Rules of Indices

How to use this page

Rigorous Notes: For this page, I’d make notes on each rule individually. I’d read its derivation to see why it works. In my rough work, I’d try the rule with new numbers and work through any variations or new ingredients I can think of. Then I’d neatly write up each rule with any notes and my own key examples.
Topic: Building every rule of indices from a single definition, then applying them in all the skills we need.
To see what this looks like, here are my notes for the first four rules.

How to support this course

I plan to finish writing up the whole course as a free online revision textbook. If you'd like to help, you can:

This is the first page where you’ll definitely encounter new concepts, especially in the application section. Time for your notes to get detailed.

Rather than memorising a list of rules, we’ll work each one out for ourselves, starting from a single definition. On this page we’ll:

Derive the rules to see why they work
See everything you need to know in the Summary of Rules
Apply the rules to exam-style questions

Deriving the Rules from First Principles

Defining Indices

Indices are a shorthand for repeated multiplication. You’ll hear the terms indices, powers and exponents used interchangeably.

Definition: For a base number $x$ with index $a$ , we define the operation as:

x^a = \underbrace{x \times x \times \dots \times x}_{a \text{ times}} \qquad \text{for } x \in \mathbb{R}, \; a \in \mathbb{N}

Note: $x \in \mathbb{R}$ translates to $x$ is a number from the set of real numbers (any number on the number line, including ones whose decimals never end, such as $\pi$ or $\sqrt{2}$ ). Then $a \in \mathbb{N}$ means that $a$ is a natural number, or a counting number, or a positive integer e.g. $\{1, 2, 3, \dots\}$ .

For clarification, the index (or power, or exponent) is the small raised number $a$ , and $x$ is the base. The whole operation is called raising $x$ to the power $a$ , so we read $2^3$ as “2 raised to the power 3”.

Starting with this definition, we can algebraically deduce other rules.

Powers of Products

Firstly, recall that a product is the result of a multiplication, such as $2 \times 3$ . Now, our first deduction uses only the definition and the associative and commutative properties of multiplication, which we covered in section The Classic Operations.

Key Example:

\begin{aligned} (2 \times 3)^3 &= (2 \times 3) \times (2 \times 3) \times (2 \times 3) \\ &= (2 \times 2 \times 2) \times (3 \times 3 \times 3) \\ &= 2^3 \times 3^3 \end{aligned}

Note: From the first line to the second, we drop the brackets (associative) and reorder the factors (commutative).

Consider: How does $(2x)^3$ expand in the same way?

Conclusion: The same regrouping works for any two real numbers, raised to any natural number power:

\begin{aligned} (xy)^n &= \underbrace{xy \times xy \times \dots \times xy}_{n \text{ times}} \\ &= (\underbrace{x \times \dots \times x}_{n \text{ times}}) \times (\underbrace{y \times \dots \times y}_{n \text{ times}}) \\ &= x^n y^n \qquad \text{for } x, y \in \mathbb{R}, \; n \in \mathbb{N} \end{aligned}

Powers of Quotients

Recall that a quotient is the result of a division, such as $\dfrac{2}{3} = 2 \div 3$ . Quotients work the same way, because we multiply fractions by multiplying the numerators together and the denominators together.

Key Example:

\begin{aligned} \left(\frac{2}{3}\right)^3 &= \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \\ &= \frac{2 \times 2 \times 2}{3 \times 3 \times 3} \\ &= \frac{2^3}{3^3} \end{aligned}

Consider: How does $\left(\dfrac{4}{3}\right)^4$ expand in the same way?

Conclusion: The same works for any fraction, as long as the denominator is not zero:

\begin{aligned} \left(\frac{x}{y}\right)^n &= \underbrace{\frac{x}{y} \times \frac{x}{y} \times \dots \times \frac{x}{y}}_{n \text{ times}} \\ &= \frac{\overbrace{x \times \dots \times x}^{n \text{ times}}}{\underbrace{y \times \dots \times y}_{n \text{ times}}} \\ &= \frac{x^n}{y^n} \qquad \text{for } x, y \in \mathbb{R}, \; y \neq 0, \; n \in \mathbb{N} \end{aligned}

Multiplying Indices

Key Example:

\begin{aligned} 2^5 \times 2^3 &= (\underbrace{2 \times 2 \times 2 \times 2 \times 2}_{5 \text{ times}}) \times (\underbrace{2 \times 2 \times 2}_{3 \text{ times}}) \\ &= \underbrace{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}_{5 + 3 = 8 \text{ times}} \\ &= 2^8 \end{aligned}

Note: Why does this work? We’re using the associative property of multiplication to drop the brackets, knowing it remains equal. This property is outlined in The Classic Operations.

