Indices and Surds – A Level Maths Year 1

In $a^n$ the number $a$ is the base and $n$ is the index (or power, or exponent). At GCSE the index was usually a positive whole number. At A Level the index can be any rational number — negative, zero or a fraction — and the same three laws still hold.

The three laws (same base $a$)

$a^m \times a^n = a^{m+n}$ · $a^m \div a^n = a^{m-n}$ · $\left(a^m\right)^n = a^{mn}$

Law	In words	Example
$a^m \times a^n = a^{m+n}$	Multiplying: add the indices	$x^3 \times x^5 = x^8$
$a^m \div a^n = a^{m-n}$	Dividing: subtract the indices	$x^7 \div x^2 = x^5$
$\left(a^m\right)^n = a^{mn}$	Power of a power: multiply	$\left(x^4\right)^3 = x^{12}$
$(ab)^n = a^n b^n$	A power applies to every factor	$\left(2x\right)^3 = 8x^3$

The laws only work with the same base. $2^3 \times 3^2$ cannot be simplified to a single power, and $x^2 + x^3$ is not $x^5$ — the laws are for multiplying and dividing, never for adding.

Worked Example 1 — Simplifying with several laws

Simplify $\dfrac{\left(2x^3\right)^4}{8x^5}$.

①The power applies to the $2$ as well: $\left(2x^3\right)^4 = 2^4 x^{12} = 16x^{12}$.

②Divide the numbers and subtract the indices: $\dfrac{16}{8}\,x^{12-5}$.

③$= 2x^7$

The definitions below are not new rules to memorise blindly — each is forced by the laws in Section 1. For example $a^3 \div a^3 = a^0$ by the division law, but the answer is obviously $1$, so $a^0$ must be $1$.

The definitions ($a \neq 0$)

$a^0 = 1$ · $a^{-n} = \dfrac{1}{a^n}$ · $a^{\frac{1}{n}} = \sqrt[n]{a}$ · $a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m$

Reading a fractional index. In $a^{\frac{m}{n}}$ the denominator is the root and the numerator is the power. A negative sign means "one over". Take the root first — it keeps the numbers small.

Worked Example 2 — Evaluating without a calculator

Evaluate (a) $27^{-\frac{2}{3}}$ (b) $\left(\dfrac{16}{81}\right)^{\frac{3}{4}}$.

①(a) Cube root first: $\sqrt[3]{27} = 3$.

②Square it: $3^2 = 9$. The negative index means reciprocal: $27^{-\frac{2}{3}} = \dfrac{1}{9}$.

③(b) Fourth roots of top and bottom: $\sqrt[4]{16} = 2$ and $\sqrt[4]{81} = 3$.

④Cube: $\left(\dfrac{2}{3}\right)^3 = \dfrac{8}{27}$.

Worked Example 3 — Writing in the form $ax^p + bx^q$

Write $\dfrac{3x^2 - 2\sqrt{x}}{x}$ in the form $ax^p + bx^q$.

①Write the root as a power: $\sqrt{x} = x^{\frac{1}{2}}$.

②Split the fraction: $\dfrac{3x^2}{x} - \dfrac{2x^{\frac12}}{x}$.

③Subtract indices: $3x^{2-1} - 2x^{\frac12 - 1} = 3x - 2x^{-\frac{1}{2}}$.

Why this matters: this exact skill is the first step of almost every differentiation and integration question. You cannot differentiate a fraction term by term — you must rewrite it as a sum of powers of $x$ first.

The coefficient is not raised to the power unless it is inside the bracket. $\dfrac{1}{2x^3} = \dfrac{1}{2}x^{-3}$, but $\dfrac{1}{(2x)^3} = \dfrac{1}{8}x^{-3}$. Also, $2x^{-1} = \dfrac{2}{x}$, not $\dfrac{1}{2x}$.

Same base, same index

If $a^p = a^q$ (with $a > 0$, $a \neq 1$) then $p = q$

Worked Example 4 — Writing both sides with a common base

Solve $4^{x+1} = 8^x$.

①Both $4$ and $8$ are powers of $2$: $4 = 2^2$ and $8 = 2^3$.

②$\left(2^2\right)^{x+1} = \left(2^3\right)^x \;\Rightarrow\; 2^{2x+2} = 2^{3x}$

③Equate indices: $2x + 2 = 3x$, so $x = 2$.

④Check: $4^3 = 64$ and $8^2 = 64$ ✓

Worked Example 5 — Undoing a fractional power

Solve $x^{\frac{3}{2}} = 8$.

①Raise both sides to the reciprocal power $\tfrac{2}{3}$: $x = 8^{\frac{2}{3}}$.

②$\sqrt[3]{8} = 2$, and $2^2 = 4$, so $x = 4$.

③Check: $4^{\frac32} = \left(\sqrt{4}\right)^3 = 2^3 = 8$ ✓

Equations that need logarithms, such as $3^x = 20$, are covered in the Exponentials and Logarithms topic. Here the numbers are chosen so a common base exists.

