}{b})^a$ is the same as ${\sqrt[b]{x}}^a$ and $(x^a)^\frac{1}{b} = \sqrt[b]{x^a}$ ; both give the same result so it doesn’t matter whether you do the root first and then the integer power, or the integer power and then the root – though generally it’s easiest to take the root first.

Now evaluate:

$8^\frac{2}{3}$
$16^\frac{3}{4}$
$10000^\frac{3}{2}$
$(\frac{27}{64})^\frac{2}{3}$
$(\frac{1}{9})^\frac{5}{2}$

(Check answers)

This “flower power” meme might help you to remember how to deal with fractional indices:

Flower power for fractional indices - the root is at the bottom — Fractional power flower

Adding in negative indices

Remember, any number to the power of $-1$ is its own reciprocal. It doesn’t actually matter whether you do the reciprocal bit first or the fractional indices bit first, but I’d usually suggest getting rid of the negative powers as early as possible.

Here’s an example:

$(\frac{4}{9})^{-\frac{1}{2}} \hspace{12}= (\frac{9}{4})^\frac{1}{2} \hspace{12} = \frac{9^\frac{1}{2}}{4^\frac{1}{2}} \hspace{12}= \frac{\sqrt{9}}{\sqrt{4}} \hspace{12} = \frac{3}{2}$

Now here are some for you to have a go at:

$4^{-\frac{1}{2}}$
$9^{-\frac{3}{2}}$
$(\frac{1}{16})^{-\frac{3}{2}}$
$(\frac{5}{9})^{-\frac{3}{2}}$
$10000^{-\frac{3}{4}}$
$(\frac{27}{64})^{-\frac{2}{3}}$
$0.01^{-\frac{3}{2}}$

And of course the example in the image at the beginning of this article should make sense to you now too!

(Check answers)

Mixing it up with algebra

Now have a go at these questions involving a mixture of negative and fractional indices combined with algebraic expressions. Some of them are quite challenging – take care to double-check your working as it’s very easy to make mistakes!

Express $\sqrt{x}^5$ in index form.
Expand and simplify $(x + y^\frac{1}{2})^2$ , giving your answer in surd form (i.e. using surds rather than indices).
Given that $n = 8$ , evaluate $\frac{3}{n^{-\frac{2}{3}}}$
Given that $x = 4$ , evaluate $(16x)^{-\frac{3}{2}}$ .
Express $3\sqrt[3]{y} \times (8y)^\frac{4}{3}$ in surd form.
Simplify $2\sqrt{x} \div (4x)^\frac{5}{2}$ .

(Check answers)

If you got all of those right then you’ve done exceptionally well!

Here’s a great codebreaker activity from JustMaths that I use with my Higher Tier and A-level students to practise negative and fractional indices (it downloads as a PDF).

And that covers fractional indices!

I hope you’ve found this post helpful. If you have then please share it, and also please take a look at the rest of my site to see what other help I can offer.

🤞 Key points to remember (both Tiers):

A positive integer power tells you how many of the base you are multiplying together.

To raise a fraction to a power, raise both the numerator and the denominator to that power.

1st law of indices: $x^a \times x^b = x^{a + b}$ (if the terms are multiplied then you add the powers)

2nd law of indices: $x^a \div x^b = x^{a - b}$ (if the terms are divided then you subtract the powers)

3rd law of indices: $(x^a)^b = x^{ab}$ (if it’s a power of a power then you multiply the powers)

You can only apply the laws of indices if the bases are the same.

Anything to the power of zero is 1.

A negative power means a reciprocal: if it’s a whole number then put 1 over it; if it’s a fraction then flip it (which amounts to the same thing).

A fractional power means a root; the root is the bottom bit so the denominator tells you what root you need (square, cube, 4th, etc.). You also need to raise the value to the power of the numerator – but of course this will only have an effect if the numerator is not 1.

Answers

How do fractional indices work?

$9^\frac{1}{2} = \sqrt{9} = 3$
$64^\frac{1}{3} = \sqrt[3]{64} = 4$
$10000^\frac{1}{4} = \sqrt[4]{10000} = 10$
$(\frac{9}{16})^\frac{1}{2} = \sqrt{\frac{9}{16}} = \frac{\sqrt{9}}{\sqrt{16}} = \frac{3}{4}$
$(\frac{3}{4})^\frac{1}{2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2}$ (Yes, you will sometimes end up with surds!)

