Indices

Topic

Course Support Current PowerPoint version

More questions on this topic: GCSE Algebra: Indices (178) IGCSE Algebra: Indices (174)

AQA

AQA · Old spec

June 2025 Paper 1 Q13

AQACurrent spec1 markDrawing & Using Graphs Indices

13 Here are four sketch graphs.

Circle the letter of the sketch graph that represents \(\quad y = 5^x\) [1 mark]

Four sketch graphs labelled A, B, C and D

Answer	Mark	Comments
C	B1

June 2025 Paper 1 Q8

AQACurrent spec1 markIndices

8 Work out \(\quad \left(\dfrac{x^{\frac{9}{2}}}{x^{\frac{1}{2}}}\right)^3 \quad\) as a power of \(x\).

Circle your answer. [1 mark]

\(x^7\)
\(x^{12}\)
\(x^{15}\)
\(x^{27}\)

Answer	Mark	Comments
\(x^{12}\)	B1

June 2025 Paper 2 Q7

AQACurrent spec5 marksIndices

(a)

\(2^{5a + 1} = 2^a\)

Work out the value of \(a\). [2 marks]

(b)

\(3^{-7} \div 3^c = \left(3^{-3}\right)^{2c}\)

Work out the value of \(c\). [3 marks]

Mark scheme (a)

Answer	Mark	Comments
\(5a + 1 = a\) or \(2^{5a - a + 1} = 1\) or \(2^{5a - a + 1} = 2^0\)	M1	oe linear equation eg \(4a = -1\) or oe index equation in the form \(2^{\mathrm{f}(a)} = 1\) or \(2^{\mathrm{f}(a)} = 2^0\)
\(-\dfrac{1}{4}\) or \(-0.25\)	A1	oe value

Additional guidance

Only \(\dfrac{2^{5a + 1}}{2^a} = 1\)

Mark scheme (b)

Answer	Mark	Comments
\(3^{-7 - c}\) or \(3^{-6c}\) or \(-7 - c\) and \(-6c\)	M1	oe eg \(\dfrac{1}{3^{7 + c}}\) or \(\dfrac{1}{3^{6c}}\) or \(3^{-3 \times 2c}\) implied by a correct equation equating powers of 3 eg \(3^{-7} = 3^{-5c}\)
\(-7 - c = -6c\)	M1dep	oe linear equation eg \(5c = 7\)
\(\dfrac{7}{5}\) or 1.4	A1	oe value condone \(\dfrac{-7}{-5}\)

Additional guidance

\(\dfrac{1}{3^7} \div 3^c\) needs further processing for M1
\(\dfrac{1}{(3^3)^{2c}}\) needs further processing for M1

June 2024 Paper 1 Q21

AQACurrent spec5 marksIndices Solving Quadratics

\[\left(2^{x + 3}\right)^{4x + 1} = 8^{x - 1}\]

Work out the possible values of \(x\). [5 marks]

Mark scheme

Answer	Mark	Comments
\(2^{(4x + 1)(x + 3)}\) or \(2^{4x^2 + 13x + 3}\)	M1	lhs
\(\left(2^3\right)^{x - 1}\) or \(2^{3(x - 1)}\) or \(2^{3x - 3}\)	M1	rhs
\(4x^2 + 12x + x + 3 = 3x - 3\) or \(4x^2 + 10x + 6\ (= 0)\) or \(2x^2 + 5x + 3\ (= 0)\)	M1dep	oe with expanded brackets dependent on both M marks \(4x^2 + 12x + x + 3\) implies first M mark \(3x - 3\) implies second M mark
\((2x + 3)(x + 1)\ (= 0)\) or if quadratic formula used \(\dfrac{-5 \pm \sqrt{5^2 - 24}}{4}\) or better	M1dep	oe factorised
\(-1\) and \(-1\dfrac{1}{2}\) or \(\dfrac{-3}{2}\) or \(-1.5\)	A1

Additional guidance

Both answers correct without working will get all marks	M4A1
Only one solution correct without working scores nothing	M0A0
No oe for A mark
Possible to make both sides powers of 8. Mark to the equivalent of this MS

June 2024 Paper 2 Q16

AQACurrent spec3 marksIndices

16 Simplify \(\quad \dfrac{\sqrt{x^3}\left(\sqrt{x^3} + x^3\right)}{\sqrt{x}}\)

Give your answer in the form \(\quad x^{\frac{a}{b}} + x^c \quad\) where \(a\), \(b\) and \(c\) are integers. [3 marks]

Mark scheme

Answer	Mark	Comments
\(x^{\frac{5}{2}} + x^4\)	B3	B2 \(x^{\frac{5}{2}}\ (+ \ldots)\) or \((\ldots +)\ x^4\) B1 correct expansion of numerator eg \(\sqrt{x^3} \times \sqrt{x^3} + \sqrt{x^3} \times x^3\) or \(x^3 + x^{\frac{9}{2}}\) or correct expression with fraction eliminated eg \(\sqrt{x^2}\left(\sqrt{x^3} + x^3\right)\) or \(x\left(\sqrt{x^3} + x^3\right)\) or \(\sqrt{x^3}\left(\sqrt{x^2} + x^{\frac{5}{2}}\right)\)

Additional guidance

B1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
B1 may be awarded after multiplication by \(\dfrac{\sqrt{x}}{\sqrt{x}}\) eg \(\dfrac{\sqrt{x^3}\left(\sqrt{x^3} + x^3\right)}{\sqrt{x}} \times \dfrac{\sqrt{x}}{\sqrt{x}} = \dfrac{\sqrt{x^4}\left(\sqrt{x^3} + x^3\right)}{x}\) \(\dfrac{\sqrt{x^7} + x^3\sqrt{x^4}}{x}\) (B1 at this stage for correct expansion of numerator)	B1
Check the answer line and award B3 or B2 if possible Otherwise check the working for B2 or B1
Condone \(x^4 + x^{\frac{5}{2}}\)	B3
Accept oe values for fractional powers for up to B2
Accept eg \(x^{\frac{10}{4}} + x^4\)	B3

June 2024 Paper 2 Q14

AQACurrent spec3 marksIndices

14 \(P\) is the point on the graph \(\quad y = 3^{-x} \quad\) that has \(x\)-coordinate \(-2\)

\(Q\) is a point on the graph \(\quad y = 1152 \times \left(\dfrac{1}{2}\right)^x\)

\[y\text{-coordinate of } P = y\text{-coordinate of } Q\]

Work out the \(x\)-coordinate of \(Q\). [3 marks]

Mark scheme

Answer	Mark	Comments
\(3^2\) or 9	M1	oe value for \(y\) eg \(3^{--2}\)
\(\left(\dfrac{1}{2}\right)^x = \dfrac{\text{their } 9}{1152}\) or \(\dfrac{1}{2^x} = \dfrac{1}{128}\)	M1dep	oe equation with exponential term isolated eg \(2^{-x} =\) [0.0078, 0.007813] or \(2^x = 128\)
7	A1

Additional guidance

M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
Missing brackets must be recovered eg1 \(\dfrac{1}{2}^{x} = \dfrac{9}{1152}\) with no further correct work eg2 \(\dfrac{1}{2}^{x} = \dfrac{9}{1152}\) with answer 7	M1M0A0 M2A1
Logarithms may be seen eg \(\log\left(\dfrac{1}{2}\right)^x = \log\dfrac{9}{1152}\) or \((x =)\ \log_{0.5}\dfrac{1}{128}\)	M1M1

June 2024 Paper 2 Q8

AQACurrent spec3 marksDifferentiation Indices

\[y = \frac{1}{2}x^2 + \frac{3}{4x^4}\]

Work out \(\quad \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\)

Give your answer in the form \(\quad a + bx^n \quad\) where \(a\), \(b\) and \(n\) are integers. [3 marks]

Mark scheme

Answer	Mark	Comments
\(1 + 15x^{-6}\)	B3	B2 answer \((\ldots +)\ 15x^{-6}\) or answer \((\ldots +)\ \dfrac{15}{x^6}\) or \(x\) and \(-3x^{-5}\) B1 \(x\) or \(-3x^{-5}\) SC2 \(1 - 15x^{-6}\)

