Maths · 1 Number

1.7 Indices I

Before you start

Square numbers up to $15^{2} = 225$ and the cube numbers $1, 8, 27, 64, 125$ and $1000$ , with their square roots and cube roots.
What a reciprocal is: the reciprocal of $5$ is $\frac{1}{5}$ and the reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$ (subtopic 1.1).
Adding, subtracting and multiplying negative integers, e.g. $3 - (- 2) = 5$ and $(- 2) \times (- 3) = 6$ .
Multiplying and dividing fractions, e.g. $1 \div \frac{1}{4} = 4$ .

Syllabus objectives covered (2)

Understand and use indices (positive, zero, negative and fractional).
Understand and use the rules of indices.

Part 1 of 7

What an index tells you

In $2^{5}$ , the $2$ is the base and the small raised $5$ is the index. You will also hear it called the power or the exponent. The index counts how many copies of the base are multiplied together.
$2^{5} = 2 \times 2 \times 2 \times 2 \times 2 = 32$
Index form is shorthand for repeated multiplication. $3 \times 3 \times 3 \times 3 \times 5 \times 5$ has four 3s and two 5s, so it is $3^{4} \times 5^{2}$ . Count each base on its own. Different bases stay separate.
An index of 1 means one copy: $7^{1} = 7$ . You do not normally write the 1, but you need it when you use the rules later, because $7$ is the same as $7^{1}$ .
The index applies only to what is immediately before it. In $- 3^{2}$ the index is on the 3 alone, so it means $- (3 \times 3) = - 9$ . In ${(- 3)}^{2}$ the bracket puts the index on $- 3$ , so it means $(- 3) \times (- 3) = 9$ .
A negative base with an even index gives a positive answer. With an odd index it gives a negative answer. So ${(- 2)}^{4} = 16$ but ${(- 2)}^{3} = - 8$ , because each pair of negatives multiplies to a positive and an odd index leaves one negative over.
The index is not a multiplier. $2^{5}$ is $32$ , not $2 \times 5 = 10$ . Write the multiplication out once if you feel unsure, and the mistake cannot happen.

Part 2 of 7

Multiplying and dividing powers of the same base

Write $2^{3} \times 2^{4}$ out in full. That is three 2s followed by four 2s, which is seven 2s in a row. So $2^{3} \times 2^{4} = 2^{7}$ .
$2^{3} \times 2^{4} = (2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2) = 2^{7}$
That gives the first rule. When you multiply powers of the same base, add the indices. The base does not change.
$a^{m} \times a^{n} = a^{m + n}$
Now divide. $2^{5} \div 2^{2}$ has five 2s on top and two 2s underneath. Each 2 underneath cancels one 2 on top, which leaves three 2s. So $2^{5} \div 2^{2} = 2^{3}$ .
$\frac{2^{5}}{2^{2}} = \frac{2 \times 2 \times 2 \times 2 \times 2}{2 \times 2} = 2 \times 2 \times 2 = 2^{3}$
That gives the second rule. When you divide powers of the same base, subtract the indices. The order matters: the index of the number being divided, minus the index of the number you divide by.
$a^{m} \div a^{n} = a^{m - n}$
Both rules need the same base. $2^{3} \times 3^{2}$ is $8 \times 9 = 72$ , and it cannot be written as one power. Do not add the indices, and do not multiply the bases.
The base is never multiplied. $2^{3} \times 2^{4}$ is $2^{7} = 128$ , not $4^{7}$ . Check with the values: $8 \times 16 = 128$ .

Part 3 of 7

A power of a power, and powers of brackets

${(2^{3})}^{2}$ means $2^{3}$ squared, that is $2^{3} \times 2^{3}$ . By the first rule that is $2^{3 + 3} = 2^{6}$ .
Two copies of three 2s is six 2s. So when you raise a power to a power, multiply the indices.
${(a^{m})}^{n} = a^{m n}$
So ${(2^{3})}^{2} = 2^{6} = 64$ . The trap is to add and get $2^{5}$ . Ask yourself which shape you have. Two powers multiplied together means add. A bracket with an index outside it means multiply.
A power outside a bracket applies to every factor inside it. ${(2 \times 5)}^{3} = 2^{3} \times 5^{3} = 8 \times 125 = 1000$ , which matches $10^{3} = 1000$ .
For a fraction, the power applies to the top and to the bottom.
${(\frac{2}{3})}^{2} = \frac{2^{2}}{3^{2}} = \frac{4}{9}$
This works for a product, not for a sum. ${(2 + 3)}^{2} = 5^{2} = 25$ , but $2^{2} + 3^{2} = 4 + 9 = 13$ . Work out what is inside an addition bracket first.