Observations:

We notice that we can just add the exponents directly because we are collecting the total number of identical factors.
In the above example, we’re counting how many times the factor of $2$ repeats. Hence, combining a group of $5$ identical items with a group of $3$ identical items, the total count is $5 + 3 = 8$ identical items.

Consider: What if we swap the order to $2^3 \times 2^5$ ? Can you see how the commutative property of multiplication, and of addition, both tell us we still get $2^8$ ?

Conclusion: Extending this logic to any base number we arrive at another fundamental rule:

\begin{aligned} x^a \times x^b &= (\underbrace{x \times x \times \dots \times x}_{a \text{ times}}) \times (\underbrace{x \times x \times \dots \times x}_{b \text{ times}}) \\ &= \underbrace{x \times x \times \dots \times x}_{a+b \text{ times}} \\ &= x^{a+b} \qquad \text{for } x \in \mathbb{R}, \; a,b \in \mathbb{N} \end{aligned}

Dividing Indices

Key Example 1:

\begin{aligned} 2^5 \div 2^3 &= \frac{\overbrace{2 \times 2 \times 2 \times 2 \times 2}^{5 \text{ times}}}{\underbrace{2 \times 2 \times 2}_{3 \text{ times}}} \\ &= 2 \times 2 \\ &= 2^2 \end{aligned}

Note: Why does this work? Since any non-zero number divided by itself equals $1$ ( $\dfrac{2}{2} = 1$ ), we can cancel out matching factors of $2$ . Cancelling is outlined in The Classic Operations.

Key Example 2:

\begin{aligned} 2^3 \div 2^5 &= \frac{\overbrace{2 \times 2 \times 2}^{3 \text{ times}}}{\underbrace{2 \times 2 \times 2 \times 2 \times 2}_{5 \text{ times}}} \\ &= \frac{1}{2 \times 2} \\ &= \frac{1}{2^2} \end{aligned}

Note: Why does this lead to a different result? Division is non-commutative, $2^5 \div 2^3 \neq 2^3 \div 2^5$ . This time the denominator has more factors, so the leftover factors stay on the bottom.

Observations: By cancelling out common factors in the numerator and denominator, we are always left with the difference in the total number of factors:

When the top power is larger ( $a > b$ ), the remaining factors stay in the numerator.
When the bottom power is larger ( $a < b$ ), the remaining factors stay in the denominator. Here $a - b$ is negative, which we haven’t defined yet.
When the powers are equal ( $a = b$ ), everything cancels, leaving $1$ . Here $a - b = 0$ , which we also haven’t defined yet.

Conclusion: Extending this logic to any non-zero base number $x$ , we establish another fundamental rule, for now when $a > b$ :

\begin{aligned} x^a \div x^b &= \frac{\overbrace{x \times x \times \dots \times x}^{a \text{ times}}}{\underbrace{x \times x \times \dots \times x}_{b \text{ times}}} \\ &= x^{a-b} \qquad \text{for } x \in \mathbb{R}, \; x \neq 0, \; a,b \in \mathbb{N}, \; a > b \end{aligned}

Zero Index

We’ve defined what happens when the index $a \in \mathbb{N}$ . However, we need to determine what happens when $a \in \{ 0, -1, -2, -3, \text{ ... } \}$ . We’ll start with $0$ .

Key Example 1: Consider the expansion of these powers of $2$ .

\begin{array}{rrll} & 2^3 &= 2 \times 2 \times 2 = 8 & \\ \scriptsize \times 2 \; \Big\uparrow & & & \Big\downarrow \; \scriptsize \div 2 \\ & 2^2 &= 2 \times 2 = 4 & \\ \scriptsize \times 2 \; \Big\uparrow & & & \Big\downarrow \; \scriptsize \div 2 \\ & 2^1 &= 2 & \end{array}

Observations:

When the power increases by $1$ , we multiply by the base number ( $2$ ) one more time.
We can extend this logic the other way. When we decrease the power by $1$ $1$ , we do the inverse operation, dividing by $2$ $2$ .
- e.g. to move down from the top row, we divide by $2$ . $2^3 \div 2 = \dfrac{2 \times 2 \times 2}{2} = 2 \times 2 = 2^2$
- One more step down from $2^1$ gives $2 \div 2 = 1$ , so the pattern says $2^0 = 1$ .

