N3, N6, N7 - Powers, roots and the order of operations

Use powers, roots and integer indices to calculate exactly. Brackets and operation order tell you what an expression means; inverse operations and familiar powers help you check the result.

Priority and inverse operations

Can changing just one sign turn $18 - 2 \times (3 + 1)^2 \div 4$ from $10$ into $16$ ? You need an agreed order before the question has a definite answer.

You already know

Addition and subtraction are inverse operations, as are multiplication and division by a non-zero number. A reciprocal reverses multiplication: $12 \times \frac14 = 12 \div 4 = 3$ .

First work inside brackets, starting with the innermost pair. Next evaluate powers and roots. Then perform multiplication and division from left to right. Finally perform addition and subtraction from left to right. Each pair has equal priority: multiplication does not always precede division, and addition does not always precede subtraction.

A fraction bar groups the entire numerator and denominator. Thus $\frac{6+10}{2 \times 4} = \frac{16}{8} = 2$ . On a calculator, enter $(6+10) \div (2 \times 4)$ , with both pairs of brackets. Similarly, $2\sqrt{49} = 2 \times 7 = 14$ : the root is evaluated before the multiplication.

Worked example 1: following operation priority

Question. Work out $18 - 2 \times (3 + 1)^2 \div 4$ .

Plan. Evaluate the bracket, then its power, then the multiplication and division, then subtract.

\begin{aligned} 18 - 2 \times (3+1)^2 \div 4 &= 18 - 2 \times 4^2 \div 4 && \text{bracket first} \\ &= 18 - 2 \times 16 \div 4 && \text{evaluate the square} \\ &= 18 - 32 \div 4 && \text{multiply before moving right} \\ &= 18 - 8 = 10 && \text{divide, then subtract} \end{aligned}

Why not calculate $18 - 2$ first? That subtraction is outside the product and has lower priority.

Answer. $10$ .

Check. The subtracted amount is $8$ ; $10 + 8 = 18$ reverses the final operation.

Change only the plus sign inside the bracket to a minus sign. The order of the steps stays the same.

Worked example 2: changing one operation

Question. Work out $18 - 2 \times (3 - 1)^2 \div 4$ .

Plan. Use the same order, with the new bracket value.

\begin{aligned} 18 - 2 \times (3-1)^2 \div 4 &= 18 - 2 \times 2^2 \div 4 && \text{bracket first} \\ &= 18 - 2 \times 4 \div 4 && \text{evaluate the square} \\ &= 18 - 8 \div 4 && \text{multiply from the left} \\ &= 18 - 2 = 16 && \text{divide, then subtract} \end{aligned}

Why does one changed sign matter so much? It changes the bracket before squaring; the square changes from $16$ to $4$ .

Answer. $16$ .

Check. $16 + 2 = 18$ , and the subtracted amount $2$ is smaller than the previous $8$ .

To undo a chain, reverse its order as well as its operations. Starting with $7$ , multiply by $3$ , then add $5$ , to get $26$ . To recover $7$ , subtract $5$ first, then divide by $3$ .

Watch out

Wrong: $24 \div 6 \times 2 = 24 \div 12 = 2$ . Right: $24 \div 6 \times 2 = 4 \times 2 = 8$ . Multiplication and division have equal priority, so work left to right.

Change only the starting number in the second example.

Positive, zero and negative indices

A power records repeated multiplication; its index counts factors, not additions.

Base and index

In $3^4 = 3 \times 3 \times 3 \times 3 = 81$ , the base is $3$ and the index, or exponent, is $4$ . A positive integer index tells you how many copies of the base are multiplied.

For example, $3^4$ is not $3 \times 4$ .

Recognise familiar powers so that you can also recognise their roots. Each step across a row multiplies by that row's base.

Base	First power	Square	Cube	Fourth power	Fifth power
$2$	$2$	$4$	$8$	$16$	$32$
$3$	$3$	$9$	$27$	$81$	$243$
$4$	$4$	$16$	$64$	$256$	$1024$
$5$	$5$	$25$	$125$	$625$	$3125$

Moving backwards divides by the base. Continuing $2^2 = 4$ , $2^1 = 2$ gives $2^0 = 1$ , $2^{-1} = \frac12$ and $2^{-2} = \frac14$ . The negative index therefore gives a reciprocal.

