Work out
(a) $5^3$ [1]
(b) $\sqrt{144} + \sqrt[3]{64}$ [1]
(c) $2^4 \times 3^2$ [1]
- (a) $125$B1
- (b) $16$B1
- (c) $144$B1
- Answer: (a) 125 (b) 16 (c) 144
- Examiner tip: $5^3$ means 5 × 5 × 5, not 5 × 3.
(a) Write $7^5 \times 7^3$ as a single power of 7 [1]
(b) Write $\dfrac{(3^4)^2}{3^5}$ as a single power of 3 [1]
(c) Simplify $\dfrac{x^7}{x^{-2}}$ [1]
- (a) $7^8$B1
- (b) $3^3$B1
- (c) $x^9$B1
- Answer: (a) $7^8$ (b) $3^3$ (c) $x^9$
- Examiner tip: Dividing powers means subtracting indices: 7 − (−2) = 9, not 5.
Write down the value of
(a) $8^0$ [1]
(b) $4^{-2}$ [1]
(c) $25^{\frac{1}{2}}$ [1]
- (a) $1$B1
- (b) $\frac{1}{16}$ (accept $0.0625$)B1
- (c) $5$B1
- Answer: (a) 1 (b) $\frac{1}{16}$ (c) 5
- Examiner tip: A negative index means a reciprocal, not a negative number: $4^{-2} = \frac{1}{16}$, not −16.
Work out the value of $\left(\dfrac{8}{27}\right)^{-\frac{2}{3}}$
Give your answer as a fraction. [2]
- for one correct step, e.g. $\left(\frac{27}{8}\right)^{\frac{2}{3}}$, or the cube root of $\frac{8}{27}$ or $\frac{27}{8}$ evaluated as $\frac{2}{3}$ or $\frac{3}{2}$, or $\left(\frac{4}{9}\right)^{-1}$ seenM1
- $\frac{9}{4}$ (accept $2\frac{1}{4}$)A1
- Answer: $\frac{9}{4}$
- Examiner tip: Deal with the three parts of the index separately: minus → flip, denominator 3 → cube root, numerator 2 → square.
(a) Solve $2^x = \dfrac{1}{32}$ [1]
(b) Solve $9^y = 27$ [2]
- (a) $x = -5$B1
- (b) for writing both sides as powers of 3, e.g. $3^{2y} = 3^3$, or $27 = 9^{\frac{3}{2}}$M1
- (b) $y = \frac{3}{2}$ (accept $1.5$)A1
- Answer: (a) −5 (b) $\frac{3}{2}$
- Examiner tip: Write both sides with the same base first – 9 and 27 are both powers of 3.
$a$ and $b$ are integers such that
$$2^a \times 4^b = 128 \qquad\text{and}\qquad \frac{3^a}{9^b} = \frac{1}{3}$$
Find the value of $a$ and the value of $b$. [4]
- for $2^{a + 2b} = 2^7$ or $a + 2b = 7$M1
- for $3^{a - 2b} = 3^{-1}$ or $a - 2b = -1$M1
- for a correct method to solve their two linear equations, e.g. adding to get $2a = 6$M1
- $a = 3$ and $b = 2$A1
- Answer: a = 3, b = 2
- Examiner tip: Rewrite 4 as $2^2$ and 9 as $3^2$ so that indices can be compared; check your values in both equations.
Show that $16^{\frac{3}{4}} \times 27^{-\frac{1}{3}} = \dfrac{8}{3}$ [2]
- for $16^{\frac{3}{4}} = 8$ or $27^{-\frac{1}{3}} = \frac{1}{3}$M1
- for both evaluated and $8 \times \frac{1}{3} = \frac{8}{3}$ shownA1
- Answer: $8 \times \frac{1}{3} = \frac{8}{3}$
- Examiner tip: In a 'show that' you must write the value of each part (8 and 1/3) – just restating 8/3 scores 0.
Here is a list of numbers.
$$16 \quad 27 \quad 32 \quad 36 \quad 64 \quad 81 \quad 100 \quad 125$$
From the list, write down
(a) the number that is both a square number and a cube number [1]
(b) the number that is a power of 2 but is not a square number [1]
(c) Write 125 as a power of 5 [1]
- (a) 64 caoB1
- (b) 32 caoB1
- (c) $5^3$ (do not accept $5 \times 5 \times 5$ or $3^5$)B1
- Answer: (a) 64 (b) 32 (c) $5^3$
- Examiner tip: In (b), 16 and 64 are powers of 2 too, but they are also square numbers ($4^2$ and $8^2$), so only 32 fits.
(a) Use your calculator to work out $$\frac{\sqrt{53.29} + 2.6^3}{4.1^2 - 3.8}$$ Write down all the figures on your calculator display. [2]
(b) Write your answer to part (a) correct to 3 significant figures. [1]
- (a) for 24.876 or 13.01 seen, or $\sqrt{53.29}=7.3$ and $2.6^3 = 17.576$ both seenM1
- (a) 1.912067640… (allow 1.91206764 or better)A1
- (b) 1.91 (ft their answer to (a) if more than 3 sf given)B1ft
- Answer: (a) 1.912067640… (b) 1.91
- Examiner tip: Put brackets round the whole numerator and the whole denominator; typing it straight in divides only part of the expression.