Key Example 2: The division rule agrees. Working out $2^3 \div 2^3$ both ways:

\begin{aligned} 2^3 \div 2^3 &= \frac{2 \times 2 \times 2}{2 \times 2 \times 2} = 1 \\ 2^3 \div 2^3 &= 2^{3-3} = 2^0 \end{aligned}

Note: Our definition only covered $a \in \mathbb{N}$ , so $2^0$ had no meaning yet. We choose $2^0 = 1$ , because it fits the pattern and keeps the division rule working.

Conclusion: The same works for any base except $0$ , because dividing by $0$ is undefined.

x^0 = 1 \quad \text{for } x \in \mathbb{R}, \; x \neq 0

Negative Indices

Key Example 1: We can take those observations and use them to expand beyond positive integer powers. We know that to decrease the power by 1 we divide the row above by the base number.

\begin{array}{rrll} & 2^3 &= 2 \times 2 \times 2 = 8 & \\ \scriptsize \times 2 \; \Big\uparrow & & & \Big\downarrow \; \scriptsize \div 2 \\ & 2^2 &= 2 \times 2 = 4 & \\ \scriptsize \times 2 \; \Big\uparrow & & & \Big\downarrow \; \scriptsize \div 2 \\ & 2^1 &= 2 & \\ \scriptsize \times 2 \; \Big\uparrow & & & \Big\downarrow \; \scriptsize \div 2 \\ & 2^0 &= 1 & \\ \scriptsize \times 2 \; \Big\uparrow & & & \Big\downarrow \; \scriptsize \div 2 \\ & 2^{-1} &= \frac{1}{2} & \\ \scriptsize \times 2 \; \Big\uparrow & & & \Big\downarrow \; \scriptsize \div 2 \\ & 2^{-2} &= \frac{1}{2^2} = \frac{1}{4} & \\ \scriptsize \times 2 \; \Big\uparrow & & & \Big\downarrow \; \scriptsize \div 2 \\ & 2^{-3} &= \frac{1}{2^3} = \frac{1}{8} & \end{array}

Observations:

Recall also that dividing by $2$ is the same as $\times \dfrac{1}{2}$ hence this multiplies the denominator by $2$

Key Example 2: The division rule agrees again. Working out $2^3 \div 2^5$ both ways:

\begin{aligned} 2^3 \div 2^5 &= \frac{2 \times 2 \times 2}{2 \times 2 \times 2 \times 2 \times 2} = \frac{1}{2^2} \\ 2^3 \div 2^5 &= 2^{3-5} = 2^{-2} \end{aligned}

Question: Consider a base number of $3$ and the same powers. What operations are you doing to increase or decrease the power by $1$ ?

\begin{aligned} 3^3 &= \\ 3^2 &= \\ 3^1 &= \\ 3^0 &= \\ 3^{-1} &= \\ 3^{-2} &= \\ 3^{-3} &= \end{aligned}

We observe this same behaviour for different bases. We just need to note that $x \neq 0$ , because dividing by $0$ is undefined.

\begin{array}{rrll} & x^3 &= x \times x \times x & \\ \scriptsize \times x \; \Big\uparrow & & & \Big\downarrow \; \scriptsize \div x \\ & x^2 &= x \times x & \\ \scriptsize \times x \; \Big\uparrow & & & \Big\downarrow \; \scriptsize \div x \\ & x^1 &= x & \\ \scriptsize \times x \; \Big\uparrow & & & \Big\downarrow \; \scriptsize \div x \\ & x^0 &= 1 & \\ \scriptsize \times x \; \Big\uparrow & & & \Big\downarrow \; \scriptsize \div x \\ & x^{-1} &= \frac{1}{x} & \\ \scriptsize \times x \; \Big\uparrow & & & \Big\downarrow \; \scriptsize \div x \\ & x^{-2} &= \frac{1}{x^2} & \\ \scriptsize \times x \; \Big\uparrow & & & \Big\downarrow \; \scriptsize \div x \\ & x^{-3} &= \frac{1}{x^3} \end{array}

Conclusion: Hence, we observe this rule:

x^{-n} = \frac{1}{x^n} \quad \text{for } x \in \mathbb{R}, \; x \neq 0, \; n \in \mathbb{N}

Note: With this, the division rule $x^a \div x^b = x^{a-b}$ now works for all $a, b \in \mathbb{N}$ , whichever is larger.