Additional guidance

Answer must be in the required form
eg \(1 + 15x^{-6}\) in working with answer \(1 + \dfrac{15}{x^6}\)	B2
Accept equivalents for \(x\) eg \(\dfrac{1}{2} \times 2x^{2-1}\)
Accept equivalents for \(-3x^{-5}\) eg \(-\dfrac{3}{x^5}\) or \(\dfrac{3}{4} \times -4 \times x^{-4-1}\)
\(\ldots\) may be incorrect or blank
Condone \(15x^{-6} + 1\)	B3
Condone \(1 + -15x^{-6}\) or \(1 + -\dfrac{15}{x^6}\)	SC2
Condone eg \(y = x - 3x^{-5}\)	B2

June 2024 Paper 1 Q5

AQACurrent spec3 marksDifferentiation Indices

\[y = \frac{4x^3 + x^7}{x^4}\]

Work out \(\quad \dfrac{\mathrm{d}y}{\mathrm{d}x}\) [3 marks]

Mark scheme

Answer	Mark	Comments
\(4x^{-1}\) or \(\dfrac{4}{x}\) or \(x^3\)	M1	either term correct (ignore additional terms)
\(-4x^{-2}\) or \(3x^2\)	A1	a correct term differentiated correctly (must be just two terms before differentiation) accept \(-\dfrac{4}{x^2}\)
\(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) -4x^{-2} + 3x^2\)	A1	both terms correct accept \(-\dfrac{4}{x^2}\)

Additional guidance

Quotient rule or product rule used to achieve correct simplified answer	M1A2
Quotient rule or product rule used with incorrect or unsimplified answer	M0A0
\(\dfrac{-4 + 3x^4}{x^2}\)	M1A2
Penalise additional incorrect further working eg \(\dfrac{-1}{4x^2}\) for the final A mark

June 2024 Paper 1 Q1

AQACurrent spec2 marksIndices

1 Work out the value of \(\quad \sqrt{\dfrac{t}{20}} \quad\) where \(\quad t = 2.42 \times 10^3\) [2 marks]

Mark scheme

Answer	Mark	Comments
2420 or 121 or \(1.21 \times 10^2\)	M1	doesn’t need to be seen in a correct expression
11	A1	accept \(\pm 11\)

Additional guidance

Condone \(t = 11\)	M1A1
Condone \(1.1 \times 10^1\) or \(1.1 \times 10\)	M1A1
\(\dfrac{1210}{10}\) or \(\dfrac{242}{2}\)	M1
\(\dfrac{\sqrt{121}}{10}\) whilst incorrect would gain the method mark for 121	M1

June 2023 Paper 2 Q15

AQACurrent spec4 marksFunctions Indices

15 The function g is given by \(\quad \mathrm{g}(x) = a \times b^x \quad\) where \(a\) and \(b\) are constants.

The domain of the function is \(\quad -1 \leqslant x \leqslant 2\)

\(P\left(0, \dfrac{1}{2}\right)\) and \(Q\left(1, \dfrac{3}{2}\right)\) are points on the graph \(\quad y = \mathrm{g}(x)\)

Graph of the increasing curve y = g(x) for x from −1 to 2, passing through P on the y-axis and Q at x = 1

Not drawn accurately

Work out the range of the function. [4 marks]

Mark scheme

Answer	Mark	Comments
\(\dfrac{1}{2} = a \times b^0\) or \(a = \dfrac{1}{2}\) or \(\dfrac{3}{2} = a \times b^{(1)}\)	M1	oe eg \(a \times 1 = 0.5\) or \(ab = 1.5\)
(\(a =\)) \(\dfrac{1}{2}\) and (\(b =\)) 3	A1	oe values may be implied eg \(y = \dfrac{1}{2} \times 3^x\)
their \(\dfrac{1}{2} \times (\text{their } 3)^{-1}\) or \(\dfrac{1}{2} \times \dfrac{1}{3}\) or \(\dfrac{1}{6}\) and their \(\dfrac{1}{2} \times (\text{their } 3)^2\) or \(\dfrac{1}{2} \times 3^2\) or \(\dfrac{9}{2}\)	M1	oe neither their \(\dfrac{1}{2}\) nor their 3 can be 0 or 1 \(\dfrac{1}{6}\) and \(\dfrac{9}{2}\) is M1A1M1 ignore other calculations or values
\(\dfrac{1}{6} \leqslant \mathrm{g}(x) \leqslant \dfrac{9}{2}\) or both of \(\mathrm{g}(x) \geqslant \dfrac{1}{6} \qquad \mathrm{g}(x) \leqslant \dfrac{9}{2}\)	A1ft	oe eg \(\left[\dfrac{1}{6}, \dfrac{9}{2}\right]\) only ft M1A0M1

Additional guidance

Ignore words if answer given as separate inequalities eg1 \(\mathrm{g}(x) \geqslant \dfrac{1}{6}\) or \(\mathrm{g}(x) \leqslant \dfrac{9}{2}\) eg2 \(\mathrm{g}(x) \geqslant \dfrac{1}{6}\) and \(\mathrm{g}(x) \leqslant 4.5\)	M1A1M1A1 M1A1M1A1
\(a = \dfrac{1}{2}\) and \(b = 2\) \(\dfrac{1}{4} \leqslant \mathrm{g}(x) \leqslant 2\)	M1A0 M1A1ft
Condone \(\mathrm{g}(x)\) replaced by eg \(y\) or g or g\(x\) or \(\mathrm{f}(x)\) or f or f\(x\) or \(a \times b^x\)
Do not allow \(\mathrm{g}(x)\) to be replaced by \(x\) for the final mark eg \(\dfrac{1}{6} \leqslant x \leqslant \dfrac{9}{2}\)	M1A1M1A0
Equivalent inequalities may be seen eg \(\dfrac{9}{2} \geqslant \mathrm{g}(x) \geqslant \dfrac{1}{6}\)	M1A1M1A1
Inequality symbols must be correct eg \(\dfrac{1}{6} \lt \mathrm{g}(x) \leqslant \dfrac{9}{2}\) or \(\dfrac{1}{6} \leqslant \mathrm{g}(x) \geqslant \dfrac{9}{2}\)	M1A1M1A0
\(\dfrac{1}{6} \leqslant \mathrm{g}(x) \leqslant \dfrac{9}{2}\) in working with list of integers on answer line	M1A1M1A0
Only a list of integers	M0A0M0A0
Allow 0.16(…) or 0.17 for \(\dfrac{1}{6}\)
Correct answer followed by incorrect further work eg \(\dfrac{1}{6} \leqslant \mathrm{g}(x) \leqslant \dfrac{9}{2}\) and \(\left(\dfrac{9}{2} - \dfrac{1}{6} =\right) \dfrac{13}{3}\)	M1A1M1A0

June 2023 Paper 1 Q15

AQACurrent spec4 marksIndices Solving Quadratics

15 Solve \(\quad \left(x^{\frac{1}{2}} - x^{\frac{3}{2}}\right)^2 = x^2 + x\) [4 marks]

Mark scheme

Answer	Mark	Comments
\(x - x^2 - x^2 + x^3 \quad (= x^2 + x)\)	M1	allow error in 1 term but must not have missing terms (could be seen in a grid) if they go straight to \(x - 2x^2 + x^3\) then M1 if they go straight to \(x - kx^2 + x^3\) where \(k\) is not 2 then this could be more than one error so M0
\(x^3 - 3x^2 = 0\) or \(x^3 = 3x^2\)	M1dep	no ft
\(x^2(x - 3)(= 0)\)	M1	factorising their equation fully and correctly (must have \(x^3\) plus/minus other term(s) in \(x^n\) where n\(\neq\)0).
(\(x =\)) 0 and (\(x =\)) 3	A1	condone \(x = 0\) \(x = 0\) \(x = 3\)

