Maths · 2 Algebra

2.4 Indices II

Before you start

Indices I (subtopic 1.7): the rules $a^{m} \times a^{n} = a^{m + n}$ , $a^{m} \div a^{n} = a^{m - n}$ and ${(a^{m})}^{n} = a^{m n}$ , used with numbers.
Zero, negative and fractional indices with numbers, e.g. $5^{0} = 1$ , $2^{- 3} = \frac{1}{8}$ and $8^{\frac{2}{3}} = 4$ .
Powers of 2, 3 and 5 on sight: $2, 4, 8, 16, 32, 64$ , then $3, 9, 27, 81, 243$ , then $5, 25, 125, 625$ .
Multiplying and dividing fractions, e.g. $\frac{3}{4} \times 8 = 6$ and $\frac{3}{4} \div 6 = \frac{1}{8}$ .
Solving a linear equation such as $x + 4 = 2 x + 2$ .

Syllabus objectives covered (2)

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

Part 1 of 6

Letters follow the same rules as numbers

In Indices I the base was a number. Now the base is a letter. Nothing new happens: $x^{3}$ still means $x \times x \times x$ , so the rules still hold, for the same reasons.
$x^{3} \times x^{4} = (x \times x \times x) \times (x \times x \times x \times x) = x^{7}$
So the three rules are true for any base $x$ . Every question in this subtopic uses them, so learn them in this form.
$x^{m} \times x^{n} = x^{m + n} x^{m} \div x^{n} = x^{m - n} {(x^{m})}^{n} = x^{m n}$
A term such as $4 x^{3}$ has two parts. The coefficient is the number in front, 4. The power is $x^{3}$ . The index belongs to the $x$ only: $4 x^{3}$ means $4 \times x^{3}$ , not ${(4 x)}^{3}$ .
To multiply two terms, deal with each part on its own. Multiply the coefficients. Add the indices of the letter.
$4 x^{3} \times 5 x^{2} = (4 \times 5) \times x^{3 + 2} = 20 x^{5}$
The trap here is to mix up the two jobs. You multiply 4 and 5 to get 20. You do not add them. You add the indices 3 and 2 to get 5. You do not multiply them.
To divide, divide the coefficients and subtract the indices.
$\frac{12 x^{7}}{3 x^{2}} = \frac{12}{3} \times x^{7 - 2} = 4 x^{5}$
Different letters are different bases. Each letter keeps its own index. Collect the powers of $a$ together, and the powers of $b$ together.
$2 a^{3} b \times 3 a b^{4} = 6 a^{3 + 1} b^{1 + 4} = 6 a^{4} b^{5}$
Notice that $a$ on its own is $a^{1}$ . If you forget the hidden 1, you add 0 in its place and your answer has one power too few.

Part 2 of 6

A power of a bracket

A power outside a bracket applies to every factor inside it, and that includes the coefficient. Raise the coefficient to the power. Multiply the letter's index by the power.
${(3 x^{4})}^{2} = 3^{2} \times {(x^{4})}^{2} = 9 x^{8}$
The trap here is to write ${(3 x^{4})}^{2} = 3 x^{8}$ . Write the bracket out once to see why it is wrong: $(3 x^{4}) (3 x^{4}) = 9 x^{8}$ . The 3 appears twice, so it is squared.
With several letters, every letter's index is multiplied by the power. A negative index is multiplied in the same way.
${(2 a^{3} b^{- 1})}^{3} = 2^{3} a^{3 \times 3} b^{- 1 \times 3} = 8 a^{9} b^{- 3}$
With a fraction, raise the top and the bottom. The number on the bottom is raised too.
${(\frac{3 x^{2}}{2})}^{3} = \frac{3^{3} {(x^{2})}^{3}}{2^{3}} = \frac{27 x^{6}}{8}$
This rule is for products and quotients only. ${(x + y)}^{2}$ is not $x^{2} + y^{2}$ . Check with $x = 1$ and $y = 2$ : ${(1 + 2)}^{2} = 9$ , but $1^{2} + 2^{2} = 5$ .