Power of a Power

Now, what happens when we raise a power to another power?

Key Example:

\begin{aligned} (2^5)^3 &= 2^5 \times 2^5 \times 2^5 \\ &= 2^{5 + 5 + 5} \\ &= 2^{5 \times 3} \\ &= 2^{15} \end{aligned}

Note: Why does this work? Cubing means multiplying three copies of $2^5$ together, so we can use the rule from Multiplying Indices. Then adding $5$ three times is the same as $5 \times 3$ , because multiplication is repeated addition (The Classic Operations).

Observations:

Each bracket holds $5$ factors of $2$ , and there are $3$ brackets, so there are $5 \times 3 = 15$ factors of $2$ in total.

Consider: How many factors of $3$ are there in $(3^2)^4$ ?

Conclusion: Extending this logic to any base number we arrive at another fundamental rule:

\begin{aligned} (x^a)^b &= \underbrace{x^a \times x^a \times \dots \times x^a}_{b \text{ times}} \\ &= x^{\overbrace{a + a + \dots + a}^{b \text{ times}}} \\ &= x^{ab} \qquad \text{for } x \in \mathbb{R}, \; a,b \in \mathbb{N} \end{aligned}

Roots as Indices

So far every index has been a whole number. Now, what could $2^{\frac{1}{2}}$ mean? Just like with zero and negative indices, we choose the meaning that keeps our rules working.

Key Example 1: Consider how

2^{\frac{1}{2}} \times 2^{\frac{1}{2}} = 2^{\frac{1}{2}+\frac{1}{2}} = 2^1 = 2

So $2^{\frac{1}{2}}$ is a number that multiplies by itself to make $2$ . Both $\sqrt{2}$ and $-\sqrt{2}$ do that, so we choose the positive one:

2^{\frac{1}{2}} = \sqrt{2}

Key Example 2:

2^{\frac{1}{3}} \times 2^{\frac{1}{3}} \times 2^{\frac{1}{3}} = 2^{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}} = 2^1 = 2

\therefore 2^{\frac{1}{3}} = \sqrt[3]{2}

Note: Why is there no choice this time? Only one real number cubes to make $2$ . Even roots, like square roots, always have a positive and a negative candidate, and we always take the positive one.

Consider: How many copies of $2^{\frac{1}{4}}$ would you need to multiply together to get back to $2^1$ ? What does that tell you $2^{\frac{1}{4}}$ must be?

Conclusion: Extending this logic to any positive base number we arrive at another fundamental rule:

x^{\frac{1}{b}} = \sqrt[b]{x} \qquad \text{for } x \in \mathbb{R}, \; x > 0, \; b \in \mathbb{N}

Note: We stick to positive bases because, for example, $(-4)^{\frac{1}{2}}$ isn’t a real number.

Fractional Indices

Now, what about a fractional index where the numerator isn’t $1$ ?

Key Example: Consider $8^{\frac{4}{3}}$ . Since $\dfrac{4}{3} = \dfrac{1}{3} \times 4$ , we can use Power of a Power to split $8^{\frac{4}{3}}$ into two steps, in either order:

\begin{aligned} 8^{\frac{4}{3}} &= \left(8^{\frac{1}{3}}\right)^4 = 2^4 = 16 \\ 8^{\frac{4}{3}} &= \left(8^4\right)^{\frac{1}{3}} = \sqrt[3]{4096} = 16 \end{aligned}

Observations:

Why are we free to choose the order? In Power of a Power the indices multiply, and multiplication is commutative, so $\left(x^a\right)^b = x^{ab} = x^{ba} = \left(x^b\right)^a$ . Taking the root first or the power first must give the same answer.
Both orders give $16$ , but taking the root first keeps the numbers small.

Question: Work out $27^{\frac{2}{3}}$ both ways. Which is easier?

Conclusion: Hence we build the rule:

x^{\frac{a}{b}} = \left(\sqrt[b]{x}\right)^a = \sqrt[b]{x^a} \qquad \text{for } x \in \mathbb{R}, \; x > 0, \; a \in \mathbb{Z}, \; b \in \mathbb{N}

Summary of Rules

Here’s every rule we’ve derived, in one place. If you’re unsure about any of them, click its name to go back to why it works.