Back to this section

Numerator greater than 1

$8^\frac{2}{3} = \sqrt[3]{8}^2 = 2^2 = 4$
$16^\frac{3}{4} = \sqrt[4]{16}^3 = 2^3 = 8$
$10000^\frac{3}{2} = \sqrt{10000}^3 = 100^3 = 1,000,000$
$(\frac{27}{64})^\frac{2}{3} = \sqrt[3]{\frac{27}{64}}^2 = (\frac{\sqrt[3]27}{\sqrt[3]64})^2 = (\frac{3}{4})^2 = (\frac{3^2}{4^2}) = \frac{9}{16}$
$(\frac{1}{9})^\frac{5}{2} = \sqrt{\frac{1}{9}}^5 = (\frac{1}{3})^5 = \frac{1}{243}$

Back to this section

Adding in negative indices

$4^{-\frac{1}{2}} = \frac{1}{4^\frac{1}{2}} = \frac{1}{\sqrt{4}} = \frac{1}{2}$
$9^{-\frac{3}{2}} = (\frac{1}{9})^\frac{3}{2} = ((\frac{1}{9})^\frac{1}{2})^3} = \frac{\sqrt{1}^3}{\sqrt{9}^3} = \frac{1}{3^3} = \frac{1}{27}$
$(\frac{1}{16})^{-\frac{3}{2}} = 16^\frac{3}{2} = ((\frac{1}{16})^\frac{1}{2})^3} = \sqrt{16}^3 = 4^3 = 64$
$(\frac{5}{9})^{-\frac{3}{2}} = (\frac{9}{5})^\frac{3}{2} = \sqrt{\frac{9}{5}}^3 = \frac{\sqrt{9}^3}{\sqrt{5}^3} = \frac{3^3}{\sqrt{5}^3} = \frac{27}{5\sqrt{5}}$
$10000^{-\frac{3}{4}} = \frac{1}{10000^\frac{3}{4}} = \frac{1}{\sqrt[4]{10000}^3} = \frac{1}{10^3} = \frac{1}{1000} (= 0.001)$
$(\frac{27}{64})^{-\frac{2}{3}} = (\frac{64}{27})^\frac{2}{3} = \sqrt[3]{\frac{64}{27}} = \frac{\sqrt[3]{64}^2}{\sqrt[3]{27}^2} = \frac{4^2}{3^2} = \frac{16}{9}$
$0.01^{-\frac{3}{2}} = (\frac{1}{100})^{-\frac{3}{2}} = 100^\frac{3}{2} = \sqrt{100}^3 = 10^3 = 1000$

Back to this section

Mixing it up with algebra

1. $\sqrt{x}^5 = (x^\frac{1}{2})^5 = x^\frac{5}{2}$

2. $(x + y^\frac{1}{2})^2 = (x + y^\frac{1}{2})(x + y^\frac{1}{2})$
$= x^2 + xy^\frac{1}{2} + xy^\frac{1}{2} + (y^\frac{1}{2})^2$
$= x^2 + 2x\sqrt{y} + y$

3. $\frac{3}{n^{-\frac{2}{3}}} = \frac{3}{8^{-\frac{2}{3}}}$
$= 3 \times 8^\frac{2}{3} = 3 \times \sqrt[3]{8}^2$
$= 3 \times 2^2 = 3 \times 4 = 12$

4. $(16x)^{-\frac{3}{2}} = (16 \times 4)^{-\frac{3}{2}}$
$= 64^{-\frac{3}{2}} = (\frac{1}{64})^\frac{3}{2}$
$= \sqrt{\frac{1}{64}}^3 = (\frac{1}{8})^3 = \frac{1}{512}$

5. $3\sqrt[3]{y} \times (8y)^\frac{4}{3}$
$= 3 \times y^\frac{1}{3} \times 8^\frac{4}{3} \times y^\frac{4}{3}$
$= 3 \times \sqrt[3]{8}^4 \times y^{\frac{1}{3}+\frac{4}{3}}$
$= 3 \times 2^4 \times y^\frac{5}{3}$
$= 3 \times 16 \times \sqrt[3]{y}^5 = 48\sqrt[3]{y}^5$

6. $2\sqrt{x} \div (4x)^\frac{5}{2}$
$= \frac{2x^\frac{1}{2}}{(4x)^\frac{5}{2}}$
$= \frac{2x^\frac{1}{2}}{4^\frac{5}{2} \times x^\frac{5}{2}}$
$= \frac{2}{\sqrt{4}^5}\times x^{\frac{1}{2}-\frac{5}{2}}$
$= \frac{2}{2^5} \times x^{-\frac{4}{2}}$
$= 2^{1-5} \times x^{-2}$
$= 2^{-4} \times x^{-2} = \frac{1}{2^4} \times\frac{1}{x^2} = \frac{1}{16x^2}$ .

Back to this section

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Fractional indices: Essential skills for GCSE and A-level Maths

This article is Part 2 of a two-part series.

The laws of indices

How do fractional indices work?

Your turn: Work out the values of:

What if the numerator is greater than 1?

Now evaluate:

Adding in negative indices

Mixing it up with algebra

🤞 Key points to remember (both Tiers):

Answers

How do fractional indices work?

Numerator greater than 1

Adding in negative indices

Mixing it up with algebra

One Comment

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This article is Part 2 of a two-part series.

The laws of indices

How do fractional indices work?

Your turn: Work out the values of:

What if the numerator is greater than 1?

Now evaluate:

Adding in negative indices

Mixing it up with algebra

🤞 Key points to remember (both Tiers):

Answers

How do fractional indices work?

Numerator greater than 1

Adding in negative indices

Mixing it up with algebra

One Comment

Leave a Reply Cancel reply

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