Zero and negative integer indices

For $a \ne 0$ and a positive integer $n$ ,

a^0 = 1 \qquad a^{-n} = \frac1{a^n}

Not on the formula sheet: learn it.

The non-zero condition matters: zero has no reciprocal. Do not use this rule to assign a value to $0^0$ or a negative power of zero.

Worked example 3: a positive power

Question. Work out $5^3$ .

Plan. Multiply three copies of five.

\begin{aligned} 5^3 &= 5 \times 5 \times 5 && \text{three equal factors} \\ &= 25 \times 5 = 125 && \text{multiply in stages} \end{aligned}

Why not $5 \times 3$ ? The index counts copies in a product; it is not an ordinary multiplier.

Answer. $125$ .

Check. $125 \div 5 = 25 = 5^2$ , so one backward step agrees with the power pattern.

Change only the sign of the index. The base remains positive.

Worked example 4: a negative index

Question. Work out $5^{-3}$ . Give an exact answer.

Plan. Evaluate the positive power, then take its reciprocal.

\begin{aligned} 5^{-3} &= \frac1{5^3} && \text{negative index means reciprocal} \\ &= \frac1{5 \times 5 \times 5} && \text{expand the positive power} \\ &= \frac1{125} && \text{evaluate the denominator} \end{aligned}

Why is the answer positive? The base is positive; a minus sign in the index tells you to invert, not to negate.

Answer. $\frac1{125}$ .

Check. $125 \times \frac1{125} = 1$ , as reciprocal values must give.

A negative base is a separate issue. Brackets decide whether the sign is included: $(-3)^2 = 9$ , whereas $-3^2 = -(3^2) = -9$ . An even power of a negative base is positive because its negative factors pair up; an odd power is negative. Thus $(-2)^3 = -8$ and $(-2)^{-3} = -\frac18$ .

Watch out

Wrong: $5^{-3} = -125$ . Right: $5^{-3} = \frac1{125}$ . A negative index and a negative base do different jobs.

Change only the base in the negative-index example.

Roots and the meaning of the square-root sign

A root reverses a power: it asks which number, raised to that power, gives the input.

Principal square root

For $a \ge 0$ , $\sqrt a$ means the non-negative number whose square is $a$ . Thus $\sqrt{49} = 7$ , although both $7$ and $-7$ solve $x^2 = 49$ .

No real number squares to a negative number, so $\sqrt{-9}$ has no real value.

Even-root symbols also choose the non-negative root: $\sqrt[4]{81} = 3$ . Odd roots can be negative and have just one real value: $\sqrt[3]{-125} = -5$ , because $(-5)^3 = -125$ . Similarly, $\sqrt[5]{32} = 2$ .

Roots of decimals use exactly the same meaning as roots of whole numbers.

Worked example 5: the root of a decimal

Question. Work out $\sqrt{0.09}$ .

Plan. Find the non-negative decimal whose square is $0.09$ .

\begin{aligned} 0.3^2 &= \frac3{10} \times \frac3{10} && \text{write tenths as fractions} \\ &= \frac9{100} = 0.09 && \text{multiply} \\ \sqrt{0.09} &= 0.3 && \text{choose the non-negative root} \end{aligned}

Why not $0.03$ ? Its square has four decimal places: $0.03^2 = 0.0009$ .

Answer. $0.3$ .

Check. Squaring $0.3$ returns the original input exactly.

Change only the decimal scale of the input: divide it by $100$ . The square root is divided by $10$ .

Worked example 6: a smaller decimal input

Question. Work out $\sqrt{0.0009}$ .

Plan. Test $0.03$ , keeping both decimal scale factors when squaring.

\begin{aligned} 0.03^2 &= \frac3{100} \times \frac3{100} && \text{write hundredths as fractions} \\ &= \frac9{10000} = 0.0009 && \text{multiply} \\ \sqrt{0.0009} &= 0.03 && \text{choose the non-negative root} \end{aligned}

Why does the root have two decimal places here? Squaring $\frac3{100}$ squares its denominator as well as its numerator.

Answer. $0.03$ .

Check. $0.03 \times 0.03 = 0.0009$ . The answer is positive and below $0.1$ .

For fractions, $\sqrt{\frac4{25}} = \frac25$ , checked by $(\frac25)^2 = \frac4{25}$ . Evaluate familiar roots before combining them: $\sqrt{64} + \sqrt[3]{27} = 8 + 3 = 11$ .