Section	Rule	In symbols
5.1.1	Defining Indices	$x^a = \underbrace{x \times \dots \times x}_{a \text{ times}}$ for $a \in \mathbb{N}$
5.1.2	Powers of Products	$(xy)^n = x^n y^n$
5.1.3	Powers of Quotients	$\left(\dfrac{x}{y}\right)^n = \dfrac{x^n}{y^n}$
5.1.4	Multiplying Indices	$x^a \times x^b = x^{a+b}$
5.1.5	Dividing Indices	$x^a \div x^b = x^{a-b}$
5.1.6	Zero Index	$x^0 = 1$
5.1.7	Negative Indices	$x^{-n} = \dfrac{1}{x^n}$
5.1.8	Power of a Power	$(x^a)^b = x^{ab}$
5.1.9	Roots as Indices	$x^{\frac{1}{b}} = \sqrt[b]{x}$
5.1.10	Fractional Indices	$x^{\frac{a}{b}} = \left(\sqrt[b]{x}\right)^a = \sqrt[b]{x^a}$

Note: Many of our deductions used natural numbers, because counting factors only works with whole numbers. We then chose meanings for zero, negative and fractional indices that keep the same rules working, which is why you can use them for any rational index.

Applying the Rules

These questions were carefully chosen to test the core skills of this topic: how the rules combine, and how they show up in real exam papers. Make sure you understand each one, and you’ll be well prepared.

When we’re solving problems, you want to consider every rule in the summary as a tool you might be expected to use.

Every rule works for any rational index: whole, negative or fractional. Just be careful with:

the definition, which only works for $a \in \mathbb{N}$ , as we can’t multiply $x$ by itself $-2$ or $\dfrac{1}{2}$ times;
a base of $0$ where we’d divide by it, e.g. $0^{-1}$ ;
a negative base with fractional indices, e.g. $(-4)^{\frac{1}{2}}$ isn’t a real number.

Negative Fractional Indices

Key Example 1: Work out $8^{-\frac{2}{3}}$ .

\begin{aligned} 8^{-\frac{2}{3}} &= \frac{1}{8^{\frac{2}{3}}} \\ &= \frac{1}{\left(\sqrt[3]{8}\right)^2} \\ &= \frac{1}{2^2} \\ &= \frac{1}{4} \end{aligned}

Note: We deal with the minus sign first using Negative Indices, then the fraction using Fractional Indices, taking the root first to keep the numbers small.

Key Example 2: Work out $4^{-2.5}$ .

\begin{aligned} 4^{-2.5} &= 4^{-\frac{5}{2}} \\ &= \frac{1}{4^{\frac{5}{2}}} \\ &= \frac{1}{\left(\sqrt{4}\right)^5} \\ &= \frac{1}{2^5} \\ &= \frac{1}{32} \end{aligned}

Note: First, we write the decimal as a fraction. Because the indices multiply (Power of a Power), we can then deal with the minus, the denominator (the root) and the numerator (the power) in any order.

Key Question 1: Work out $81^{-0.75}$ .

Worked solution

Write the decimal index as a fraction first: $0.75 = \dfrac{3}{4}$ .

\begin{aligned} 81^{-0.75} &= 81^{-\frac{3}{4}} \\ &= \frac{1}{81^{\frac{3}{4}}} \\ &= \frac{1}{\left(\sqrt[4]{81}\right)^3} \\ &= \frac{1}{3^3} \\ &= \frac{1}{27} \end{aligned}

Key Question 2: Given that $a = 16^{-\frac{3}{4}}$ and $b = 8^{-\frac{2}{3}}$ , find the exact value of $\dfrac{a}{b}$ .

Worked solution

Work out each one first:

\begin{aligned} a &= 16^{-\frac{3}{4}} = \frac{1}{\left(\sqrt[4]{16}\right)^3} = \frac{1}{2^3} = \frac{1}{8} \\ b &= 8^{-\frac{2}{3}} = \frac{1}{\left(\sqrt[3]{8}\right)^2} = \frac{1}{2^2} = \frac{1}{4} \end{aligned}

Then divide:

\frac{a}{b} = \frac{1}{8} \div \frac{1}{4} = \frac{1}{8} \times 4 = \frac{1}{2}

Flipping a Fraction

Key Example 1: Work out $\left(\dfrac{2}{3}\right)^{-2}$ .

\begin{aligned} \left(\frac{2}{3}\right)^{-2} &= \frac{1}{\left(\frac{2}{3}\right)^2} \\ &= 1 \div \frac{2^2}{3^2} \\ &= 1 \times \frac{3^2}{2^2} \\ &= \frac{9}{4} \end{aligned}

Note: We use Negative Indices to move the power to the bottom, then Powers of Quotients. Dividing by a fraction is the same as multiplying by its reciprocal (the fraction flipped upside down), so the fraction flips.