A surd is an irrational root such as $\sqrt{2}$ or $\sqrt{15}$. Leaving answers in surd form keeps them exact; a decimal would be rounded. Questions that say "give your answer in exact form" or "in the form $a + b\sqrt{c}$" expect surds, and a calculator decimal earns no credit.

The two rules

$\sqrt{ab} = \sqrt{a}\,\sqrt{b}$ · $\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$ (for $a, b > 0$)

There is no rule for adding. $\sqrt{9 + 16} = \sqrt{25} = 5$, but $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$. So $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$.

Worked Example 6 — Simplifying and collecting

Simplify $\sqrt{12} + \sqrt{75}$.

①Find the largest square factor of each: $12 = 4 \times 3$ and $75 = 25 \times 3$.

②$\sqrt{12} = 2\sqrt{3}$ and $\sqrt{75} = 5\sqrt{3}$.

③They are now "like terms": $2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3}$.

Worked Example 7 — Expanding brackets with surds

Expand and simplify $\left(3 + \sqrt{2}\right)\left(5 - 2\sqrt{2}\right)$.

①Every term times every term: $15 - 6\sqrt{2} + 5\sqrt{2} - 2\left(\sqrt{2}\right)^2$.

②$\left(\sqrt{2}\right)^2 = 2$, so the last term is $-4$.

③$= 15 - 4 - \sqrt{2} = 11 - \sqrt{2}$

Rationalising means rewriting a fraction so that there is no surd in the denominator. You multiply top and bottom by the same thing, so the value does not change.

Denominator	Multiply top and bottom by	Why it works
$\sqrt{b}$	$\sqrt{b}$	$\sqrt{b}\times\sqrt{b} = b$
$a + \sqrt{b}$	$a - \sqrt{b}$	$(a+\sqrt{b})(a-\sqrt{b}) = a^2 - b$
$a - \sqrt{b}$	$a + \sqrt{b}$	difference of two squares again
$\sqrt{a} - \sqrt{b}$	$\sqrt{a} + \sqrt{b}$	$(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}) = a - b$

The conjugate. $a - \sqrt{b}$ is called the conjugate of $a + \sqrt{b}$. Multiplying a pair of conjugates is the difference of two squares, and the surd terms cancel exactly.

Worked Example 8 — Two rationalisations

Rationalise (a) $\dfrac{6}{\sqrt{3}}$ (b) $\dfrac{4}{3 - \sqrt{5}}$.

①(a) $\dfrac{6}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$

②(b) Multiply by the conjugate: $\dfrac{4\left(3 + \sqrt{5}\right)}{\left(3 - \sqrt{5}\right)\left(3 + \sqrt{5}\right)}$

③Denominator: $9 - 5 = 4$.

④$\dfrac{4\left(3+\sqrt{5}\right)}{4} = 3 + \sqrt{5}$

Worked Example 9 — Answer in the form $a + b\sqrt{c}$

Express $\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}$ in the form $a + b\sqrt{3}$, where $a$ and $b$ are integers.

①Multiply top and bottom by $2 + \sqrt{3}$.

②Denominator: $4 - 3 = 1$.

③Numerator: $\left(2 + \sqrt{3}\right)^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$.

④So $a = 7$, $b = 4$. Check: $\dfrac{3.732}{0.268} \approx 13.93$ and $7 + 4(1.732) \approx 13.93$ ✓

Worked Example 10 — An equation with surd coefficients

Solve $x\sqrt{8} = x\sqrt{2} + 3$, giving $x$ with a rational denominator.

①$\sqrt{8} = 2\sqrt{2}$, so $2\sqrt{2}\,x - \sqrt{2}\,x = 3$.

②$\sqrt{2}\,x = 3$, so $x = \dfrac{3}{\sqrt{2}}$.

③Rationalise: $x = \dfrac{3\sqrt{2}}{2}$.

Multiply / divide

Add / subtract the indices (same base only).

Power of a power

$\left(a^m\right)^n = a^{mn}$; applies to every factor in the bracket.

Zero index

$a^0 = 1$ for any $a \neq 0$.

Negative index

$a^{-n} = \dfrac{1}{a^n}$ — reciprocal, not negative.

Fractional index

$a^{\frac{m}{n}}$: denominator is the root, numerator the power.

Equations

Write both sides with the same base, then equate indices.

Simplifying surds

Pull out the largest square factor: $\sqrt{50} = 5\sqrt{2}$.

No adding rule

$\sqrt{a+b} \neq \sqrt{a}+\sqrt{b}$.

Rationalising

Multiply by $\sqrt{b}$, or by the conjugate $a \mp \sqrt{b}$.

Exact form

Leave surds in; a decimal is not exact.

Answer these without a calculator — that is how indices and surds are usually tested.

Question 1

Evaluate (a) $64^{\frac{2}{3}}$ (b) $32^{-\frac{3}{5}}$ (c) $\left(\dfrac{9}{4}\right)^{-\frac{1}{2}}$.

▶ Show solution

(a) $\sqrt[3]{64} = 4$, then $4^2 = 16$.

(b) $\sqrt[5]{32} = 2$, then $2^3 = 8$; the negative index gives $\dfrac{1}{8}$.