Part 3 of 6

Zero and negative indices on a letter

The meanings from Indices I do not change. For any $x \neq 0$ :
$x^{0} = 1 x^{- n} = \frac{1}{x^{n}}$
So $x^{- 3} = \frac{1}{x^{3}}$ . A negative index puts the power on the bottom of a fraction, with a positive index. It does not make anything negative.
Now look closely at $5 x^{- 2}$ . The index $- 2$ belongs to the $x$ only, so only the $x$ goes to the bottom. The 5 stays on top.
$5 x^{- 2} = 5 \times \frac{1}{x^{2}} = \frac{5}{x^{2}}$
Compare ${(5 x)}^{- 2}$ . The bracket puts the index on the 5 as well, so the whole bracket goes to the bottom.
${(5 x)}^{- 2} = \frac{1}{{(5 x)}^{2}} = \frac{1}{25 x^{2}}$
You also need to go the other way. A power of $x$ on the bottom of a fraction can come to the top with a negative index. This is how you write an expression in the form $k x^{n}$ , one number times one power of $x$ .
$\frac{3}{x^{4}} = 3 x^{- 4} \frac{1}{4 x^{2}} = \frac{1}{4} x^{- 2}$
In $\frac{1}{4 x^{2}}$ the 4 is on the bottom too, but it has no index of its own. It stays a number on the bottom, so the coefficient is $\frac{1}{4}$ . It is not $4 x^{- 2}$ .
Watch the zero index inside a term. $7 x^{0} = 7 \times 1 = 7$ , because the index is on the $x$ only. But ${(7 x)}^{0} = 1$ , because the index is on the whole bracket.
Negative indices obey the three rules. Add and subtract them as ordinary integers, and remember that subtracting a negative adds.
$x^{- 3} \times x^{5} = x^{2} \frac{x^{2}}{x^{- 4}} = x^{2 - (- 4)} = x^{6}$

Part 4 of 6

Fractional indices: roots of letters

An index of $\frac{1}{2}$ is a square root, and an index of $\frac{1}{3}$ is a cube root. The reason is the multiplication rule: $x^{\frac{1}{2}} \times x^{\frac{1}{2}} = x^{1} = x$ , so $x^{\frac{1}{2}}$ is the number that multiplies by itself to give $x$ .
$x^{\frac{1}{2}} = \sqrt{x} x^{\frac{1}{3}} = \sqrt[3]{x} x^{\frac{m}{n}} = {(\sqrt[n]{x})}^{m}$
The rules work on indices, not on root signs. So change every root into an index before you start. Learn to see these on sight.
$\sqrt{x} = x^{\frac{1}{2}} \frac{1}{\sqrt{x}} = x^{- \frac{1}{2}} x \sqrt{x} = x^{1} \times x^{\frac{1}{2}} = x^{\frac{3}{2}} \sqrt[3]{x^{2}} = x^{\frac{2}{3}}$
Fractional indices add and subtract like any other fractions. Use a common denominator.
$x^{\frac{1}{2}} \times x^{\frac{1}{3}} = x^{\frac{3}{6} + \frac{2}{6}} = x^{\frac{5}{6}}$
A fractional power of a bracket works on every factor. Take the root of the coefficient. Multiply each index by the fraction.
${(16 x^{8})}^{\frac{1}{2}} = 16^{\frac{1}{2}} \times x^{8 \times \frac{1}{2}} = 4 x^{4}$
With a negative fractional power of a bracket, do the three jobs one at a time and write each one down. The minus sign gives the reciprocal. The denominator 3 gives the cube root, ${(27 x^{6})}^{\frac{1}{3}} = 3 x^{2}$ . The numerator 2 gives the square.
${(27 x^{6})}^{- \frac{2}{3}} = \frac{1}{{(27 x^{6})}^{\frac{2}{3}}} = \frac{1}{{(3 x^{2})}^{2}} = \frac{1}{9 x^{4}}$
To undo a fractional power in an equation, raise both sides to the reciprocal power. This works because ${(x^{\frac{3}{2}})}^{\frac{2}{3}} = x^{\frac{3}{2} \times \frac{2}{3}} = x^{1} = x$ .
$x^{\frac{3}{2}} = 8 \Rightarrow x = 8^{\frac{2}{3}} = {(\sqrt[3]{8})}^{2} = 4$
Check: $4^{\frac{3}{2}} = {(\sqrt{4})}^{3} = 2^{3} = 8$ . Correct.