The table shows some powers of 2.
| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| $2^n$ | 2 | 4 | 8 | 16 | 32 | 64 | 128 | 256 | 512 | 1024 | 2048 | 4096 |
Use the table to work out
(a) $32 \times 128$ [2]
(b) $\sqrt[3]{512}$ [1]
(c) $2048 \div 64$ [1]
- (a) for $2^5 \times 2^7$ or $2^{5+7}$ or $2^{12}$M1
- (a) 4096 caoA1
- (b) 8 (from $512 = 2^9$, $2^{9 \div 3} = 2^3$)B1
- (c) 32 (accept $2^5$)B1
- Answer: (a) 4096 (b) 8 (c) 32
- Examiner tip: Convert each number to a power of 2 first, then add, subtract or divide the indices; finish by reading the answer back from the table.
Simplify fully
(a) $(2a^3)^4$ [2]
(b) $\dfrac{12x^5y^2}{4x^2y}$ [2]
- (a) for $16a^k$ or $ka^{12}$, $k \neq 0$B1
- (a) $16a^{12}$ fully correct (scores B2)B1
- (b) for two of the three parts $3$, $x^3$, $y$ correct in a productB1
- (b) $3x^3y$ fully correct (scores B2)B1
- Answer: (a) $16a^{12}$ (b) $3x^3y$
- Examiner tip: The power applies to the number as well: $2^4 = 16$, not $8$.
(a) Kai writes $$3^4 \times 3^2 = 9^6$$ Kai is wrong. Explain the mistake Kai has made and write down the correct answer as a power of 3. [1]
(b) Mia says $$2^3 + 2^3 = 2^6$$ Is Mia correct? You must show your working. [2]
- (a) explains that the bases should not be multiplied (the base stays as 3) and gives $3^6$C1
- (b) for evaluating both sides, e.g. $8 + 8 = 16$ and $2^6 = 64$; or for $2^3 + 2^3 = 2 \times 2^3 = 2^4$M1
- (b) No, with correct supporting values, e.g. $16 \neq 64$ or $2^4 \neq 2^6$C1
- Answer: (a) bases multiplied instead of kept; $3^6$ (b) No: $2^3+2^3 = 16 = 2^4$, but $2^6 = 64$
- Examiner tip: The index laws only work for multiplying and dividing powers of the same base, never for adding them.
(a) Find the two consecutive whole numbers that $\sqrt[3]{90}$ lies between. Show how you decide. [2]
(b) Without using a calculator, show that $\sqrt{51}$ is closer to 7 than it is to 8. [2]
- (a) for $4^3 = 64$ and $5^3 = 125$ seen (or one cube either side of 90)M1
- (a) 4 and 5A1
- (b) for $7.5^2 = 56.25$ seen (or compares 51 with the square of the midpoint 7.5), or for a square just above 51 that is below $7.5^2$, e.g. $7.2^2 = 51.84$M1
- (b) for a complete argument, e.g. $51 < 56.25$ so $\sqrt{51} < 7.5$, so it is closer to 7 (or $51 < 51.84$ so $\sqrt{51} < 7.2 < 7.5$). Comparing only $51-49=2$ with $64-51=13$ scores M0C0C1
- Answer: (a) 4 and 5 (b) $7.5^2 = 56.25 > 51$ so $\sqrt{51} < 7.5$
- Examiner tip: Comparing 51 with 49 and 64 is not enough in (b): you must test the halfway point 7.5.
(a) Write $\dfrac{4^n \times 8}{\sqrt{2}}$ as a single power of 2. [3]
(b) Simplify fully $\left(\dfrac{27x^6}{y^3}\right)^{-\frac{2}{3}}$ [3]
- (a) for $4^n = 2^{2n}$ or $8 = 2^3$M1
- (a) for $\sqrt{2} = 2^{\frac{1}{2}}$ used as the denominatorM1
- (a) $2^{2n + \frac{5}{2}}$ oe, e.g. $2^{\frac{4n+5}{2}}$A1
- (b) for inverting, e.g. $\left(\frac{y^3}{27x^6}\right)^{\frac{2}{3}}$, or for $27^{\frac{2}{3}} = 9$ seenM1
- (b) for $x^{-4}$ or $x^4$ in the denominator, or $y^2$ in the numeratorM1
- (b) $\dfrac{y^2}{9x^4}$ (accept $\frac{1}{9}x^{-4}y^2$)A1
- Answer: (a) $2^{2n+\frac{5}{2}}$ (b) $\dfrac{y^2}{9x^4}$
- Examiner tip: Apply the index to every part of the fraction: the 27, the $x^6$ and the $y^3$ all need the $-\frac{2}{3}$ power.
The diagram shows a cube.
The volume of the cube is $64x^{12}$ cm³, where $x > 0$.
(a) Show that the total surface area of the cube is $96x^8$ cm². [3]
(b) The total surface area of the cube is 1536 cm².
Work out the volume of the cube. [3]
- (a) for $\sqrt[3]{64x^{12}}$ or $(64x^{12})^{\frac{1}{3}}$M1
- (a) side length $4x^4$A1
- (a) for $6 \times (4x^4)^2 = 6 \times 16x^8 = 96x^8$ with all steps shownC1
- (b) for $96x^8 = 1536$ or $1536 \div 6$ $(= 256)$M1
- (b) for $x^8 = 16$ leading to $x^4 = 4$ (or $x = \sqrt{2}$), or side length $= 16$ (from $\sqrt{256}$ or $4x^4 = 16$)M1
- (b) 4096 (cm³)A1
- Answer: (a) side $= 4x^4$, area $= 6 \times 16x^8 = 96x^8$ (b) 4096 cm³
- Examiner tip: In (b) a quick route is $1536 \div 6 = 256$ per face, so the side is 16 cm and the volume is $16^3$.
Still losing marks on this topic?
Book a free introductory session and we'll find exactly which step is costing you.