Part 4 of 7

Zero and negative indices

Use the division rule on $2^{3} \div 2^{3}$ . Subtracting the indices gives $2^{3 - 3} = 2^{0}$ . But any number divided by itself is 1. Both answers must be the same, so $2^{0} = 1$ .
The same argument works for any base except zero. So $7^{0} = 1$ , $1000^{0} = 1$ and ${(\frac{3}{4})}^{0} = 1$ . An index of 0 does not give 0.
$a^{0} = 1 (a \neq 0)$
Now try $2^{3} \div 2^{5}$ . Subtracting gives $2^{3 - 5} = 2^{- 2}$ . Cancelling gives $\frac{2 \times 2 \times 2}{2 \times 2 \times 2 \times 2 \times 2} = \frac{1}{2 \times 2}$ . So $2^{- 2} = \frac{1}{2^{2}} = \frac{1}{4}$ .
So a negative index means 'take the reciprocal'. It does not make the number negative.
$a^{- n} = \frac{1}{a^{n}}$
A pattern shows the same thing. Each time the index goes down by 1, you divide by the base once more. Below $2^{0}$ you keep dividing by 2, so you get fractions, not negative numbers.
$2^{3} = 8, 2^{2} = 4, 2^{1} = 2, 2^{0} = 1, 2^{- 1} = \frac{1}{2}, 2^{- 2} = \frac{1}{4}, 2^{- 3} = \frac{1}{8}$
So $7^{- 2} = \frac{1}{7^{2}} = \frac{1}{49}$ and $10^{- 3} = \frac{1}{1000} = 0.001$ . Give the answer as an exact fraction or an exact decimal, not a rounded decimal.
For a fraction, the negative index turns it upside down. Then you apply the positive index to the top and the bottom.
${(\frac{2}{3})}^{- 2} = {(\frac{3}{2})}^{2} = \frac{9}{4}$
Negative indices obey all three rules. Add and subtract them as ordinary integers, and watch the signs. $2^{- 3} \times 2^{4} = 2^{- 3 + 4} = 2^{1} = 2$ , and $3^{2} \div 3^{- 4} = 3^{2 - (- 4)} = 3^{6}$ , because subtracting a negative adds.
A division can produce a negative index without warning: $2^{3} \div 2^{4} = 2^{- 1} = \frac{1}{2}$ . That is a correct answer. Leave it as $2^{- 1}$ if index form is asked for, or give $\frac{1}{2}$ if the value is asked for.

Part 5 of 7

Fractional indices are roots

What could $9^{\frac{1}{2}}$ mean? Use the multiplication rule. Two copies of it multiply to $9^{\frac{1}{2} + \frac{1}{2}} = 9^{1} = 9$ .
$9^{\frac{1}{2}} \times 9^{\frac{1}{2}} = 9^{1} = 9$
So $9^{\frac{1}{2}}$ is the number that gives 9 when you multiply it by itself. That is the square root: $9^{\frac{1}{2}} = \sqrt{9} = 3$ .
In the same way, three copies of $8^{\frac{1}{3}}$ multiply to $8^{1} = 8$ . So $8^{\frac{1}{3}}$ is the cube root of 8, which is 2. In general, an index of $\frac{1}{n}$ means the $n$ th root.
$a^{\frac{1}{n}} = \sqrt[n]{a}$
So $81^{\frac{1}{2}} = \sqrt{81} = 9$ , and $16^{\frac{1}{4}} = \sqrt[4]{16} = 2$ , because $2^{4} = 16$ . If the root is not a whole number, leave it as a root: $6^{\frac{1}{2}} = \sqrt{6}$ .
To find a fourth or fifth root without a calculator, you need a few powers ready: $2^{4} = 16$ , $2^{5} = 32$ , $2^{6} = 64$ , $3^{4} = 81$ , $3^{5} = 243$ . For example, $32^{\frac{1}{5}} = 2$ because $2^{5} = 32$ .
For a fraction, take the root of the top and of the bottom separately.
${(\frac{4}{25})}^{\frac{1}{2}} = \frac{\sqrt{4}}{\sqrt{25}} = \frac{2}{5}$
A fractional index does not divide the base. $16^{\frac{1}{2}}$ is $4$ , not $16 \div 2 = 8$ . Check: $4 \times 4 = 16$ , but $8 \times 8 = 64$ .