Part 5 of 6

Terms with fraction coefficients and mixed indices

Some terms have a fraction for a coefficient and a negative or fractional index. The method does not change: coefficients with coefficients, indices with indices. Do each one on its own line.
For $4 x^{- 3} \times \frac{5}{6} x^{\frac{1}{2}}$ , first multiply the coefficients: $4 \times \frac{5}{6} = \frac{20}{6} = \frac{10}{3}$ .
Then add the indices: $- 3 + \frac{1}{2} = - \frac{6}{2} + \frac{1}{2} = - \frac{5}{2}$ . Put the two parts together.
$4 x^{- 3} \times \frac{5}{6} x^{\frac{1}{2}} = \frac{10}{3} x^{- \frac{5}{2}}$
Simplify the coefficient fully. $\frac{20}{6}$ is not finished, because 20 and 6 share a factor of 2. $\frac{10}{3}$ is finished.
Division works the same way. For $\frac{2}{3} x^{\frac{1}{2}} \div 4 x^{- 1}$ , divide the coefficients. Dividing by 4 is the same as multiplying by $\frac{1}{4}$ , so $\frac{2}{3} \div 4 = \frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6}$ .
Then subtract the indices, in order: the index you divide into, minus the index you divide by. $\frac{1}{2} - (- 1) = \frac{3}{2}$ .
$\frac{2}{3} x^{\frac{1}{2}} \div 4 x^{- 1} = \frac{1}{6} x^{\frac{3}{2}}$
A negative index in the answer is not a sign of an error. $\frac{10}{3} x^{- \frac{5}{2}}$ and $\frac{10}{3 x^{\frac{5}{2}}}$ are the same expression. If the question asks for a particular form, such as the form $k x^{n}$ , give that form.

Part 6 of 6

Equations with the unknown in the index

In $2^{x} = 32$ the unknown is the index. You cannot add, subtract or divide to get $x$ on its own. Instead, write both sides as powers of the same base: $32 = 2^{5}$ , so $2^{x} = 2^{5}$ .
If two powers of the same base are equal, their indices are equal. So $x = 5$ . This is true for any positive base except 1, so it works for 2, 3, 5 and every other base in this lesson.
To choose the base, find the smallest number that every number in the equation is a power of. For 4, 8 and 16 that is 2. For 9, 27 and 81 it is 3. For 25 and 125 it is 5.
For $8^{x} = 2$ , write $8 = 2^{3}$ . Then $8^{x} = {(2^{3})}^{x} = 2^{3 x}$ , by the power-of-a-power rule. The right-hand side is $2 = 2^{1}$ .
$2^{3 x} = 2^{1} \Rightarrow 3 x = 1 \Rightarrow x = \frac{1}{3}$
A fraction or a root on one side becomes a negative or a fractional index. $\frac{1}{3} = 3^{- 1}$ and $\sqrt{3} = 3^{\frac{1}{2}}$ . So $9^{x} = \frac{1}{3}$ becomes $3^{2 x} = 3^{- 1}$ , which gives $2 x = - 1$ and $x = - \frac{1}{2}$ .
When the index is an expression, put it in a bracket before you multiply. The 2 multiplies the whole index, not only the $x$ .
$9^{x + 1} = {(3^{2})}^{x + 1} = 3^{2 (x + 1)} = 3^{2 x + 2}$
If one side has two powers multiplied, combine them first with the multiplication rule: $2^{x} \times 4 = 2^{x} \times 2^{2} = 2^{x + 2}$ . You need one power on each side before you can compare indices.
Then compare the indices and solve the linear equation that is left. For $3^{x + 4} = 9^{x + 1}$ , the right-hand side is $3^{2 x + 2}$ .
$3^{x + 4} = 3^{2 x + 2} \Rightarrow x + 4 = 2 x + 2 \Rightarrow x = 2$
Check by putting your answer into the original equation. With $x = 2$ : $3^{6} = 729$ and $9^{3} = 729$ . Both sides match.
This method needs both sides to be powers of one base. An equation such as $2^{x} = 10$ cannot be written that way, because 10 is not a power of 2. That kind of equation is solved with logarithms (subtopic 3.7) or with a graph on your calculator.