Additional guidance

First M mark could be done in root form but would still need to end up with \(x - x^2 - x^2 + x^3 \quad (= x^2 + x)\) for M1
Candidates may divide by \(x\) at some point during the method. If they do this and state that \(x = 0\) is a solution at this point, the method can be continued to a correct outcome for full marks or \(x^2 - 3x = 0\) or \(x^2 = 3x\) would be M1dep and \(x(x - 3)(= 0)\) would be the third M mark eg \(x - x^2 - x^2 + x^3 = x^2 + x\) \(1 - x - x + x^2 = x + 1\) with \(x = 0\) \(x^2 - 3x = 0\) If they do not state \(x = 0\) is a solution and continue to a correct solution then award full marks but if it’s not stated and they don’t get to the correct solution then no further marks available as soon as the division is done eg \(x - x^2 - x^2 + x^3 = x^2 + x\) \(1 - x - x + x^2 = x + 1\) (no mention of \(x = 0\) being a possible solution) \(x^2 - 3x = 0\)	M1 M1dep M1 M0dep
If no other marks awarded then \(x = 0\) or \(x = 3\) seen on answer line (or the last thing they have written)	SC1
\(x = 0\) and \(x = 3\) not from incorrect working	M3 A1

June 2023 Paper 2 Q13

AQACurrent spec6 marksIndices

(a) Circle the expression equivalent to \(\quad 3x^{-7}\) [1 mark]

\(-\dfrac{3}{x^7}\)
\(-\dfrac{1}{3x^7}\)
\(\dfrac{1}{3x^7}\)
\(\dfrac{3}{x^7}\)

(b) Simplify fully \(\quad \dfrac{12w^8}{\left(4w^3\right)^2}\) [2 marks]

Work out the least possible values of \(c\) and \(d\). [3 marks]

Answer	Mark	Comments
\(\dfrac{3}{x^7}\)	B1

Mark scheme (b)

Answer	Mark	Comments
\(\dfrac{3w^2}{4}\) or \(0.75w^2\)	B2	B1 unsimplified equivalent fraction with brackets processed eg \(\dfrac{12w^8}{16w^6}\) or \(\dfrac{12w^2}{16}\) or \(\dfrac{6w^2}{8}\) or \(\dfrac{3w^5}{4w^3}\) SC1 \(\dfrac{3w^3}{4}\) or \(0.75w^3\) or \(3w^2\)

Additional guidance

B1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
\(\dfrac{3}{4}w^2\) is equivalent to \(\dfrac{3w^2}{4}\) etc

Mark scheme (c)

Answer	Mark	Comments
\(c = 6\) and \(d = 5\)	B3	B2 \(y^{\frac{5k}{6k}}\) or \(\sqrt[6k]{y^{5k}}\) or \(\left(y^{5k}\right)^{\frac{1}{6k}}\) or \(\left(y^{\frac{1}{6k}}\right)^{5k}\) where \(k\) is an integer \(\geqslant 1\) or \(\left(\dfrac{1}{2} + \dfrac{1}{3} =\right) \dfrac{5}{6}\) oe fractions or \(\left(\dfrac{c}{2} + \dfrac{c}{3} =\right) \dfrac{5c}{6}\) oe fractions or \(5c = 6d\) oe equation B1 \(y^{\frac{1}{2}}\) or \(y^{\frac{1}{3}}\) or \(y^{\frac{c}{2}}\) or \(y^{\frac{c}{3}}\) or \(\left(y^c\right)^{\frac{1}{2}}\) or \(\left(y^c\right)^{\frac{1}{3}}\) or \(y^{\frac{d}{c}}\) or \(\left(y^{\frac{1}{c}}\right)^d\) or \(\left(y^d\right)^{\frac{1}{c}}\) or \(\dfrac{1}{2} + \dfrac{1}{3}\) oe sum of fractions or \(\dfrac{c}{2} + \dfrac{c}{3}\) oe sum of fractions SC2 \(c = 6k\) and \(d = 5k\) where \(k\) is an integer \(\geqslant 2\)

Additional guidance

Up to B2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
\(\dfrac{3}{6} + \dfrac{2}{6} = \dfrac{5}{6}\)	B2
\(\dfrac{3}{6} + \dfrac{2}{6}\)	B1
\(\dfrac{10}{12}\) or \(y^{\frac{10}{12}}\)	B2
\(c = 18\) and \(d = 15\)	SC2
\(\dfrac{1}{2}\) may be 0.5
Only trialling numerical values for \(y\) is B0 or B3

June 2022 Paper 2 Q22

AQACurrent spec4 marksIndices Solving Quadratics

22 Using powers of 2 or otherwise, work out the non-zero value of \(x\) for which

\[\left(16^x\right)^x = \dfrac{1}{2^{3x}}\]

You must show your working. [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1 Uses powers of 2
\((16^x =)\; 2^{4x}\) or \(((16^x)^x =)\; (2^4)^{x^2}\)	M1	implied by \(((16^x)^x =)\; 2^{4x^2}\) may be implied by 3rd M1
\(((16^x)^x =)\; 2^{4x^2}\)	M1dep	implied by \(2^{4x^2 + 3x}\) may be implied by 3rd M1
Correct quadratic equation or correct linear equation or correct equation involving indices with the same base	M1dep	eg \(4x^2 = -3x\) or \(4x^2 + 3x = 0\) or \(4x = -3\) or \(2^{4x^2} = 2^{-3x}\) or \(2^{4x^2 + 3x} = 2^0\) do not allow if the equation is from incorrect working do not allow if the only equation is \(x = -\dfrac{3}{4}\)
M3 and \(-\dfrac{3}{4}\)	A1	oe ignore inclusion of answer 0
Alternative method 2 Uses powers of 16
\(((16^x)^x =)\; 16^{x^2}\) or \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{\left(16^{\frac{1}{4}}\right)^{3x}}\)	M1	implied by \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{16^{\frac{3x}{4}}}\) or \(\left(\dfrac{1}{2^{3x}} =\right) 16^{-\frac{3x}{4}}\) may be implied by 3rd M1
\(((16^x)^x =)\; 16^{x^2}\) and \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{16^{\frac{3x}{4}}}\)	M1dep	oe eg \(((16^x)^x =)\; 16^{x^2}\) and \(\left(\dfrac{1}{2^{3x}} =\right) 16^{-\frac{3x}{4}}\) may be implied by 3rd M1
Correct quadratic equation or correct linear equation or correct equation involving indices with the same base	M1dep	eg \(x^2 = -\dfrac{3}{4}x\) or \(4x^2 + 3x = 0\) or \(16^{x^2} = 16^{-\frac{3x}{4}}\) do not allow if the equation is from incorrect working do not allow if the only equation is \(x = -\dfrac{3}{4}\)
M3 and \(-\dfrac{3}{4}\)	A1	oe ignore inclusion of answer 0
Alternative method 3 Uses powers of 4
\((16^x =)\; 4^{2x}\) or \(((16^x)^x =)\; (4^2)^{x^2}\) or \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{\left(4^{\frac{1}{2}}\right)^{3x}}\)	M1	implied by \(((16^x)^x =)\; 4^{2x^2}\) or \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{4^{\frac{3x}{2}}}\) or \(\left(\dfrac{1}{2^{3x}} =\right) 4^{-\frac{3}{2}x}\) may be implied by 3rd M1
\(((16^x)^x =)\; 4^{2x^2}\) and \(\left(\dfrac{1}{2^{3x}} =\right) \dfrac{1}{4^{\frac{3x}{2}}}\)	M1dep	oe \(((16^x)^x =)\; 4^{2x^2}\) and \(\left(\dfrac{1}{2^{3x}} =\right) 4^{-\frac{3}{2}x}\) may be implied by 3rd M1
Correct quadratic equation or correct linear equation or correct equation involving indices with the same base	M1dep	eg \(2x^2 = -\dfrac{3}{2}x\) or \(4x^2 + 3x = 0\) or \(4^{2x^2} = 4^{-\frac{3}{2}x}\) do not allow if the equation is from incorrect working do not allow if the only equation is \(x = -\dfrac{3}{4}\)
M3 and \(-\dfrac{3}{4}\)	A1	oe ignore inclusion of answer 0
Alternative method 4 Takes the \(x\)th root of each side and uses powers of 2
\((16^x =)\; 2^{4x}\) or \(16^x = \left(\dfrac{1}{2^{3x}}\right)^{\frac{1}{x}}\)	M1	oe eg \(16^x = \sqrt[x]{\dfrac{1}{2^{3x}}}\) or \(16^x = \dfrac{1}{2^3}\) or \(16^x = 2^{-3}\) may be implied by 3rd M1
\(2^{4x} = \left(\dfrac{1}{2^{3x}}\right)^{\frac{1}{x}}\)	M1dep	oe eg \(2^{4x} = \dfrac{1}{2^3}\) may be implied by 3rd M1
Correct quadratic equation or correct linear equation or correct equation involving indices with the same base	M1dep	eg \(4x = -3\) or \(2^{4x} = 2^{-3}\) do not allow if the equation is from incorrect working do not allow if the only equation is \(x = -\dfrac{3}{4}\)
M3 and \(-\dfrac{3}{4}\)	A1	oe ignore inclusion of answer 0