Part 6 of 7

Indices such as 2/3 and −2/3

The index $\frac{2}{3}$ is $\frac{1}{3} \times 2$ . By the power-of-a-power rule, $8^{\frac{2}{3}} = {(8^{\frac{1}{3}})}^{2}$ . So the denominator gives the root and the numerator gives the power.
$a^{\frac{m}{n}} = {(\sqrt[n]{a})}^{m}$
So $8^{\frac{2}{3}} = {(\sqrt[3]{8})}^{2} = 2^{2} = 4$ .
You can also do the power first: $8^{\frac{2}{3}} = \sqrt[3]{8^{2}} = \sqrt[3]{64} = 4$ . The answer is the same, but the numbers are bigger. Take the root first, because it keeps the numbers small enough to work in your head.
A negative fractional index does three jobs. The minus sign means reciprocal. The denominator means root. The numerator means power. Do them one at a time, and write each one down.
For $8^{- \frac{2}{3}}$ : the reciprocal gives $\frac{1}{8^{\frac{2}{3}}}$ . The root gives $\sqrt[3]{8} = 2$ . The power gives $2^{2} = 4$ . So $8^{- \frac{2}{3}} = \frac{1}{4}$ .
$8^{- \frac{2}{3}} = \frac{1}{8^{\frac{2}{3}}} = \frac{1}{{(\sqrt[3]{8})}^{2}} = \frac{1}{2^{2}} = \frac{1}{4}$
With a fraction as the base, turn it over for the minus sign. Then take the root and the power of the top and the bottom separately.
${(\frac{16}{81})}^{- \frac{3}{4}} = {(\frac{81}{16})}^{\frac{3}{4}} = {(\frac{3}{2})}^{3} = \frac{27}{8}$
Write each stage on its own line. If your final answer is wrong, a correct stage such as $\sqrt[3]{8} = 2$ on the page is still a valid method that can be credited.

Part 7 of 7

Writing numbers as powers of the same base

The rules need one base. When the bases are different but related, rewrite them first. These are worth knowing on sight.
$4 = 2^{2}, 8 = 2^{3}, 16 = 2^{4}, 32 = 2^{5} 9 = 3^{2}, 27 = 3^{3}, 81 = 3^{4} 25 = 5^{2}, 125 = 5^{3}$
For example, write $8 \times 4^{3}$ as a single power of 2. $8 = 2^{3}$ , and $4^{3} = {(2^{2})}^{3} = 2^{6}$ . So $8 \times 4^{3} = 2^{3} \times 2^{6} = 2^{9}$ . Check: $8 \times 64 = 512 = 2^{9}$ .
A unit fraction becomes a negative index: $\frac{1}{8} = \frac{1}{2^{3}} = 2^{- 3}$ and $\frac{1}{27} = 3^{- 3}$ . A root becomes a fractional index: $\sqrt{2} = 2^{\frac{1}{2}}$ .
To find a missing index, write both sides as powers of the same base. For a base such as 2, 3 or 5, two powers are equal only when their indices are equal. So you can compare the indices.
For $2^{n} = \frac{1}{32}$ : $\frac{1}{32} = \frac{1}{2^{5}} = 2^{- 5}$ , so $n = - 5$ .
For $9^{n} = 27$ : $9^{n} = {(3^{2})}^{n} = 3^{2 n}$ and $27 = 3^{3}$ . So $2 n = 3$ , and $n = \frac{3}{2}$ . A fraction is a correct value for $n$ . Check: $9^{\frac{3}{2}} = {(\sqrt{9})}^{3} = 3^{3} = 27$ .