Worked examples

Example 1. Simplify (a) $3 x^{4} y \times 5 x^{2} y^{3}$ (b) $\frac{20 m^{8}}{4 m^{- 3}}$ (c) ${(5 p^{- 2} q^{3})}^{2}$

Show the working

(a) Multiply the coefficients: $3 \times 5 = 15$ . Add the indices of each letter. $y$ on its own is $y^{1}$ .
$3 x^{4} y \times 5 x^{2} y^{3} = 15 x^{4 + 2} y^{1 + 3} = 15 x^{6} y^{4}$
(b) Divide the coefficients: $20 \div 4 = 5$ . Subtract the indices, and remember that subtracting $- 3$ adds 3.
$\frac{20 m^{8}}{4 m^{- 3}} = 5 m^{8 - (- 3)} = 5 m^{11}$
(c) The power 2 applies to every factor in the bracket. Square the coefficient: $5^{2} = 25$ . Multiply each index by 2.
${(5 p^{- 2} q^{3})}^{2} = 25 p^{- 2 \times 2} q^{3 \times 2} = 25 p^{- 4} q^{6}$
You may also write (c) as $\frac{25 q^{6}}{p^{4}}$ . It is the same expression.

Answer: (a) $15 x^{6} y^{4}$ (b) $5 m^{11}$ (c) $25 p^{- 4} q^{6}$ , which is $\frac{25 q^{6}}{p^{4}}$

Example 2. Simplify, giving each answer as a single term. (a) $5 x^{- 2} \times \frac{3}{10} x^{\frac{1}{2}}$ (b) $\frac{3}{4} x^{\frac{1}{3}} \div 6 x^{- 1}$

Show the working

(a) Coefficients first: $5 \times \frac{3}{10} = \frac{15}{10} = \frac{3}{2}$ .
Then the indices, which are added because the terms are multiplied: $- 2 + \frac{1}{2} = - \frac{4}{2} + \frac{1}{2} = - \frac{3}{2}$ .
$5 x^{- 2} \times \frac{3}{10} x^{\frac{1}{2}} = \frac{3}{2} x^{- \frac{3}{2}}$
(b) Coefficients first. Dividing by 6 is multiplying by $\frac{1}{6}$ : $\frac{3}{4} \times \frac{1}{6} = \frac{3}{24} = \frac{1}{8}$ .
Then the indices, which are subtracted because the terms are divided: $\frac{1}{3} - (- 1) = \frac{1}{3} + \frac{3}{3} = \frac{4}{3}$ .
$\frac{3}{4} x^{\frac{1}{3}} \div 6 x^{- 1} = \frac{1}{8} x^{\frac{4}{3}}$

Answer: (a) $\frac{3}{2} x^{- \frac{3}{2}}$ (b) $\frac{1}{8} x^{\frac{4}{3}}$

Example 3. Simplify (a) ${(9 x^{4})}^{\frac{3}{2}}$ (b) ${(125 x^{9})}^{- \frac{2}{3}}$ (c) ${(\frac{4}{x^{6}})}^{- \frac{1}{2}}$

Show the working

(a) The power applies to the 9 and to the $x^{4}$ . For the 9, take the square root first, then cube: $9^{\frac{3}{2}} = {(\sqrt{9})}^{3} = 3^{3} = 27$ . For the letter, multiply the indices: $4 \times \frac{3}{2} = 6$ .
${(9 x^{4})}^{\frac{3}{2}} = 27 x^{6}$
(b) Three jobs, one at a time. The minus sign gives the reciprocal. The cube root of $125 x^{9}$ is $5 x^{3}$ , because $5^{3} = 125$ and $9 \times \frac{1}{3} = 3$ . Then square it.
${(125 x^{9})}^{- \frac{2}{3}} = \frac{1}{{(125 x^{9})}^{\frac{2}{3}}} = \frac{1}{{(5 x^{3})}^{2}} = \frac{1}{25 x^{6}}$
(c) The minus sign turns the fraction over. Then the index $\frac{1}{2}$ is a square root of the top and of the bottom.
${(\frac{4}{x^{6}})}^{- \frac{1}{2}} = {(\frac{x^{6}}{4})}^{\frac{1}{2}} = \frac{x^{3}}{2}$
Check (a) with $x = 1$ : ${(9)}^{\frac{3}{2}} = 27$ and $27 \times 1^{6} = 27$ . A check with $x = 1$ tests the coefficient. A check with $x = 2$ tests the index as well.