Additional guidance

Up to M2 may be awarded for correct work with no, or incorrect answer, even if this is seen amongst multiple attempts
Allow \(2^{4 \times x \times x}\) for \(2^{4x^2}\) etc
Responses using other powers eg powers of 8 can be escalated	Escalate
Ignore simplification or conversion if correct answer seen

June 2022 Paper 2 Q15

AQACurrent spec5 marksAlgebraic Fractions Indices

15 Simplify fully \(\qquad \left(\dfrac{x}{2} + \dfrac{3x}{5}\right) \div \sqrt{\dfrac{x^6}{4}}\) [5 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1 Processes the brackets then divides
\(\dfrac{5x}{10} + \dfrac{6x}{10}\)	M1	oe valid common denominator with both numerators correct eg \(\dfrac{10x}{20} + \dfrac{12x}{20}\)
\(\dfrac{11x}{10}\)	A1	oe single term eg \(\dfrac{22x}{20}\) or \(1.1x\) may be implied eg by single term with roots evaluated that is equivalent to \(\dfrac{11}{5x^2}\)
\(\dfrac{x^{6 \div 2}}{2}\) or \(\dfrac{x^3}{2}\)	M1	may be implied eg by multiplication by \(\dfrac{2}{x^3}\)
their \(\dfrac{11x}{10} \times \dfrac{2}{x^3}\) or \(\dfrac{22x}{10x^3}\) or \(\dfrac{22}{10x^2}\) or \(\dfrac{11x}{5x^3}\) or \(\dfrac{22}{10}x^{-2}\)	M1dep	oe multiplication eg \(\dfrac{11x}{10} \times 2x^{-3}\) their \(\dfrac{11x}{10}\) can be unprocessed dep on 2nd M1
\(\dfrac{11}{5x^2}\) or \(\dfrac{11}{5}x^{-2}\) or \(2.2x^{-2}\)	A1	allow \(2\dfrac{1}{5}x^{-2}\) or \(\dfrac{2.2}{x^2}\)
Alternative method 2 Divides then expands the brackets
\(\dfrac{x^{6 \div 2}}{2}\) or \(\dfrac{x^3}{2}\)	M1	may be implied eg by multiplication by \(\dfrac{2}{x^3}\)
\(\left(\dfrac{x}{2} + \dfrac{3x}{5}\right) \times \dfrac{2}{x^3}\)	M1dep	oe multiplication eg \(\left(\dfrac{x}{2} + \dfrac{3x}{5}\right) \times 2x^{-3}\)
\(\dfrac{2x}{2x^3} + \dfrac{6x}{5x^3}\) or \(\dfrac{1}{x^2} + \dfrac{6}{5x^2}\)	M1dep	oe expansion of brackets
\(\dfrac{10x}{10x^3} + \dfrac{12x}{10x^3}\) or \(\dfrac{5}{5x^2} + \dfrac{6}{5x^2}\) or \(\dfrac{22x}{10x^3}\) or \(\dfrac{22}{10x^2}\) or \(\dfrac{11x}{5x^3}\) or \(\dfrac{22}{10}x^{-2}\)	M1dep	oe valid common denominator with both numerators correct eg \(\dfrac{10x^4}{10x^6} + \dfrac{12x^4}{10x^6}\) or \(\dfrac{22x^4}{10x^6}\) roots must be processed
\(\dfrac{11}{5x^2}\) or \(\dfrac{11}{5}x^{-2}\) or \(2.2x^{-2}\)	A1	allow \(2\dfrac{1}{5}x^{-2}\) or \(\dfrac{2.2}{x^2}\)

Additional guidance

Any single fraction with roots evaluated that is equivalent to \(\dfrac{11}{5x^2}\)	4 marks
Allow inclusion of \(\pm\) from the square root for up to 4 marks
\(\dfrac{11}{5x^2}\) in working with answer \(\dfrac{11}{5}x^2\)	4 marks
Alt 1 \(\dfrac{11x}{10}\) subsequently squared and not recovered	M1A1 M0M0A0

June 2022 Paper 1 Q13

AQACurrent spec3 marksIndices

13 Simplify fully \(\quad (6x^3y^{-2} + 9x^5y) \div 3x^2y^{-3}\) [3 marks]

Mark scheme

Answer	Mark	Comments
2 and 3	B1	coefficients
\(x\) and \(x^3\)	B1
\(y\) and \(y^4\)	B1

Additional guidance

\(2xy + 3x^3y^4\) or \(xy(2 + 3x^2y^3)\) scores B3
If no B marks awarded then \(3x^3(2y^{-2} + 3x^2y)\)
or \(3x^3y(2y^{-3} + 3x^2)\)
or \(3x^3y^{-2}(2 + 3x^2y^3)\)
or \(3x^2(2xy^{-2} + 3x^3y)\)
or \(3x^2y(2xy^{-3} + 3x^3)\)
or \(3x^2y^{-2}(2x + 3x^3y^3)\) seen in the working for the numerator
Penalise incorrect further working for the B marks

B3
SC1

June 2022 Paper 2 Q12

AQACurrent spec5 marksDifferentiation Indices

12 Work out the gradient of the curve \(\qquad y = \dfrac{12x^3 - 8x + 3}{4x^2}\)

at the point where \(\quad x = -1\)

You must show your working. [5 marks]

Mark scheme

Answer	Mark	Comments
\(3x\) or \(-2x^{-1}\) or \(0.75x^{-2}\)	M1	oe must have powers of \(x\) simplified eg \(\dfrac{12x}{4}\) or \(-\dfrac{2}{x}\) or \(\dfrac{3}{4x^2}\)
\(3x\) and \(-2x^{-1}\) and \(0.75x^{-2}\)	M1dep	oe must have powers of \(x\) simplified eg \(\dfrac{12x}{4}\) and \(-\dfrac{2}{x}\) and \(\dfrac{3}{4x^2}\)
Any one of \(3x\) and \(3(x^0)\) or \(-2x^{-1}\) and \(2x^{-2}\) or \(0.75x^{-2}\) and \(-1.5x^{-3}\)	M1	oe eg \(\dfrac{12x}{4}\) and \(\dfrac{12}{4}x^{1 - 1}\) or \(-\dfrac{2}{x}\) and \(\dfrac{2}{x^2}\) or \(-\dfrac{2}{x}\) and \(- -2x^{-2}\) or \(\dfrac{3}{4x^2}\) and \(-\dfrac{3}{2x^3}\) implies 1st M1 for the derivatives \(x\) may be \((-1)\)
At least two of \(3x\) and \(3(x^0)\) or \(-2x^{-1}\) and \(2x^{-2}\) or \(0.75x^{-2}\) and \(-1.5x^{-3}\)	M1dep	oe dep on 3rd M1 for the derivatives \(x\) may be \((-1)\)
All three terms and their derivatives correct and 6.5	A1	oe eg all three terms and their derivatives correct and \(\dfrac{13}{2}\) for the derivatives \(x\) may be \((-1)\) SC3 104

Additional guidance

Up to M4 may be awarded for correct work with no, or incorrect answer, even if this is seen amongst multiple attempts
\(\dfrac{3}{4x^2}\) seen but subsequently incorrectly simplified eg \(12x^{-2}\) (subsequent marks may be scored)	(1st) M1
Correct answer after correct use of quotient rule or product rule	M4A1
Incorrect answer after use of quotient rule or product rule	Zero
Condone \(y = 3 + 2x^{-2} \ldots\) etc
All three terms and their derivatives correct and 6.5 in working but different answer eg \(y = 6.5x \ldots\)	M4A0
SC3 is for multiplying the numerator by \(4x^{-2}\) with no subsequent errors