Worked examples

Example 1. Simplify, leaving each answer in index form. (a) $5^{4} \times 5^{3}$ (b) $3^{8} \div 3^{- 2}$ (c) ${(7^{3})}^{4}$ (d) $\frac{2^{5} \times 2^{3}}{{(2^{2})}^{3}}$

Show the working

(a) Same base, multiplied, so add the indices: $4 + 3 = 7$ .
$5^{4} \times 5^{3} = 5^{7}$
(b) Same base, divided, so subtract the indices. Subtracting $- 2$ is the same as adding 2.
$3^{8} \div 3^{- 2} = 3^{8 - (- 2)} = 3^{10}$
(c) A power of a power, so multiply the indices: $3 \times 4 = 12$ .
${(7^{3})}^{4} = 7^{12}$
(d) Simplify the top and the bottom first. The top is a multiplication: $2^{5 + 3} = 2^{8}$ . The bottom is a power of a power: $2^{2 \times 3} = 2^{6}$ . Then divide.
$\frac{2^{8}}{2^{6}} = 2^{8 - 6} = 2^{2}$
Check (d) with values: $\frac{32 \times 8}{64} = \frac{256}{64} = 4$ , and $2^{2} = 4$ .

Answer: (a) $5^{7}$ (b) $3^{10}$ (c) $7^{12}$ (d) $2^{2}$

Example 2. Without a calculator, find the value of (a) $4^{- 3}$ (b) $125^{\frac{1}{3}}$ (c) $32^{- \frac{3}{5}}$ (d) ${(\frac{9}{4})}^{- \frac{1}{2}}$

Show the working

(a) A negative index means reciprocal. Then work out the positive power: $4^{3} = 64$ .
$4^{- 3} = \frac{1}{4^{3}} = \frac{1}{64}$
(b) An index of $\frac{1}{3}$ means cube root. $5^{3} = 125$ , so the cube root of 125 is 5.
$125^{\frac{1}{3}} = \sqrt[3]{125} = 5$
(c) Three jobs, one at a time. The minus sign gives the reciprocal. The denominator 5 gives the fifth root: $\sqrt[5]{32} = 2$ , because $2^{5} = 32$ . The numerator 3 gives the cube: $2^{3} = 8$ .
$32^{- \frac{3}{5}} = \frac{1}{32^{\frac{3}{5}}} = \frac{1}{{(\sqrt[5]{32})}^{3}} = \frac{1}{2^{3}} = \frac{1}{8}$
(d) The minus sign turns the fraction over. The index $\frac{1}{2}$ is a square root, taken of the top and the bottom separately.
${(\frac{9}{4})}^{- \frac{1}{2}} = {(\frac{4}{9})}^{\frac{1}{2}} = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3}$

Answer: (a) $\frac{1}{64}$ (b) $5$ (c) $\frac{1}{8}$ (d) $\frac{2}{3}$

Example 3. Find the value of $n$ in each equation. (a) $3^{n} = \frac{1}{81}$ (b) $5^{n} \times 25 = 5^{7}$ (c) $8^{n} = 16$

Show the working

(a) $81 = 3^{4}$ , so $\frac{1}{81} = \frac{1}{3^{4}} = 3^{- 4}$ . Both sides are now powers of 3, so compare the indices.
$3^{n} = 3^{- 4} \Rightarrow n = - 4$
(b) Write 25 as a power of 5: $25 = 5^{2}$ . The left side becomes $5^{n} \times 5^{2} = 5^{n + 2}$ .
$5^{n + 2} = 5^{7} \Rightarrow n + 2 = 7 \Rightarrow n = 5$
(c) 8 is not a power of 16, but both are powers of 2. $8^{n} = {(2^{3})}^{n} = 2^{3 n}$ and $16 = 2^{4}$ .
$2^{3 n} = 2^{4} \Rightarrow 3 n = 4 \Rightarrow n = \frac{4}{3}$
Check (c): $8^{\frac{4}{3}} = {(\sqrt[3]{8})}^{4} = 2^{4} = 16$ . Correct.