Answer: (a) $27 x^{6}$ (b) $\frac{1}{25 x^{6}}$ (c) $\frac{x^{3}}{2}$

Example 4. Solve (a) $25^{x} = 125$ (b) $2^{x + 3} = 8^{x - 1}$ (c) $9^{x} \times 3 = 27^{x + 1}$

Show the working

(a) 25 and 125 are both powers of 5. $25^{x} = {(5^{2})}^{x} = 5^{2 x}$ and $125 = 5^{3}$ .
$5^{2 x} = 5^{3} \Rightarrow 2 x = 3 \Rightarrow x = \frac{3}{2}$
(b) Write 8 as $2^{3}$ , and keep the index $x - 1$ in a bracket so the 3 multiplies all of it.
$8^{x - 1} = {(2^{3})}^{x - 1} = 2^{3 (x - 1)} = 2^{3 x - 3}$
Both sides are now powers of 2, so compare the indices.
$x + 3 = 3 x - 3 \Rightarrow 6 = 2 x \Rightarrow x = 3$
Check (b): $2^{3 + 3} = 2^{6} = 64$ and $8^{3 - 1} = 8^{2} = 64$ . Correct.
(c) The left-hand side has two powers multiplied. Write each as a power of 3 and add the indices: $9^{x} \times 3 = 3^{2 x} \times 3^{1} = 3^{2 x + 1}$ . The right-hand side is $27^{x + 1} = 3^{3 (x + 1)} = 3^{3 x + 3}$ .
$2 x + 1 = 3 x + 3 \Rightarrow x = - 2$
A negative answer is allowed. Check (c): $9^{- 2} \times 3 = \frac{1}{81} \times 3 = \frac{1}{27}$ , and $27^{- 1} = \frac{1}{27}$ . Correct.

Answer: (a) $x = \frac{3}{2}$ (b) $x = 3$ (c) $x = - 2$

Where people go wrong

Mistake: Moving the coefficient as well as the power, e.g. $3 x^{- 2} = \frac{1}{3 x^{2}}$ .

Instead: Only the $x$ goes to the bottom: $3 x^{- 2} = \frac{3}{x^{2}}$ . The expression $\frac{1}{3 x^{2}}$ is a different one: it is $\frac{1}{3} x^{- 2}$ .

The index belongs to the thing immediately before it, which is $x$ . Check with $x = 1$ : $3 \times 1^{- 2} = 3$ , but $\frac{1}{3 \times 1^{2}} = \frac{1}{3}$ .

Mistake: Not raising the coefficient in a power of a bracket, e.g. ${(3 x^{2})}^{3} = 3 x^{6}$ .

Instead: Every factor in the bracket is raised: ${(3 x^{2})}^{3} = 3^{3} x^{6} = 27 x^{6}$ .

${(3 x^{2})}^{3}$ is $(3 x^{2}) (3 x^{2}) (3 x^{2})$ . There are three 3s multiplied, which is 27.

Mistake: Multiplying the indices when two terms are multiplied, e.g. $4 x^{3} \times 5 x^{2} = 20 x^{6}$ .

Instead: Add the indices: $4 x^{3} \times 5 x^{2} = 20 x^{5}$ . Multiply indices only for a power of a power, such as ${(x^{3})}^{2} = x^{6}$ .

$x^{3} \times x^{2}$ is three $x$ s followed by two $x$ s, which is five $x$ s. The coefficients are multiplied because they are ordinary numbers, not indices.

Mistake: Comparing indices before the bases are the same, or dividing, e.g. solving $4^{x} = 8$ as $x = 2$ .

Instead: Write both sides with one base first: $2^{2 x} = 2^{3}$ , so $2 x = 3$ and $x = \frac{3}{2}$ .

$4^{2} = 16$ , not 8, so $x = 2$ does not satisfy the equation. Equal powers have equal indices only when the bases are equal.

Mistake: Dropping the bracket when a base is rewritten, e.g. $9^{x + 1} = 3^{2 x + 1}$ .

Instead: Multiply the whole index: $9^{x + 1} = 3^{2 (x + 1)} = 3^{2 x + 2}$ .

The power-of-a-power rule multiplies the index $x + 1$ by 2, and that means both terms of it. With $x = 0$ : $9^{1} = 9$ and $3^{2} = 9$ , but $3^{1} = 3$ .