June 2022 Paper 2 Q10

AQACurrent spec1 markDrawing & Using Graphs Indices

10 Here are four sketch graphs.

Circle the letter of the sketch graph that represents \(\quad y = 3 \times 2^x\) [1 mark]

Four sketch graphs. A: decreasing curve crossing the y-axis at (0, 6). B: decreasing curve crossing the y-axis at (0, 3). C: increasing curve crossing the y-axis at (0, 6). D: increasing curve crossing the y-axis at (0, 3)

Answer	Mark	Comments
D	B1

November 2021 Paper 1 Q22

AQACurrent spec4 marksIndices

22 Solve \(\quad 8^x = \dfrac{2^{56} - 4^{26}}{30}\) [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1 Working with powers of 2
\(2^{3x}\) or \(2^{52}\)	B1	from \(8^x\) and \(4^{26}\)
\(2^{52}(2^4 - 1)\) or \(2^{52} \times 15\) or \(2 \times 8^x = 2^{52}\) or \(2 \times 2^{3x} = 2^{52}\)	M1	oe
\(2^{3x + 1} = 2^{52}\) or \(3x + 1 = 52\)	M1dep	oe eg \(2^{3x} = 2^{51}\)
17	A1
Alternative method 2 Working with powers of 4
\(4^{1.5x}\) or \(4^{28}\)	B1	from \(8^x\) and \(2^{56}\)
\(4^{26}(4^2 - 1)\) or \(4^{26} \times 15\) or \(2 \times 4^{1.5x} = 4^{26}\)	M1	oe
\(4^{1.5x} = \dfrac{4^{26}}{4^{0.5}}\) or \(1.5x = 26 - 0.5\)	M1dep	oe eg \(4^{1.5x} = 4^{25.5}\)
17	A1
Alternative method 3 Working with powers of 8
\(8^{\frac{56}{3}}\) or \(8^{\frac{52}{3}}\)	B1	from \(2^{56}\) and \(4^{26}\)
\(8^{\frac{52}{3}}\left(8^{\frac{4}{3}} - 1\right)\) or \(8^{\frac{52}{3}} \times 15\) or \(2 \times 8^x = 8^{\frac{52}{3}}\)	M1	oe
\(8^x = 8^{\frac{52}{3}} \div 8^{\frac{1}{3}}\) or \(x = \dfrac{52}{3} - \dfrac{1}{3}\)	M1dep	oe eg \(8^x = 8^{17}\)
17	A1

November 2021 Paper 1 Q13

AQACurrent spec3 marksDrawing & Using Graphs Indices

13 Here is a sketch of \(\quad y = k^x \quad\) where \(\quad k \gt 0\)

\(A\left(2,\ 3\frac{1}{16}\right)\) is a point on the curve.

Sketch of an increasing exponential curve y = k^x crossing the positive y-axis, with the point A (2, 3 1/16) marked on it

Not drawn accurately

(a) Work out the value of \(k\). [2 marks]

(b) \(B\) is a point on the curve with \(x\)-coordinate \(-1\)

Work out the \(y\)-coordinate of \(B\). [1 mark]

Mark scheme (a)

Answer	Mark	Comments
\(k^2 = \dfrac{49}{16}\) or \(k = \sqrt{\dfrac{49}{16}}\)	M1	oe
\(\dfrac{7}{4}\)	A1	oe

Additional guidance

\(-\dfrac{7}{4}\) or \(\pm\dfrac{7}{4}\) will not gain the A mark unless recovered

M1A0

Answer	Mark	Comments
\(\dfrac{4}{7}\)	B1ft	oe allow decimal rounded to 2dp or better ft their (a) but not if \(k = 0\) or 1

November 2021 Paper 2 Q7

AQACurrent spec4 marksIndices

(a) \(a\) is a value greater than 1

Work out the value of \(m\) for which \(\quad (a^m)^4 = (a^5)^{2m}\) [2 marks]

(b) \(w^3x^2y^5 = w^{13}x^7\)

Write \(y\) in terms of \(w\) and \(x\).

Give your answer in its simplest form. [2 marks]

Answer	Mark	Comments
\(a^{4m}\) or \(a^{10m}\) or \(4m = 10m\)	M1	oe eg \(a^{4 \times m}\)
0	A1

Mark scheme (b)

Answer	Mark	Comments
\(w^{13}x^7 \div w^3x^2\) or \(w^{10}x^5\) or \(x^2y^5 = w^{10}x^7\) or \(y^5 = \dfrac{w^{10}x^7}{x^2}\) or \(w^3y^5 = w^{13}x^5\) or \(y^5 = \dfrac{w^{13}x^5}{w^3}\)	M1	oe eg \(\dfrac{w^{13}x^7}{w^3x^2}\) may be embedded eg \(\sqrt[5]{w^{10}x^5}\)
\(w^2x^{(1)}\)	A1	oe eg \(xw^2\)

Additional guidance

\(y = w^{10}x^5\)

M1A0

Specimen 2020 Paper 1 Q18

AQACurrent spec3 marksIndices

18 Solve \(\quad x^{-\frac{2}{3}} = 12\dfrac{1}{4}\) [3 marks]

Mark scheme

Answer	Mark	Comments
\(\left(x^{-\frac{2}{3}} =\right)\ \dfrac{49}{4}\) or \(x^{\frac{2}{3}} = \dfrac{4}{49}\)	M1
\(\left(\dfrac{49}{4}\right)^{\frac{3}{2}}\) or \(\left(\dfrac{4}{49}\right)^{\frac{3}{2}}\)	M1dep	oe
\(\dfrac{8}{343}\)	A1

Specimen 2020 Paper 2 Q17

AQACurrent spec4 marksIndices Solving Quadratics

17 By multiplying both sides of the equation by \(x^{\frac{1}{2}}\)

Solve \(\quad 2x^{\frac{3}{2}} - 3x^{\frac{1}{2}} = 7x^{-\frac{1}{2}} \qquad\) for \(\quad x \gt 0\)

Give your answer to 3 significant figures.

You must show your working. [4 marks]

Mark scheme

Answer	Mark	Comments
\(2x^2 - 3x = 7\)	M1	at least two terms correct
\(2x^2 - 3x - 7\ (= 0)\)	A1	oe 3-term quadratic equation
\(\dfrac{--3 \pm \sqrt{(-3)^2 - 4 \times 2 \times -7}}{2 \times 2}\) or \(\dfrac{3}{4} \pm \sqrt{\dfrac{65}{16}}\)	M1	oe correct attempt to solve their 3-term quadratic equation
2.77	A1	2.77 and \(-1.27\) is A0

Specimen 2020 Paper 2 Q15

AQACurrent spec4 marksIndices Simultaneous Equations

15 Here is a sketch of the curve \(\quad y = ab^{-x} \quad\) where \(a\) and \(b\) are positive constants.

(0, 3) and (2, 0.48) lie on the curve.

Decreasing curve crossing the y-axis at (0, 3) and passing through (2, 0.48)

Not drawn accurately

Work out the values of \(a\) and \(b\). [4 marks]

Mark scheme

Answer	Mark	Comments
\(a = 3\)	B1
\(0.48 =\) their \(3 \times b^{-2}\)	M1	oe
\(b^2 = \dfrac{\text{their } 3}{0.48}\) or \(b^2 = 6.25\) \(\sqrt{\dfrac{\text{their } 3}{0.48}}\) or \(\sqrt{6.25}\)	M1dep	oe
\(b = 2.5\)	A1ft	ft B0M2

Specimen 2020 Paper 1 Q10

AQACurrent spec5 marksFunctions Indices

10 \(\mathrm{f}(x) = \left(\dfrac{9x}{2}\right)^{-1}\)

\(\mathrm{g}(x) = \sqrt{1 - px^3} \qquad\) where \(p\) is a constant.