Answer: (a) $n = - 4$ (b) $n = 5$ (c) $n = \frac{4}{3}$

Example 4. Show that $\frac{4^{3} \times 8^{- 1}}{2^{- 2}} = 2^{5}$ .

Show the working

The question gives you the answer, so the working is what you are showing. Every step has to be on the page, and the last line has to be exactly $2^{5}$ .
Rewrite every number as a power of 2, so that the rules can be used.
$4^{3} = {(2^{2})}^{3} = 2^{6} 8^{- 1} = {(2^{3})}^{- 1} = 2^{- 3}$
Multiply the top: add the indices.
$2^{6} \times 2^{- 3} = 2^{6 + (- 3)} = 2^{3}$
Divide by $2^{- 2}$ : subtract the indices, and remember that subtracting $- 2$ adds 2.
$\frac{2^{3}}{2^{- 2}} = 2^{3 - (- 2)} = 2^{5}$
Check with values: $64 \times \frac{1}{8} = 8$ , and $8 \div \frac{1}{4} = 32 = 2^{5}$ .

Answer: $\frac{4^{3} \times 8^{- 1}}{2^{- 2}} = \frac{2^{6} \times 2^{- 3}}{2^{- 2}} = \frac{2^{3}}{2^{- 2}} = 2^{5}$ , as required.

Where people go wrong

Mistake: Multiplying the indices when you multiply two powers, e.g. $2^{3} \times 2^{4} = 2^{12}$ .

Instead: Add the indices: $2^{3} \times 2^{4} = 2^{7}$ . Multiply the indices only for a power of a power, such as ${(2^{3})}^{4} = 2^{12}$ .

$2^{3} \times 2^{4}$ is three 2s followed by four 2s, which is seven 2s, not twelve. Writing the 2s out once shows which rule applies.

Mistake: Multiplying the bases as well, e.g. $2^{3} \times 2^{4} = 4^{7}$ .

Instead: The base stays the same: $2^{3} \times 2^{4} = 2^{7} = 128$ .

Every factor in the product is a 2, so the answer is still a power of 2. Check: $8 \times 16 = 128$ , but $4^{7} = 16 384$ .

Mistake: Reading a negative index as a negative number, e.g. $5^{- 2} = - 25$ or $5^{- 2} = - 10$ .

Instead: A negative index means reciprocal: $5^{- 2} = \frac{1}{5^{2}} = \frac{1}{25}$ .

Going down from $5^{1} = 5$ to $5^{0} = 1$ , each step divides by 5. So $5^{- 1} = \frac{1}{5}$ and $5^{- 2} = \frac{1}{25}$ . Dividing a positive number by 5 never makes it negative.

Mistake: Treating a fractional index as a division or a multiplication, e.g. $16^{\frac{1}{2}} = 8$ or $8^{\frac{2}{3}} = \frac{16}{3}$ .

Instead: The denominator is a root: $16^{\frac{1}{2}} = \sqrt{16} = 4$ and $8^{\frac{2}{3}} = {(\sqrt[3]{8})}^{2} = 4$ .

$16^{\frac{1}{2}} \times 16^{\frac{1}{2}}$ must equal $16^{1} = 16$ . Only $4$ does that, because $4 \times 4 = 16$ and $8 \times 8 = 64$ .

Mistake: Writing $7^{0} = 0$ or $7^{0} = 7$ .

Instead: Any non-zero number to the power 0 is 1: $7^{0} = 1$ .

$7^{0} = 7^{2 - 2} = 7^{2} \div 7^{2} = 49 \div 49 = 1$ . A zero index comes from dividing a number by itself.

