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Non-calculator 3 marks(a)
Simplify \(t^{ 7 } \times t^{ 6 }\)
(b)Simplify \(t^{ 12 } \div t^{ 4 }\)
(c)Simplify \((t^{ 2 })^{ 2 }\)
Show answer
(a) \(t^{ 13 }\)
(b) \(t^{ 8 }\)
(c) \(t^{ 4 }\)Multiplying powers of the same letter adds the indices.
\(t^{ 7 } \times t^{ 6 } = t^{ 7 + 6 } = t^{ 13 }\)
Dividing powers of the same letter subtracts the indices.
\(t^{ 12 } \div t^{ 4 } = t^{ 12 - 4 } = t^{ 8 }\)
A power of a power multiplies the indices.
\((t^{ 2 })^{ 2 } = t^{ 2 \times 2 } = t^{ 4 }\)
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Non-calculator 3 marks(a)
Write down the value of \(5^{0}\)
(b)Work out \(3^{ -2 }\)
Give your answer as a fraction.
(c)Write \(\frac{1}{ k^{ 3 } }\) as a power of \(k\)
Show answer
(a) \(1\)
(b) \(\frac{1}{ 9 }\)
(c) \(k^{ -3 }\)Any number other than 0, raised to the power 0, is 1.
\(5^{0} = 1\)
A negative index means one over the matching positive power.
\(3^{ -2 } = \frac{1}{ 3^{ 2 } } = \frac{1}{ 9 }\)
Moving a power from the bottom of a fraction to the top changes the sign of the index.
\(\frac{1}{ k^{ 3 } } = k^{ -3 }\)
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Non-calculator 4 marks(a)
Simplify fully \(4w^{ 2 } \times 2w^{ 3 }\)
(b)Simplify fully \(\frac{ 8w^{ 5 } }{ 4w^{ 2 } }\)
Show answer
(a) \(8w^{ 5 }\)
(b) \(2w^{ 3 }\)Multiply the numbers in front, then add the indices.
\(4 \times 2 = 8\) and \(w^{ 2 } \times w^{ 3 } = w^{ 5 }\)
\(4w^{ 2 } \times 2w^{ 3 } = 8w^{ 5 }\)
Divide the numbers in front, then subtract the indices.
\(8 \div 4 = 2\) and \(w^{ 5 } \div w^{ 2 } = w^{ 3 }\)
\(\frac{ 8w^{ 5 } }{ 4w^{ 2 } } = 2w^{ 3 }\)
- (a) M1numbers multiplied \(4 \times 2 = 8\) or indices added \(2 + 3 = 5\)
- (a) A1\(8w^{ 5 }\)
- (b) M1numbers divided \(8 \div 4 = 2\) or indices subtracted \(5 - 2 = 3\)
- (b) A1\(2w^{ 3 }\)
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Non-calculator 2 marks(a)
\(10^{ x } = \frac{1}{1000}\)
Calculate the value of \(x\).
(b)\(3^{ x } = 27\)
Calculate the value of \(x\).
Show answer
(a) \(-3\)
(b) \(3\)Count how many 10s multiply together to make 1000: it takes 3 of them, so \(10^{ 3 } = 1000\).
The number is a fraction with 1 on top, so the index is negative: a negative index means one over the positive power.
\(10^{ -3 } = \frac{1}{1000}\)
\(x = -3\)
Count how many 3s multiply together to make 27: it takes 3 of them, so \(3^{ 3 } = 27\).
The number is bigger than 1, so the index is positive.
\(3^{ 3 } = 27\)
\(x = 3\)
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Non-calculator 3 marks
Hannah and Oscar are each checking some index working.
(a)Hannah writes \((7^{ 8 })^{ 6 } = 7^{ 14 }\)
Is Hannah correct? Give a reason for your answer.
(b) Oscar wants to simplify \(7t^{ 4 } \times 5t^{ 2 }\) He says, “Add 7 and 5, then multiply 4 and 2.” His answer is \(12t^{ 8 }\) Write correct or not correct for each part of his method, and show the working that decides each one.Part of the method Correct or not correct Add 7 and 5 Multiply 4 and 2 Show answer
(a) No: a power of a power multiplies the indices, \(8 \times 6 = 48\), so \((7^{ 8 })^{ 6 } = 7^{ 48 }\)
(b) Add 7 and 5: not correct. Multiply 4 and 2: not correct.When a power is raised to a power, the indices are multiplied.
\(8 \times 6 = 48\), so \((7^{ 8 })^{ 6 } = 7^{ 48 }\)
Hannah wrote \(7^{ 14 }\), so the answer is no.
Judge each stage against the law: multiply the numbers in front, add the indices.
Numbers in front: the law gives \(7 \times 5 = 35\). Oscar chose to add, so this part is not correct.
Indices: the law gives \(t^{ 4 } \times t^{ 2 } = t^{ 6 }\), since \(4 + 2 = 6\). Oscar chose to multiply, so this part is not correct.
- (b) A1both parts judged: add 7 and 5 is not correct, multiply 4 and 2 is not correct
- (b) M1the law applied to each stage: \(7 \times 5 = 35\) and \(4 + 2 = 6\)
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Non-calculator 2 marks
Simplify fully \(\frac{ (p^{ 2 })^{ 4 } }{ p^{ 6 } }\)
Show answer
\(p^{ 2 }\)Deal with the bracket first. A power of a power multiplies the indices.