Observations:

A negative index on a fraction just flips it: $\left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^2$ . So flip first, then deal with the positive power.

Key Example 2: Work out $0.64^{-\frac{1}{2}}$ .

\begin{aligned} 0.64^{-\frac{1}{2}} &= \left(\frac{16}{25}\right)^{-\frac{1}{2}} \\ &= \left(\frac{25}{16}\right)^{\frac{1}{2}} \\ &= \frac{\sqrt{25}}{\sqrt{16}} \\ &= \frac{5}{4} \end{aligned}

Key Question 1: Work out $\left(\dfrac{27}{8}\right)^{-\frac{2}{3}}$ .

Worked solution

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

Key Question 2: Work out $2.25^{-\frac{3}{2}}$ , giving your answer as a fraction.

Worked solution

Write the decimal as a fraction first, then flip it:

\begin{aligned} 2.25^{-\frac{3}{2}} &= \left(\frac{9}{4}\right)^{-\frac{3}{2}} \\ &= \left(\frac{4}{9}\right)^{\frac{3}{2}} \\ &= \left(\frac{\sqrt{4}}{\sqrt{9}}\right)^3 \\ &= \left(\frac{2}{3}\right)^3 \\ &= \frac{8}{27} \end{aligned}

Writing Terms as Powers of x

There are questions where we need all our terms to be in the form $ax^n$ , for some numbers $a$ and $n$ .

Key Example 1: Write $\dfrac{3}{\sqrt{x}}$ in the form $ax^n$ .

\begin{aligned} \frac{3}{\sqrt{x}} &= \frac{3}{x^{\frac{1}{2}}} \\ &= 3x^{-\frac{1}{2}} \end{aligned}

Note: Only the $x$ moves to the top, so the $3$ stays as it is. Watch out: $\dfrac{3}{\sqrt{x}}$ is not $(3x)^{-\frac{1}{2}}$ .

Key Example 2: Write $\dfrac{x^2 - 4}{2\sqrt{x}}$ as a sum of powers of $x$ .

\begin{aligned} \frac{x^2 - 4}{2\sqrt{x}} &= \frac{x^2}{2x^{\frac{1}{2}}} - \frac{4}{2x^{\frac{1}{2}}} \\ &= \frac{1}{2}x^{2 - \frac{1}{2}} - 2x^{-\frac{1}{2}} \\ &= \frac{1}{2}x^{\frac{3}{2}} - 2x^{-\frac{1}{2}} \end{aligned}

Note: We split the fraction into one term per part of the numerator, then use Dividing Indices on each term.

Key Question 1: Write $\dfrac{(2x - 5)\sqrt{9x}}{3x}$ as a sum of powers of $x$ .

Worked solution

Firstly, by Powers of Products, $\sqrt{9x} = (9x)^{\frac{1}{2}} = 9^{\frac{1}{2}}x^{\frac{1}{2}} = 3x^{\frac{1}{2}}$ . So:

\begin{aligned} \frac{(2x - 5)\sqrt{9x}}{3x} &= \frac{(2x - 5) \times 3x^{\frac{1}{2}}}{3x} \\ &= \frac{(2x - 5)x^{\frac{1}{2}}}{x} \\ &= \frac{2x^{\frac{3}{2}} - 5x^{\frac{1}{2}}}{x} \\ &= \frac{2x^{\frac{3}{2}}}{x} - \frac{5x^{\frac{1}{2}}}{x} \\ &= 2x^{\frac{1}{2}} - 5x^{-\frac{1}{2}} \end{aligned}

Key Question 2: Given that $\dfrac{4x^3 - 6\sqrt{x}}{2x^2}$ can be written in the form $2x^p - 3x^q$ , find the values of $p$ and $q$ .

Worked solution

Split the fraction, then use Dividing Indices on each term:

\begin{aligned} \frac{4x^3 - 6\sqrt{x}}{2x^2} &= \frac{4x^3}{2x^2} - \frac{6x^{\frac{1}{2}}}{2x^2} \\ &= 2x^{3-2} - 3x^{\frac{1}{2}-2} \\ &= 2x^1 - 3x^{-\frac{3}{2}} \end{aligned}

So $p = 1$ and $q = -\dfrac{3}{2}$ .