Check yourself

Simplify $7 p^{3} \times 2 p^{5}$ .
Answer
Multiply the coefficients and add the indices: $14 p^{8}$ .
Simplify $\frac{18 y^{9}}{6 y^{- 2}}$ .
Answer
$18 \div 6 = 3$ and $9 - (- 2) = 11$ , so $3 y^{11}$ .
Simplify ${(5 a^{2} b^{3})}^{2}$ .
Answer
$5^{2} = 25$ , and each index is doubled: $25 a^{4} b^{6}$ .
Write $\frac{6}{x^{5}}$ in the form $k x^{n}$ .
Answer
$6 x^{- 5}$ , so $k = 6$ and $n = - 5$ .
Write $x^{2} \sqrt{x}$ as a single power of $x$ .
Answer
$x^{2} \times x^{\frac{1}{2}} = x^{\frac{5}{2}}$
Simplify ${(25 x^{6})}^{- \frac{1}{2}}$ .
Answer
${(25 x^{6})}^{\frac{1}{2}} = 5 x^{3}$ , and the minus sign gives the reciprocal: $\frac{1}{5 x^{3}}$ , which is $\frac{1}{5} x^{- 3}$ .
Solve $8^{x} = \frac{1}{4}$ .
Answer
$2^{3 x} = 2^{- 2}$ , so $3 x = - 2$ and $x = - \frac{2}{3}$ .
Solve $25^{x} = 5^{x + 3}$ .
Answer
$25^{x} = 5^{2 x}$ , so $2 x = x + 3$ and $x = 3$ . Check: $25^{3} = 15 625$ and $5^{6} = 15 625$ .

Exam questions

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

Question 1. Simplify each expression.

(a) $5 a^{4} b^{2} \times 3 a^{- 1} b^{5}$ 2 marks No calculator

Mark scheme

B2 $15 a^{3} b^{7}$ cao; B1 for two of $15$ , $a^{3}$ , $b^{7}$ correct in a single term

(b) ${(\frac{3 p^{2}}{4})}^{3}$ 2 marks No calculator

Mark scheme

B2 $\frac{27 p^{6}}{64}$ oe; B1 for $27 p^{6}$ seen as the numerator, or $\frac{27}{64}$ seen with an incorrect power of $p$

(c) $6 x^{\frac{1}{3}} \times \frac{3}{4} x^{- 2}$ 2 marks No calculator

Mark scheme

B2 $\frac{9}{2} x^{- \frac{5}{3}}$ oe, e.g. $\frac{9}{2 x^{5 / 3}}$ or $4.5 x^{- \frac{5}{3}}$ ; B1 for $\frac{9}{2}$ oe or $x^{- \frac{5}{3}}$ correct in a single term

Question 2.

(a) Write $\frac{4}{x^{3}}$ in the form $k x^{n}$ . 1 mark No calculator

Mark scheme

B1 $4 x^{- 3}$ cao

(b) Write $\frac{1}{2 \sqrt{x}}$ in the form $k x^{n}$ . 2 marks No calculator

Mark scheme

M1 $\sqrt{x} = x^{\frac{1}{2}}$ seen, or $x^{- \frac{1}{2}}$ seen
A1 $\frac{1}{2} x^{- \frac{1}{2}}$ oe, e.g. $0.5 x^{- 0.5}$

(c) Ravi writes $3 x^{- 2} = \frac{1}{3 x^{2}}$ . Explain why he is wrong. 1 mark No calculator

Mark scheme

B1 the index $- 2$ applies to $x$ only, not to the 3, so $3 x^{- 2} = \frac{3}{x^{2}}$ ; or a numerical check, e.g. with $x = 1$ , $3 x^{- 2} = 3$ but $\frac{1}{3 x^{2}} = \frac{1}{3}$

Question 3. Solve these equations.

(a) $4^{x} = \frac{1}{32}$ 2 marks No calculator

Mark scheme

M1 $2^{2 x} = 2^{- 5}$ seen, or $2 x = - 5$ seen, oe, e.g. $4^{x} = 4^{- \frac{5}{2}}$
A1 $x = - \frac{5}{2}$ oe

(b) $3^{2 x - 1} = 27^{x - 1}$ 3 marks No calculator

Mark scheme

M1 27 written as $3^{3}$ , giving $3^{3 (x - 1)}$ or $3^{3 x - 3}$ , or another correct method that makes both sides powers of one base
M1 dep on the first M1: indices equated, $2 x - 1 = 3 x - 3$ oe
A1 $x = 2$ cao

Question 4.