Given that \(\ \mathrm{f}\left(\dfrac{1}{3}\right) = \mathrm{g}\left(\dfrac{1}{3}\right)\ \) work out the value of \(p\). [5 marks]

Mark scheme

Answer	Mark	Comments
\(\dfrac{9}{2} \times \dfrac{1}{3}\) or \(\dfrac{3}{2}\) or \(\dfrac{2}{9x}\)	M1	oe
\(\dfrac{2}{3}\)	A1
their \(\dfrac{2}{3} = \sqrt{1 - p \times \left(\dfrac{1}{3}\right)^3}\)	M1dep	oe
\(\left(\text{their } \dfrac{2}{3}\right)^2 = 1 - p \times \left(\dfrac{1}{3}\right)^3\)	M1dep	oe
15	A1

Specimen 2020 Paper 2 Q10

AQACurrent spec5 marksDifferentiation Indices

\[y = \frac{6x^9 + x^8}{2x^4}\]

Work out the value of \(\quad \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} \quad\) when \(\quad x = 0.5\) [5 marks]

Mark scheme

Answer	Mark	Comments
\(\dfrac{6x^9}{2x^4} + \dfrac{x^8}{2x^4}\) or \(3x^5\) or \(\dfrac{1}{2}x^4\)	M1
\(3x^5 + \dfrac{1}{2}x^4\)	A1
\(15x^4\) or \(2x^3\)	M1dep	differentiates at least one term correctly
\(60x^3 + 6x^2\)	M1dep	differentiates their 2-term \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) correctly
9	A1

Specimen 2020 Paper 1 Q5

AQACurrent spec3 marksDifferentiation Indices

\[y = 2x^{10} - \frac{3}{x^2}\]

Work out \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) [3 marks]

Answer	Mark	Comments
\(-3x^{-2}\)	M1
\(20x^9\) or \(+6x^{-3}\)	M1
\(20x^9 + 6x^{-3}\)	A1	no additional terms

Specimen 2020 Paper 1 Q2

AQACurrent spec2 marksIndices Solving Basic Equations

2 Solve \(\quad \sqrt[3]{7x - 13} = 2\) [2 marks]

Specimen 2020 Paper 1 Q1

AQACurrent spec4 marksIndices

(a) \(\dfrac{y^6 \times y}{y^m} = y^4\)

Circle the value of \(m\). [1 mark]

\(-2\)
1.5
2
3

(b) \(a^n \times a^5 = a^5\)

Work out the value of \(n\). [1 mark]

Work out the value of \(p\). [2 marks]

June 2019 Paper 2 Q9

AQAOld spec2 marksFactorising Indices

9 Factorise fully \(\quad (p + 6)^{11} - (p + 6)^{10}\) [2 marks]

Mark scheme

Answer	Mark	Comments
\((p + 6)^{10}(p + 5)\) or \((p + 5)(p + 6)^{10}\)	B2	B1 \((p + 6)^{10}(p + 6 - 1)\) or states \(x = p + 6\) and \(x^{10}(x - 1)\) (any letter for \(x\) other than \(p\)) or correct partial factorisation eg \((p + 6)[(p + 6)^{10} - (p + 6)^9]\) or \((p + 6)^2[(p + 6)^9 - (p + 6)^8]\)

Additional guidance

Any shape of bracket may be used
\((p + 6)^{10}((p + 6) - 1)\)	B1
Missing brackets must be recovered eg \(\ p + 5\,(p + 6)^{10}\) not recovered and B1 response not seen	B0
Condone \(\ (p + 6)^{10}(p + 5\)	B2
Condone \(\ (p + 6)^{10}(p + 6 - 1\)	B1
\((p + 6)^{10}(p + 5)\) followed by expansion attempt	B1
B1 response followed by expansion attempt	B1
\((p + 6)^{10} \times (p + 5)\)	B1
Condone multiplication signs for B1 eg \(\ (p + 6)^{10} \times (p + 6 - 1)\)	B1
\((p + 6)^{11}\left[1 - \dfrac{1}{p + 6}\right]\)	B1

June 2019 Paper 1 Q5

AQAOld spec3 marksIndices Solving Basic Equations

5 Solve \(\quad \sqrt[3]{\left(2\sqrt{x} - 10\right)} = 2\) [3 marks]

Mark scheme

Answer	Mark	Comments
\(2\sqrt{x} - 10 = 2^3\) or \(2\sqrt{x} - 10 = 8\) or \(2\sqrt{x} = 18\)	M1
\(\sqrt{x} = \dfrac{2^3+10}{2}\) or \(\sqrt{x} = \dfrac{8+10}{2}\) or \(\sqrt{x} = 9\) or \(4x = 18^2\) or \(x = 9^2\)	M1dep
\(x = 81\)	A1	\(\pm 81\) scores A0

June 2019 Paper 2 Q5

AQAOld spec5 marksIndices

(a) Write \(\quad \sqrt[4]{a \times a^{-9}} \quad\) as an integer power of \(a\). [2 marks]

(b) Simplify fully \(\quad \dfrac{(4cd^2)^3}{2cd^4}\) [3 marks]

Mark scheme (a)

Answer	Mark	Comments
\(a^{-2}\)	B2	B1 applies an index law or changes fourth root to fractional or decimal power in a correct expression eg \(\sqrt[4]{a^{-8}}\) or \((a^{-8})^{\frac{1}{4}}\) or \((a^8)^{-\frac{1}{4}}\) or \(a^{-\frac{8}{4}}\) or \(\dfrac{1}{a^{\frac{8}{4}}}\) or \(\dfrac{1}{a^2}\) or \(a^{\frac{1}{4}} \times a^{-\frac{9}{4}}\) or \(a^{\frac{1}{4}} \times \dfrac{1}{a^{\frac{9}{4}}}\) or \(\left(\dfrac{1}{a^8}\right)^{\frac{1}{4}}\) or \(\sqrt[4]{\dfrac{1}{a^8}}\) or \(\sqrt[4]{a \times \dfrac{1}{a^9}}\) or \((a \times a^{-9})^{\frac{1}{4}}\) or \(\left(a \times \dfrac{1}{a^9}\right)^{\frac{1}{4}}\)

Additional guidance

\(a^{\frac{-8}{4}}\) or \(a^{\frac{8}{-4}}\)	B1
\(a^{-2}\) in working with \(-2\) on answer line	B1
\(a^{-2}\) in working with \(\dfrac{1}{a^2}\) on answer line	B1
B1 response followed by further work is still awarded B1
Allow 0.25 for \(\dfrac{1}{4}\) etc
Allow recovery of missing brackets

Mark scheme (b)

Answer	Mark	Comments
\(32c^2d^2\) or \(32(cd)^2\)	B3	B2 (numerator \(=\)) \(64c^3d^6\) or single term answer with two of 32, \(c^2\) and \(d^2\) (not in a denominator) B1 single term answer with one of 32, \(c^2\) and \(d^2\) (not in a denominator) SC2 factorised correct expression eg \(16cd(2cd)\)

Additional guidance

\(2c^2d^2\) or \(32c^2d\) or \(32c^2\) or \(\dfrac{32d^2}{c^3}\) or \(\dfrac{c^2d^2}{32}\) or \(64(cd)^2\) etc	B2
\(32c^3d\) or \(c^2\) or \(\dfrac{d^2}{c}\) or \(\dfrac{c^2d}{32}\) or \(\dfrac{32}{c^2}\) etc	B1
\(\dfrac{32c^2d^2}{1}\) or \(\dfrac{32(cd)^2}{1}\)	B2
Allow denominator of 1 in a B2 or B1 answer eg \(\dfrac{32c^2d}{1}\)	B2
Multiplication signs in a correct expression eg \(32 \times c^2 \times d^2\)	B2
Allow multiplication signs in a B2, SC2 or B1 answer eg \(32 \times c^3 \times d\)	B1
Do not accept \(2^5\) for 32 eg \(2^5c^2d\)	B1
If answer line scores B1 or B0 check working lines for possible response for up to 2 marks
\(32c^2d^2\) in working with different answer on answer line	B2