Check yourself

Work out $- 2^{4}$ and ${(- 2)}^{4}$ .
Answer
$- 2^{4} = - (2 \times 2 \times 2 \times 2) = - 16$ . ${(- 2)}^{4} = 16$ , because the bracket puts the index on $- 2$ and four negatives multiply to a positive.
Simplify $6^{9} \div 6^{4}$ , leaving your answer in index form.
Answer
$6^{9 - 4} = 6^{5}$
Simplify ${(4^{2})}^{- 3}$ , leaving your answer in index form.
Answer
Multiply the indices: $2 \times (- 3) = - 6$ , so $4^{- 6}$ .
Find the value of $3^{- 5} \times 3^{3}$ .
Answer
$3^{- 5 + 3} = 3^{- 2} = \frac{1}{9}$
Write down the value of $10^{0}$ and the value of $10^{- 3}$ .
Answer
$10^{0} = 1$ and $10^{- 3} = \frac{1}{1000} = 0.001$
Find the value of $64^{\frac{1}{2}}$ , $64^{\frac{1}{3}}$ and $64^{\frac{1}{6}}$ .
Answer
$8$ , $4$ and $2$ , because $8^{2} = 64$ , $4^{3} = 64$ and $2^{6} = 64$ .
Find the value of $25^{- \frac{3}{2}}$ .
Answer
$\frac{1}{25^{\frac{3}{2}}} = \frac{1}{{(\sqrt{25})}^{3}} = \frac{1}{5^{3}} = \frac{1}{125}$
Find $n$ when $4^{n} = \frac{1}{16}$ .
Answer
$\frac{1}{16} = \frac{1}{4^{2}} = 4^{- 2}$ , so $n = - 2$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1.

(a) Write down the value of $9^{0}$ . 1 mark No calculator

Mark scheme

B1 1 cao

(b) Find the value of $2^{- 5}$ . 1 mark No calculator

Mark scheme

B1 $\frac{1}{32}$ oe, e.g. 0.03125

(c) Find the value of $64^{- \frac{2}{3}}$ . 2 marks No calculator

Mark scheme

M1 $\frac{1}{64^{2 / 3}}$ seen, or $64^{\frac{1}{3}} = 4$ seen, or $64^{\frac{2}{3}} = 16$ seen, or ${(\frac{1}{4})}^{2}$ or $4^{- 2}$ seen, oe
A1 $\frac{1}{16}$ oe, e.g. 0.0625

(d) Find the value of ${(\frac{8}{27})}^{- \frac{1}{3}}$ . 2 marks No calculator

Mark scheme

M1 $\frac{27}{8}$ seen as the reciprocal, or $\frac{2}{3}$ seen as the cube root of $\frac{8}{27}$
A1 $\frac{3}{2}$ oe, e.g. 1.5 or $1 \frac{1}{2}$

Question 2.

(a) Simplify $11^{5} \times 11^{- 2}$ . Give your answer in index form. 1 mark No calculator

Mark scheme

B1 $11^{3}$ cao

(b) Simplify $\frac{{(3^{4})}^{5}}{3^{7}}$ . Give your answer in index form. 2 marks No calculator

Mark scheme

M1 $3^{20}$ seen, or $3^{20 - 7}$ seen
A1 $3^{13}$ cao

(c) Priya writes $2^{3} \times 3^{2} = 6^{5}$ . Explain why she is wrong. 1 mark No calculator

Mark scheme

B1 the bases are different, so the indices cannot be added (and the bases are not multiplied); or a numerical check such as $2^{3} \times 3^{2} = 72$ but $6^{5} = 7776$

Question 3. Find the value of $n$ in each equation.

(a) $5^{n} = \frac{1}{125}$ 1 mark No calculator

Mark scheme

B1 $- 3$ cao

(b) $2^{n} = 16 \times 8^{2}$ 2 marks No calculator

Mark scheme

M1 $2^{4} \times 2^{6}$ seen, or $16 \times 64 = 1024 = 2^{10}$ seen, or $n = 4 + 6$
A1 10 cao

(c) $27^{n} = 9$ 2 marks No calculator

Mark scheme

M1 $3^{3 n} = 3^{2}$ seen, or $3 n = 2$ seen
A1 $\frac{2}{3}$ oe

Question 4.