(a) Show that $\frac{{(2 x^{2})}^{3} \times x^{- 1}}{4 x^{\frac{1}{2}}}$ simplifies to $2 x^{\frac{9}{2}}$ . 3 marks No calculator

Mark scheme

M1 ${(2 x^{2})}^{3} = 8 x^{6}$ seen
M1 $8 x^{5}$ seen as the numerator, or coefficients divided and indices subtracted, e.g. $\frac{8}{4}$ and $x^{5 - \frac{1}{2}}$ seen, or the division done first, e.g. $2 x^{\frac{11}{2}} \times x^{- 1}$ seen
A1 complete correct working reaching $2 x^{\frac{9}{2}}$ with no errors; nfww

(b) Find the value of $2 x^{\frac{9}{2}}$ when $x = 4$ . 2 marks No calculator

Mark scheme

M1 $4^{\frac{9}{2}} = 512$ seen, or ${(\sqrt{4})}^{9}$ or $2^{9}$ seen
A1 1024 cao

How this is examined

Solving an equation with the unknown in the index by writing both sides as powers of one base: A 2- or 3-mark Solve, such as $4^{x - 1} = 8^{x}$ , answered without a calculator Both sides as powers of one base, with the index in a bracket: $2^{2 (x - 1)} = 2^{3 x}$ . Then the equation of indices on its own line: $2 x - 2 = 3 x$ . Then $x = - 2$ . A fraction or a negative value of $x$ is allowed, so leave it exact.
Showing that an algebraic expression with brackets and negative or fractional indices simplifies to a printed form: A 2- or 3-mark Show (that), with the simplified form printed in the question One rule per line: the bracket expanded with its coefficient raised (e.g. ${(3 x^{3})}^{2} = 9 x^{6}$ ), then the coefficients divided and the indices subtracted, ending at exactly the printed form. The printed answer with no lines leading to it shows nothing.
Multiplying or dividing two terms that have fraction coefficients and negative or fractional indices: A 2-mark Simplify of two terms, such as $\frac{3}{5} x^{\frac{1}{2}} \times 10 x^{- 2}$ One single term with the coefficient fully simplified and the index worked out: $6 x^{- \frac{3}{2}}$ . Write $\frac{3}{5} \times 10 = 6$ and $\frac{1}{2} + (- 2) = - \frac{3}{2}$ as working, so each part can be seen to be correct.
Raising a term or a fraction to a power, with every factor raised: A 1- or 2-mark Simplify, such as ${(\frac{2 y^{3}}{5})}^{2}$ Every factor raised: the coefficient ( $2^{2} = 4$ ), the letter's index multiplied ( $3 \times 2 = 6$ ) and the number on the bottom ( $5^{2} = 25$ ), giving $\frac{4 y^{6}}{25}$ . If the 2 or the 5 is not squared, the answer is not correct.
Writing an expression with a power or a root on the bottom in the form $k x^{n}$ : A 1- or 2-mark Write in the form $k x^{n}$ , such as $\frac{2}{x \sqrt{x}}$ The root changed to an index, $x \sqrt{x} = x^{\frac{3}{2}}$ , then the power moved to the top with its sign changed: $2 x^{- \frac{3}{2}}$ . A number on the bottom stays a number: $\frac{1}{3 x^{2}} = \frac{1}{3} x^{- 2}$ , not $3 x^{- 2}$ .

Words to know

Term: A single product of a number and letters, such as $5 a^{2} b$ . Terms are separated by $+$ and $-$ signs.
Coefficient: The number that multiplies the letters in a term. In $4 x^{3}$ the coefficient is 4. In $\frac{1}{4} x^{- 2}$ it is $\frac{1}{4}$ .
The form $k x^{n}$: One number $k$ times one power of $x$ . The index $n$ can be negative or a fraction, e.g. $\frac{1}{2} x^{- \frac{1}{2}}$ .
Exponential equation: An equation with the unknown in the index, such as $2^{x + 1} = 8$ .
Equating indices: Using the fact that if $a^{m} = a^{n}$ , with $a$ positive and not 1, then $m = n$ .
Reciprocal power: The power that undoes another. The power $\frac{2}{3}$ undoes the power $\frac{3}{2}$ , because ${(x^{\frac{3}{2}})}^{\frac{2}{3}} = x$ .

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