June 2018 Paper 2 Q24

AQAOld spec4 marksIndices

24 Work out the value of \(p\) when

\(9^{0.5p} \times 81 = 27^{2p - 1}\) [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1 Powers of 3
\((3^2)^{0.5p}\) or \((3^3)^{2p - 1}\) or \(3^{2 \times 0.5p + 4}\)	M1	oe powers of 3 eg \(3^p\) or \(3^{6p - 3}\) or \(3^{p + 4}\) brackets not needed if intention clear eg \(3^{2^{0.5p}}\)
\((3^2)^{0.5p}\) and \(3^4\) and \((3^3)^{2p - 1}\) or \(3^{2 \times 0.5p + 4}\) and \((3^3)^{2p - 1}\)	M1dep	oe powers of 3 eg \(3^p\) and \(3^4\) and \(3^{6p - 3}\) or \(3^{p + 4}\) and \(3^{6p - 3}\)
\(2 \times 0.5p + 4 = 3(2p - 1)\) or \(p + 4 = 6p - 3\)	M1dep	oe equation dep on M2
1.4 or \(\dfrac{7}{5}\)	A1	oe
Alternative method 2 Powers of 9
\(9^{0.5p + 2}\) or \((9^{1.5})^{2p - 1}\)	M1	oe power of 9 eg \(9^{3p - 1.5}\) brackets not needed if intention clear eg \(9^{1.5^{2p - 1}}\)
\(9^2\) and \((9^{1.5})^{2p - 1}\) or \(9^{0.5p + 2}\) and \((9^{1.5})^{2p - 1}\)	M1dep	oe powers of 9 eg \(9^2\) and \(9^{3p - 1.5}\) or \(9^{0.5p + 2}\) and \(9^{3p - 1.5}\)
\(0.5p + 2 = 1.5(2p - 1)\) or \(0.5p + 2 = 3p - 1.5\)	M1dep	oe equation dep on M2
1.4 or \(\dfrac{7}{5}\)	A1	oe
Alternative method 3 Powers of 27
\(\left(27^{\frac{2}{3}}\right)^{0.5p}\)	M1	oe power of 27 eg \(27^{\frac{2}{3} \times 0.5p}\) or \(27^{\frac{1}{3}p}\) brackets not needed if intention clear eg \(27^{\frac{2}{3}^{0.5p}}\)
\(\left(27^{\frac{2}{3}}\right)^{0.5p}\) and \(27^{\frac{4}{3}}\)	M1dep	oe powers of 27 eg \(27^{\frac{2}{3} \times 0.5p}\) and \(27^{\frac{4}{3}}\) or \(27^{\frac{1}{3}p}\) and \(27^{\frac{4}{3}}\) M2 \(\ 27^{\frac{2}{3} \times 0.5p + \frac{4}{3}}\) or \(27^{\frac{1}{3}p + \frac{4}{3}}\)
\(\dfrac{2}{3} \times 0.5p + \dfrac{4}{3} = 2p - 1\) or \(\dfrac{1}{3}p + \dfrac{4}{3} = 2p - 1\)	M1dep	oe equation dep on M2
1.4 or \(\dfrac{7}{5}\)	A1	oe
Alternative method 4 Powers of 81
\((81^{0.5})^{0.5p}\) or \((81^{0.75})^{2p - 1}\) or \(81^{0.5 \times 0.5p + 1}\)	M1	oe powers of 81 eg \(81^{0.25p}\) or \(81^{1.5p - 0.75}\) or \(81^{0.25p + 1}\) brackets not needed if intention clear eg \(81^{0.5^{0.5p}}\)
\((81^{0.5})^{0.5p}\) and \((81^{0.75})^{2p - 1}\) or \(81^{0.5 \times 0.5p + 1}\) and \((81^{0.75})^{2p - 1}\)	M1dep	oe powers of 81 eg \(81^{0.25p}\) and \(81^{1.5p - 0.75}\) or \(81^{0.25p + 1}\) and \(81^{1.5p - 0.75}\)
\(0.5 \times 0.5p + 1 = 0.75(2p - 1)\) or \(0.25p + 1 = 1.5p - 0.75\)	M1dep	oe equation dep on M2
1.4 or \(\dfrac{7}{5}\)	A1	oe

Additional guidance

Mark positively if potentially more than one scheme used
Answer 1.4	M3A1
Correct equation implies M3
Just seeing expressions not in an equation and not as powers scores zero eg Alt 1 \(\ 6p - 3\) and \(p + 4\) not in an equation and not as powers of 3	M0M0M0
Allow recovery of missing brackets
Use of logs with answer not 1.4 - escalate

June 2018 Paper 1 Q14

AQAOld spec3 marksExpanding Brackets Indices

14 Work out the value of \(\quad \left(3^{\frac{1}{2}} + 3^{\frac{3}{2}}\right)^2\)

You must show your working. [3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(3^{\frac{1}{2}} \times 3^{\frac{1}{2}} + 3^{\frac{1}{2}} \times 3^{\frac{3}{2}} + 3^{\frac{1}{2}} \times 3^{\frac{3}{2}} + 3^{\frac{3}{2}} \times 3^{\frac{3}{2}}\) or \(\sqrt{3}\sqrt{3} + \sqrt{3}\sqrt{27} + \sqrt{3}\sqrt{27} + \sqrt{27}\sqrt{27}\)	M1	oe allow an error in one term
3 or 9 or 27	M1dep
48	A1
Alternative method 2
\(\sqrt{3}\) and \(3\sqrt{3}\)	M1	\(3\sqrt{3}\) must come from correct working
\((4\sqrt{3})^2\)	M1dep
48	A1
Alternative method 3
\(\left(3^{\frac{1}{2}}\right)^2(1 + 3)^2\)	M1	oe
\(3 \times 4^2\)	M1dep	oe
48	A1

Additional guidance

Alt 1 mark scheme … likely to see a 3 (or 9 or 27) somewhere, so need to be careful that the M1 mark has been earned before awarding A1
In alt 1, for the first M1, we want to see an attempt at the full expansion of the correct terms. Probably 4 terms, but there could be 3 if they combine the middle two terms. eg \((\sqrt{3} + 27)(\sqrt{3} + 27)\) scores M0 because it ought to be \(\sqrt{27}\) not 27

June 2018 Paper 2 Q9

AQAOld spec4 marksDifferentiation Indices

9 \(y = \dfrac{2x^7 + 15x^2}{3x}\)

Work out the value of \(x\) when \(\quad \dfrac{\mathrm{d}y}{\mathrm{d}x} = 133\) [4 marks]

Mark scheme

Answer	Mark	Comments
\(\dfrac{2x^6}{3}\) or \(\dfrac{2}{3}x^6\) or \(\dfrac{15x}{3}\) or \(5x\)	M1	implied by \(\dfrac{2x^6 + \mathrm{a}}{3}\) or \(\dfrac{\mathrm{b} + 15x}{3}\) a can be numerical or algebraic b can be numerical or algebraic allow 0.66… or 0.67 for \(\dfrac{2}{3}\)
\(6 \times \dfrac{2x^5}{3}\) or \(\dfrac{12x^5}{3}\) or \(4x^5\) or \(\dfrac{15}{3}\) or 5	M1dep	correct differentiation of one correct term implied by \(\dfrac{6 \times 2x^5 + \mathrm{a}}{3}\) or \(\dfrac{\mathrm{b} + 15}{3}\)
\(4x^5 + 5 = 133\) or \(4x^5 = 128\) or \(x^5 = 32\) or \(\sqrt[5]{32}\)	A1	oe both correct terms differentiated and simplified correctly and equated to 133
2	A1

Additional guidance

\(\dfrac{14x^6 + 30x}{3}\)

Zero

June 2017 Paper 2 Q16

AQAOld spec3 marksIndices

(a) \(a = 3^{2b}\)

Circle the correct expression for \(\dfrac{1}{a}\) [1 mark]

\(3^{2b - 1}\)
\(3^{-2b}\)
\(-3^{2b}\)
\(\left(\dfrac{1}{3}\right)^{-2b}\)

(b) \(y = 5^x\)

Circle the correct expression for \(25y\) [1 mark]

\(5^{x+2}\)
\(25^x\)
\(5^{2x}\)
\(125^x\)

Circle the correct expression for \(w^3\) [1 mark]

\(8^{3m}\)
\(6^m\)
\(2^{m+3}\)
\(2^{3m}\)