Equations with the Same Base

When both sides of an equation can be written as powers of the same number, we can compare the indices.

Key Example 1: Solve $4^x = 8^{x-1}$ .

\begin{aligned} 4^x &= 8^{x-1} \\ \left(2^2\right)^x &= \left(2^3\right)^{x-1} \\ 2^{2x} &= 2^{3x-3} \\ \Rightarrow \quad 2x &= 3x - 3 \\ x &= 3 \end{aligned}

Note: We write $4$ and $8$ as powers of $2$ , then use Power of a Power. Each power of $2$ is a different number, so if two powers of $2$ are equal, their indices must be equal too. That’s what the $\Rightarrow$ shows.

Consider: Check the answer. Do $4^3$ and $8^2$ really give the same number? How could you write both as a power of $2$ ?

Key Example 2: Solve $9^x = \dfrac{1}{\sqrt{3}}$ .

\begin{aligned} 9^x &= \frac{1}{\sqrt{3}} \\ \left(3^2\right)^x &= \frac{1}{3^{\frac{1}{2}}} \\ 3^{2x} &= 3^{-\frac{1}{2}} \\ \Rightarrow \quad 2x &= -\frac{1}{2} \\ x &= -\frac{1}{4} \end{aligned}

Note: The right-hand side needs Roots as Indices and Negative Indices before it’s a single power of $3$ .

Key Question 1: Solve $27^x = \dfrac{1}{9^{x-1}}$ .

Worked solution

Write both sides as powers of $3$ :

\begin{aligned} 27^x &= \frac{1}{9^{x-1}} \\ \left(3^3\right)^x &= \frac{1}{\left(3^2\right)^{x-1}} \\ 3^{3x} &= \frac{1}{3^{2x-2}} \\ 3^{3x} &= 3^{-(2x-2)} \\ 3^{3x} &= 3^{2-2x} \end{aligned}

Compare the indices:

\begin{aligned} \Rightarrow \quad 3x &= 2 - 2x \\ 5x &= 2 \\ x &= \frac{2}{5} \end{aligned}

Key Question 2: Given that $2^x \times 4^y = \dfrac{1}{2\sqrt{2}}$ , express $y$ in terms of $x$ . (June 2019, Paper 2, Q1)

Worked solution

Write the left-hand side as a single power of $2$ :

\begin{aligned} 2^x \times 4^y &= 2^x \times \left(2^2\right)^y \\ &= 2^x \times 2^{2y} \\ &= 2^{x + 2y} \end{aligned}

Then the right-hand side:

\begin{aligned} \frac{1}{2\sqrt{2}} &= \frac{1}{2^1 \times 2^{\frac{1}{2}}} \\ &= \frac{1}{2^{\frac{3}{2}}} \\ &= 2^{-\frac{3}{2}} \end{aligned}

Compare the indices and rearrange:

\begin{aligned} 2^{x + 2y} &= 2^{-\frac{3}{2}} \\ \Rightarrow \quad x + 2y &= -\frac{3}{2} \\ 2y &= -\frac{3}{2} - x \\ y &= -\frac{3}{4} - \frac{x}{2} \end{aligned}

Undoing a Power

To undo a power, raise both sides to the reciprocal power. It works because of Power of a Power: $\left(x^{\frac{3}{2}}\right)^{\frac{2}{3}} = x^{\frac{3}{2} \times \frac{2}{3}} = x^1 = x$ .

Key Example 1: Solve $x^{\frac{3}{2}} = 8$ .

\begin{aligned} x^{\frac{3}{2}} &= 8 \\ \left(x^{\frac{3}{2}}\right)^{\frac{2}{3}} &= 8^{\frac{2}{3}} \\ x &= \left(\sqrt[3]{8}\right)^2 \\ x &= 4 \end{aligned}

Note: $x^{\frac{3}{2}}$ contains $\sqrt{x}$ , so $x$ can’t be negative and there’s only one answer. An even power with no root is different: $x^4 = 81$ gives $x = \pm 3$ , as both $3^4$ and $(-3)^4$ are $81$ .

Key Example 2: Solve $5x^{\frac{3}{4}} - 2 = 38$ .