(a) Show that $\frac{16^{\frac{3}{4}} \times 2^{- 3}}{4^{- \frac{1}{2}}} = 2$ . 3 marks No calculator

Mark scheme

M1 $16^{\frac{3}{4}} = 8$ or $2^{3}$ seen
M1 $4^{- \frac{1}{2}} = \frac{1}{2}$ or $2^{- 1}$ seen, or dividing by $4^{- \frac{1}{2}}$ shown as multiplying by $4^{\frac{1}{2}} = 2$ , oe
A1 complete correct working leading to 2, e.g. $8 \times \frac{1}{8} = 1$ and $1 \div \frac{1}{2} = 2$ , or $2^{3 - 3 - (- 1)} = 2^{1}$ ; nfww

How this is examined

Showing that an expression with mixed powers of 2, 4, 8 and 16 equals a given value: A 2- or 3-mark Show (that), done without a calculator, with the answer printed in the question Each number rewritten as a power of one base (e.g. $16^{\frac{3}{4}} = 2^{3}$ ) or worked out as a value (e.g. $4^{- \frac{1}{2}} = \frac{1}{2}$ ), on separate lines that end exactly at the given answer. Writing the given answer with no lines leading to it does not show it.
Finding the value of a number with a negative fractional index, such as $27^{- \frac{2}{3}}$ , without a calculator: A 1- or 2-mark Find the value of, with a whole-number or fraction base One correct stage on the page for the method: the reciprocal $\frac{1}{27^{\frac{2}{3}}}$ , the root $\sqrt[3]{27} = 3$ , or $27^{\frac{2}{3}} = 9$ . Then the exact answer $\frac{1}{9}$ . A rounded decimal such as $0.11$ is not exact.
Finding a missing index by writing both sides as powers of the same prime: A Find the value of $n$ in two or three short parts, 1 or 2 marks each, such as $2^{n} = \frac{1}{8}$ or $9^{n} = 27$ Both sides written with one base, e.g. $3^{2 n} = 3^{3}$ , or the index equation $2 n = 3$ seen. Then $n = \frac{3}{2}$ . A fraction or a negative value of $n$ is allowed, so do not round it to a whole number.
Using the three rules to simplify a product, quotient or power of powers: A 1- or 2-mark Simplify, which may ask for the answer in index form Indices added for $\times$ , subtracted for $\div$ , and multiplied for a power of a power, with the base unchanged. A correct intermediate power such as $3^{20}$ shows the method. If index form is asked for, the final answer is one power, such as $3^{13}$ .
The value of a zero or a negative whole-number index: A 1-mark Write down or Find the value of, such as the value of $5^{0}$ or $2^{- 1}$ $5^{0} = 1$ , not 0 or 5. $2^{- 1} = \frac{1}{2}$ , not $- 2$ . Give a fraction or an exact decimal.

Words to know

Base: The number being multiplied repeatedly. In $2^{5}$ the base is 2.
Index (power, exponent): The raised number that says how many copies of the base are multiplied together. In $2^{5}$ the index is 5. The plural is indices.
Index form: A number written as a base with an index, e.g. $3^{4} \times 5^{2}$ instead of $3 \times 3 \times 3 \times 3 \times 5 \times 5$ .
Rules of indices: For the same base: $a^{m} \times a^{n} = a^{m + n}$ , $a^{m} \div a^{n} = a^{m - n}$ and ${(a^{m})}^{n} = a^{m n}$ .
Zero index: Any non-zero number to the power 0 is 1: $a^{0} = 1$ .
Negative index: Means the reciprocal: $a^{- n} = \frac{1}{a^{n}}$ , e.g. $7^{- 2} = \frac{1}{49}$ .
Fractional index: The denominator gives a root and the numerator gives a power: $a^{\frac{m}{n}} = {(\sqrt[n]{a})}^{m}$ , e.g. $8^{\frac{2}{3}} = 4$ .
nth root: $\sqrt[n]{a}$ is the number that gives $a$ when $n$ copies of it are multiplied together, e.g. $\sqrt[4]{16} = 2$ . For a square root, $\sqrt{a}$ means the positive one.
Reciprocal: 1 divided by a number. A number times its reciprocal is 1, e.g. the reciprocal of $\frac{3}{2}$ is $\frac{2}{3}$ .

All Maths lessons Next: 1.8 Standard form