(c) The negative index flips the fraction: $\left(\dfrac{4}{9}\right)^{\frac12} = \dfrac{2}{3}$.

Question 2

Simplify $\dfrac{6x^5y^2}{4x^2y^{-1}}$.

▶ Show solution

Numbers: $\dfrac{6}{4} = \dfrac{3}{2}$.

$x$: $x^{5-2} = x^3$. $y$: $y^{2-(-1)} = y^3$.

$\dfrac{3}{2}x^3y^3$

Question 3

Write $\dfrac{x^2 + 4}{\sqrt{x}}$ as a sum of powers of $x$.

▶ Show solution

Divide each term by $x^{\frac12}$:

$\dfrac{x^2}{x^{\frac12}} + \dfrac{4}{x^{\frac12}} = x^{\frac{3}{2}} + 4x^{-\frac{1}{2}}$

Question 4

Simplify $\sqrt{50} - \sqrt{18} + \sqrt{8}$.

▶ Show solution

$\sqrt{50} = 5\sqrt{2}$, $\sqrt{18} = 3\sqrt{2}$, $\sqrt{8} = 2\sqrt{2}$.

$5\sqrt{2} - 3\sqrt{2} + 2\sqrt{2} = 4\sqrt{2}$

Question 5

Expand and simplify $\left(2\sqrt{3} - 1\right)^2$.

▶ Show solution

$\left(2\sqrt{3}\right)^2 - 2\left(2\sqrt{3}\right)(1) + 1^2$

$= 4 \times 3 - 4\sqrt{3} + 1 = 13 - 4\sqrt{3}$

Question 6

Rationalise the denominator of $\dfrac{10}{\sqrt{7} - \sqrt{2}}$, simplifying your answer.

▶ Show solution

Multiply top and bottom by $\sqrt{7} + \sqrt{2}$.

Denominator: $7 - 2 = 5$.

$\dfrac{10\left(\sqrt{7} + \sqrt{2}\right)}{5} = 2\sqrt{7} + 2\sqrt{2}$

Question 7

Solve $9^x = 27^{x-1}$.

▶ Show solution

Base $3$: $\left(3^2\right)^x = \left(3^3\right)^{x-1}$, so $3^{2x} = 3^{3x-3}$.

$2x = 3x - 3$, giving $x = 3$.

Check: $9^3 = 729$ and $27^2 = 729$ ✓

Question 8

Express $\dfrac{5 + \sqrt{3}}{2 - \sqrt{3}}$ in the form $a + b\sqrt{3}$, where $a$ and $b$ are integers.

▶ Show solution

Multiply top and bottom by $2 + \sqrt{3}$. The denominator becomes $4 - 3 = 1$.

Numerator: $(5 + \sqrt{3})(2 + \sqrt{3}) = 10 + 5\sqrt{3} + 2\sqrt{3} + 3$.

$= 13 + 7\sqrt{3}$, so $a = 13$, $b = 7$.

Question 9

(a) Given $y = 2^x$, write $4^x$ and $2^{x+3}$ in terms of $y$. (b) Hence solve $4^x - 2^{x+3} + 16 = 0$.

▶ Show solution

(a) $4^x = \left(2^2\right)^x = \left(2^x\right)^2 = y^2$, and $2^{x+3} = 2^x \times 2^3 = 8y$.

(b) The equation becomes $y^2 - 8y + 16 = 0$, i.e. $(y - 4)^2 = 0$.

So $y = 4$, meaning $2^x = 4 = 2^2$, and $x = 2$.

Check: $16 - 32 + 16 = 0$ ✓

Question 10

A rectangle has sides of length $\left(3 + \sqrt{5}\right)$ cm and $\left(3 - \sqrt{5}\right)$ cm. Giving every answer in exact form, find:

(a) its area (b) its perimeter (c) the length of its diagonal, in the form $k\sqrt{7}$ (d) the ratio of the longer side to the shorter side, in the form $\dfrac{a + b\sqrt{5}}{2}$.

▶ Show solution

(a) Difference of two squares: $(3 + \sqrt{5})(3 - \sqrt{5}) = 9 - 5 = 4$ cm$^2$.

(b) $2(3 + \sqrt{5}) + 2(3 - \sqrt{5}) = 12$ cm — the surds cancel.

(c) $(3 + \sqrt{5})^2 = 14 + 6\sqrt{5}$ and $(3 - \sqrt{5})^2 = 14 - 6\sqrt{5}$.

By Pythagoras, $d^2 = 28$, so $d = \sqrt{28} = 2\sqrt{7}$ cm.

(d) $\dfrac{3 + \sqrt{5}}{3 - \sqrt{5}} \times \dfrac{3 + \sqrt{5}}{3 + \sqrt{5}} = \dfrac{14 + 6\sqrt{5}}{4} = \dfrac{7 + 3\sqrt{5}}{2}$

Check: $\dfrac{5.236}{0.764} \approx 6.85$ and $\dfrac{7 + 6.708}{2} \approx 6.85$ ✓