Firstly, get the power on its own:

\begin{aligned} 5x^{\frac{3}{4}} &= 40 \\ x^{\frac{3}{4}} &= 8 \end{aligned}

Then raise both sides to the power $\dfrac{4}{3}$ :

\begin{aligned} x &= 8^{\frac{4}{3}} \\ &= \left(\sqrt[3]{8}\right)^4 \\ &= 2^4 \\ &= 16 \end{aligned}

Key Question 1: Solve $2x^{-\frac{3}{2}} = 54$ .

Worked solution

Get the power on its own, then raise both sides to the power $-\dfrac{2}{3}$ :

\begin{aligned} x^{-\frac{3}{2}} &= 27 \\ x &= 27^{-\frac{2}{3}} \\ &= \frac{1}{\left(\sqrt[3]{27}\right)^2} \\ &= \frac{1}{3^2} \\ &= \frac{1}{9} \end{aligned}

Key Question 2: The height above the sea, $h$ metres, of a sea bird $t$ seconds after it leaves its nest is modelled by $h = 20 - 0.3t^{1.5}$ , where $h \geq 0$ and $0 \leq t \leq T$ . Find, according to the model, the limitation on the value of $T$ . (extracted from June 2025, Paper 1, Q6)

Worked solution

The model stops when the bird reaches the sea, so $h = 0$ when $t = T$ :

\begin{aligned} 0 &= 20 - 0.3T^{1.5} \\ 0.3T^{1.5} &= 20 \\ T^{1.5} &= \frac{200}{3} \end{aligned}

Since $1.5 = \dfrac{3}{2}$ , raise both sides to the power $\dfrac{2}{3}$ :

T = \left(\frac{200}{3}\right)^{\frac{2}{3}} = 16.4 \text{ (3 s.f.)}

So $T \leq 16.4$ : the model only works for the first $16.4$ seconds.

Hidden Quadratics

Some equations with indices are quadratics in disguise. Once you spot one power that is the square of another, you can factorise just like at GCSE.

Key Example 1: Solve $4^x - 5 \times 2^x + 4 = 0$ .

Firstly, $4^x = \left(2^2\right)^x = \left(2^x\right)^2$ , so:

\left(2^x\right)^2 - 5 \times 2^x + 4 = 0

To make the quadratic easier to see, let $u = 2^x$ :

\begin{aligned} u^2 - 5u + 4 &= 0 \\ (u - 1)(u - 4) &= 0 \end{aligned}

So $u = 1$ or $u = 4$ . Now swap $2^x$ back in:

2^x = 1 \Rightarrow x = 0 \qquad 2^x = 4 \Rightarrow x = 2

Note: The key step uses Power of a Power in reverse. Multiplication is commutative, so $\left(2^2\right)^x = 2^{2x} = \left(2^x\right)^2$ . Then $2^x = 1$ gives $x = 0$ by Zero Index.

Key Example 2: Solve $9^x - 2 \times 3^x - 3 = 0$ .

Since $9^x = \left(3^x\right)^2$ , let $u = 3^x$ :

\begin{aligned} u^2 - 2u - 3 &= 0 \\ (u - 3)(u + 1) &= 0 \end{aligned}

So $u = 3$ or $u = -1$ . Swapping back:

3^x = 3 \Rightarrow x = 1 \qquad 3^x = -1 \text{ has no solutions}

Note: A positive base raised to any power is always positive, so $3^x$ can never equal $-1$ . Always check each answer for $u$ before you swap back.

Key Question 1: Solve $4^x - 2^{x+1} - 8 = 0$ .

Worked solution

Firstly, $4^x = \left(2^x\right)^2$ and $2^{x+1} = 2^x \times 2^1 = 2 \times 2^x$ . So let $u = 2^x$ :

\begin{aligned} u^2 - 2u - 8 &= 0 \\ (u - 4)(u + 2) &= 0 \end{aligned}

So $u = 4$ or $u = -2$ . Swapping back:

2^x = 4 \Rightarrow x = 2 \qquad 2^x = -2 \text{ has no solutions}

Key Question 2: Solve $r^{10} - 4r^5 + 3 = 0$ . (extracted from October 2020, Paper 2, Q15)

Worked solution

Since $r^{10} = \left(r^5\right)^2$ , let $u = r^5$ :

\begin{aligned} u^2 - 4u + 3 &= 0 \\ (u - 1)(u - 3) &= 0 \end{aligned}

So $u = 1$ or $u = 3$ . Swapping back:

r^5 = 1 \Rightarrow r = 1 \qquad r^5 = 3 \Rightarrow r = 3^{\frac{1}{5}}