June 2017 Paper 1 Q13

AQAOld spec3 marksIndices

13 Solve \(\qquad x^{-\frac{1}{4}} = 0.2\) [3 marks]

Mark scheme

Answer	Mark	Comments
Handling the negative power first ... \(\dfrac{1}{x^{\frac{1}{4}}} = 0.2\) or \(\dfrac{1}{x^{\frac{1}{4}}} = \dfrac{1}{5}\) or \(\dfrac{1}{0.2} = x^{\frac{1}{4}}\) or \(5 = x^{\frac{1}{4}}\) or \(\dfrac{1}{\sqrt[4]{x}} = 0.2\) or \(\sqrt[4]{x} = \dfrac{1}{0.2}\)	M1	Handling the 4th root first ... \(x^{-1} = 0.2^4\) or \(x = 0.2^{-4}\)
All of these are valid 2nd steps following any of the above 1st steps \(\dfrac{1}{x} = 0.2^4\) or \(\dfrac{1}{x} = \dfrac{1}{5^4}\) or \(x = 5^4\) or \(x = \dfrac{1}{0.2^4}\) or \(x = \left(\dfrac{1}{0.2}\right)^4\)	M1
(\(x\) =) 625	A1

Additional guidance

The two method marks are for handling the negative power and for handling the 4th root ... and an error in one of the 1st steps does not mean that you cannot give credit for a correct 2nd step

eg \(-\sqrt[4]{x} = 0.2\) (incorrect) followed by \(x = 0.2^4 = 0.0016\) scores M0 M1

\(-\sqrt[4]{x} = 0.2\) (incorrect) followed by \(-x = 0.2^4 = 0.0016\), \(x = -0.0016\) is M0 M0 because the 4th power of \(-\sqrt[4]{x}\) would be a positive quantity

\(\dfrac{1}{0.0016}\) scores M1 M1 A0 0.0016 or \(\dfrac{1}{625}\) will score M1 M0 A0 unless other valid working seen

June 2017 Paper 2 Q4

AQAOld spec2 marksIndices

4 Solve \(\qquad \dfrac{56}{\sqrt[3]{x}} = 4\) [2 marks]

Mark scheme

Answer	Mark	Comments
\(\left(\dfrac{56}{4}\right)^3\) or \(14^3\) or \(4^3x = 56^3\) or \(64x = 175\,616\) or \(\dfrac{56^3}{x} = 4^3\)	M1	oe oe equation in \(x^{(1)}\) or \(\dfrac{1}{x^{(1)}}\)
2744	A1

Additional guidance

\(\sqrt[3]{x} = \dfrac{56}{4}\) or \(\sqrt[3]{x} = 14\) with no correct further work	M0
\(56x^{-\frac{1}{3}} = 4\)	M0
Solving \(\dfrac{56}{3x} = 4\)	M0
Answer \(14^3\) with 2744 not seen in working	M1A0
Embedded solution	M1A0

June 2016 Paper 2 Q15

AQAOld spec4 marksIndices Solving Basic Equations

(a) Solve \(\quad \dfrac{2}{5}\sqrt{x} = 1\) [2 marks]

(b) Solve \(\quad x^3 = 5x^2\) [2 marks]

Mark scheme (a)

Answer	Mark	Comments
\(\sqrt{x} = \dfrac{5}{2}\) or \(\sqrt{x} = 2.5\) or \(\sqrt{x} = \dfrac{1}{0.4}\) or \(\dfrac{4}{25}x = 1\) or \(4x = 25\)	M1	oe Must have \(\sqrt{x} = \ldots\) or have eliminated \(\sqrt{\phantom{x}}\) with no errors
\(\dfrac{25}{4}\) or 6.25	A1	oe

Additional guidance

\(2\sqrt{x} = 5\) (no further correct work)	M0
Allow unprocessed values for M1 eg \(\dfrac{4}{25}x = 1^2\) or \(\left(\dfrac{2}{5}\right)^2x = 1\)	M1
\(\sqrt{x}\) may be seen as \(x^{\frac{1}{2}}\) or \(x^{0.5}\)

Mark scheme (b)

Answer	Mark	Comments
\(x^2(x - 5)\ (= 0)\) or \(x^2(5 - x)\ (= 0)\) or \((x =)\ 0\) or \((x =)\ 5\)	M1	oe factorisation eg1 \((x^2 - 0)(x - 5)\) eg2 \(x(x^2 - 5x)\)
0 and 5 with no other solutions	A1

Additional guidance

For A1, the word ‘and’ is not needed If their answer has both 0 and 5 with no other solutions award M1 A1 eg 0, 5 with no other solutions	M1 A1
0, 5, \(-5\)	M1 A0
Either or both solutions seen embedded	M1 A0

June 2016 Paper 1 Q10

AQAOld spec4 marksIndices

10 \(\sqrt[4]{x} = 2 \ \) and \(\ y^{-2} = 25\)

\(x \gt 0\) and \(y \lt 0\)

Work out the value of \(\dfrac{x}{y}\) [4 marks]

Mark scheme

Answer	Mark	Comments
\(x = 2^4\) or \(x = 16\)	B1
\(\dfrac{1}{y^2} = 25\) or \(y^2 = \dfrac{1}{25}\) or \(25y^2 = 1\) or \(y = \sqrt{\dfrac{1}{25}}\) or \(y = \left(\dfrac{1}{25}\right)^{\frac{1}{2}}\)	M1
(\(y\) =) \(-\dfrac{1}{5}\) or \(\pm\dfrac{1}{5}\) or \(\dfrac{1}{5}\)	A1	oe eg 0.2
\(-80\)	A1

Additional guidance

Condone \(4^2\) for the 1st M1

80 seen is likely to be 3 marks ... but check

(\(y\) =) \(-\dfrac{1}{5}\) or \(\pm\dfrac{1}{5}\) or \(\dfrac{1}{5}\) with no incorrect working seen implies M1 A1

(\(y\) =) \(-\dfrac{1}{5}\) or \(\pm\dfrac{1}{5}\) or \(\dfrac{1}{5}\) from clearly incorrect working scores M0 A0

June 2016 Paper 1 Q8

AQAOld spec3 marksIndices Solving Basic Equations

8 Solve \(\qquad (3 - \sqrt{x})^{\frac{1}{3}} = -2\) [3 marks]

Mark scheme

Answer	Mark	Comments
\((-2)^3\) or \(-8\) seen	B1
\(-\sqrt{x} =\) (their \(-8\)) \(- 3\) or \(-\sqrt{x} = -11\) or \(\sqrt{x} = 11\)	M1
121	A1

Additional guidance

\(-2^3\) (no brackets) is B0 unless \(-8\) seen

For M1 it must say \(\sqrt{x} = \ldots\ldots\) or \(-\sqrt{x} = \ldots\ldots\) Note: ... (their \(-8\)) cannot be \(-2\)

... and it must be correct manipulation from their \(-8\)

eg \(\quad 3 - \sqrt{x} = (-2)^3\) or \(3 - \sqrt{x} = -8 \qquad\) B1
\(\sqrt{x} = -11 \qquad\) M0 (error in manipulating terms)
\(x = 121 \qquad\) A0 (correct answer from wrong working)

June 2016 Paper 2 Q7

AQAOld spec2 marksIndices

7 Simplify fully \(\quad \left(\dfrac{2}{3}x^3y\right)^3\) [2 marks]

Mark scheme

Answer	Mark	Comments
\(\dfrac{8}{27}x^9y^3\) or \(\dfrac{8x^9y^3}{27}\)	B2	oe B1 Two of the three components correct and simplified

Additional guidance

Allow multiplication signs for B2 and B1
Allow \(0.\dot{2}9\dot{6}\) or \(0.\dot{2}\dot{9}\dot{6}\) as a correct component
\(0.296x^9y^3\)	B1
\(\dfrac{8}{27}x^9y^3\) followed by incorrect further work (only penalise B2 responses)	B1
\(8x^9y^3 \div 27\)	B1
\(\left(\dfrac{2}{3}\right)^3x^9y^3\)	B1
\(\dfrac{8}{27}x^9\)	B1
\(8x^9 \times 27y^3\)	B1
\(\dfrac{8}{27}x^9 + y^3\)	B0