Watch out

A root covers only what is under its bar. $\sqrt{9+16} = 5$ , but $\sqrt9 + \sqrt{16} = 7$ . A square root does not distribute over addition.

Keep the decimal scale from the second example and change the non-zero digit.

Combining powers with the same base

Index laws count or cancel equal factors without writing every factor out.

For example, $2^3 \times 2^4$ contains seven factors of two, so it equals $2^7$ . Cancelling two factors in $\frac{2^5}{2^2}$ leaves $2^3$ . In $(2^3)^4$ , four groups of three factors give $2^{12}$ .

Index laws

For integer indices,

a^m a^n = a^{m+n} \qquad \frac{a^m}{a^n} = a^{m-n} \qquad (a^m)^n = a^{mn}

Use a non-zero base for quotients or non-positive indices. The product and quotient rules combine powers of the same base.

Not on the formula sheet: learn it.

These rules include zero and negative indices: $\frac{3^2}{3^2} = 3^0 = 1$ , while $\frac{3^2}{3^5} = 3^{-3} = \frac1{27}$ .

Worked example 7: multiplying matching bases

Question. Work out $5^{-2} \times 5^4$ .

Plan. Add the indices, then evaluate the resulting power.

\begin{aligned} 5^{-2} \times 5^4 &= 5^{-2+4} && \text{same base, so add indices} \\ &= 5^2 = 25 && \text{evaluate the power} \end{aligned}

Why can we combine these powers? Both bases are $5$ ; the exponents describe factors of the same number.

Answer. $25$ .

Check. $5^{-2} = \frac1{25}$ and $5^4 = 625$ ; $625 \div 25 = 25$ .

Change only multiplication to division. Now the denominator's factors are removed, so subtract its index.

Worked example 8: dividing matching bases

Question. Work out $5^{-2} \div 5^4$ . Give an exact answer.

Plan. Subtract the denominator's index, then use the negative-index rule.

\begin{aligned} 5^{-2} \div 5^4 &= 5^{-2-4} && \text{same base, so subtract indices} \\ &= 5^{-6} && \text{combine the signed indices} \\ &= \frac1{5^6} = \frac1{15625} && \text{take the reciprocal} \end{aligned}

Why is the result much smaller? Dividing $\frac1{25}$ by $625$ makes its denominator $25 \times 625$ .

Answer. $\frac1{15625}$ .

Check. $\frac1{15625} \times 625 = \frac1{25}$ , recovering the dividend.

The power-of-a-power rule also gives $(2^{-1})^3 = 2^{-3} = \frac18$ . For integer powers, $(ab)^n = a^n b^n$ ; for example, $(2 \times 3)^2 = 2^2 \times 3^2 = 36$ .

For non-negative inputs, $\sqrt{ab} = \sqrt a\sqrt b$ , and $\sqrt{a/b} = \sqrt a/\sqrt b$ when $b > 0$ . Thus $\sqrt{25 \times 16} = 5 \times 4 = 20$ , and $\frac{3^4}{3^2} + \sqrt{81} = 9 + 9 = 18$ . These are product rules, not addition rules.

Watch out

Wrong: $2^3 + 2^4 = 2^7$ . Right: $8 + 16 = 24$ . Adding indices describes multiplying powers, not adding them. Different bases, such as $2^3 \times 3^4$ , cannot be combined by adding indices either.

Powers can create surprisingly large changes from small inputs: $2^{10} = 1024$ , already more than a thousand. Clear notation lets you reason about them without losing track of the operations.

Quick Recap

Brackets come first, then powers and roots, then multiplication/division, then addition/subtraction.
A positive integer index counts factors; a negative index gives a reciprocal.
A square-root symbol gives one non-negative value; a squared equation may have two real solutions.
Index laws combine matching bases by counting or cancelling factors.

Yes: changing $(3+1)$ to $(3-1)$ changes its square from $16$ to $4$ , so the opening expression changes from $10$ to $16$ .

Last Accessed

Papers 1, 2 and 3

N3, N6, N7 - Powers, roots and the order of operations

Priority and inverse operations

Positive, zero and negative indices

Roots and the meaning of the square-root sign

Combining powers with the same base

Quick Recap

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