\((p^{ 2 })^{ 4 } = p^{ 2 \times 4 } = p^{ 8 }\)
Now divide, which subtracts the indices.
\(p^{ 8 } \div p^{ 6 } = p^{ 8 - 6 } = p^{ 2 }\)
- M1multiplying the indices in the bracket \(p^{ 8 }\)
- A1subtracting the indices \(p^{ 2 }\)
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Non-calculator 3 marks
Work out the value of \((2 \times 2^{ 3 }) \div (2^{ 2 } \times 2^{ 4 })\)
Give your answer as a decimal.
Show answer
\(0.25\)A number written on its own has index 1, so \(2 = 2^{ 1 }\).
With every index shown, the calculation is \((2^{ 1 } \times 2^{ 3 }) \div (2^{ 2 } \times 2^{ 4 })\)
Multiplying adds the indices and dividing subtracts them. On the top: \(1 + 3 = 4\). On the bottom: \(2 + 4 = 6\).
The top is divided by the bottom, so subtract: \(4 - 6 = -2\), which gives \(2^{ -2 }\).
A negative index means one over the positive power.
\(2^{ -2 } = \frac{1}{ 2^{ 2 } } = \frac{1}{ 4 } = 0.25\)
- M1unwritten index 1 used, \(2 = 2^{ 1 }\)
- M1indices combined on top and bottom, \(4\) and \(6\), then subtracted to give \(2^{ -2 }\)
- A1value \(0.25\)
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Non-calculator 5 marks
Give each answer as a single power of the prime number named in the part.
(a)Write \(16 \times 2^{ 7 } \times 2^{ 3 }\) as a single power of 2.
(b)Simplify \(32^{ 3 } \times 16^{ 4 }\)
Give your answer in the form \(2^{ m }\), where \(m\) is an integer.
Show answer
(a) \(2^{ 14 }\)
(b) \(2^{ 31 }\)16 is a power of 2, so write it that way: \(16 = 2^{ 4 }\).
\(2^{ 4 } \times 2^{ 7 } \times 2^{ 3 }\)
The bases are now all 2, so multiplying adds the indices and dividing subtracts them.
\(2^{ 4 + 7 } \times 2^{ 3 } = 2^{ 11 } \times 2^{ 3 } = 2^{ 11 + 3 } = 2^{ 14 }\)
The rules for indices need the same base, so write both bases as powers of 2.
\(32 = 2^{ 5 }\) and \(16 = 2^{ 4 }\)
\(32^{ 3 } = (2^{ 5 })^{ 3 } = 2^{ 5 \times 3 } = 2^{ 15 }\)
\(16^{ 4 } = (2^{ 4 })^{ 4 } = 2^{ 4 \times 4 } = 2^{ 16 }\)
For a product the indices are added.
\(2^{ 15 } \times 2^{ 16 } = 2^{ 15 + 16 } = 2^{ 31 }\)
- (a) M1\(16\) written as \(2^{ 4 }\)
- (a) A1indices combined, \(2^{ 14 }\)
- (b) M1both bases written as powers of 2, \(32 = 2^{ 5 }\) and \(16 = 2^{ 4 }\)
- (b) M1indices in each power multiplied, \(2^{ 15 }\) and \(2^{ 16 }\)
- (b) A1indices added, \(2^{ 31 }\)
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Non-calculator 2 marks
\(16t^{ 6 } = ( at^{ 3 } )^{ b }\)
where \(a\) and \(b\) are positive integers.
Work out the value of \(a\) and the value of \(b\).
Show answer
\(a = 4\) and \(b = 2\)Raising a power to a power multiplies the indices, so the power of \(t\) on the left is \(3 \times b\). Matching the powers of \(t\):
\(3b = 6\)
\(b = 6 \div 3 = 2\)
The number \(a\) is raised to the same power, so matching the numbers:
\(a^{ 2 } = 16\)
\(a = 4\) because \(4^{ 2 } = 16\)
- B1b = 2, from 3b = 6
- B1a = 4, from a^2 = 16
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Non-calculator 4 marks(a)
Simplify fully \(6c^{ 4 }d^{ 3 } \times 3c^{ 3 }d^{ 3 }\)
(b)Simplify fully \(\frac{ 36c^{ 8 }d^{ 7 } }{ 6c^{ 4 }d^{ 3 } }\)
Show answer
(a) \(18c^{ 7 }d^{ 6 }\)
(b) \(6c^{ 4 }d^{ 4 }\)Deal with the numbers and each letter separately.
\(6 \times 3 = 18\)
\(c^{ 4 } \times c^{ 3 } = c^{ 7 }\) and \(d^{ 3 } \times d^{ 3 } = d^{ 6 }\)
Putting them together: \(18c^{ 7 }d^{ 6 }\)
Divide the numbers, then subtract the indices letter by letter.
\(36 \div 6 = 6\)
\(c^{ 8 } \div c^{ 4 } = c^{ 4 }\) and \(d^{ 7 } \div d^{ 3 } = d^{ 4 }\)
Putting them together: \(6c^{ 4 }d^{ 4 }\)
- (a) M1two of: numbers \(18\), \(c^{ 7 }\), \(d^{ 6 }\)
- (a) A1\(18c^{ 7 }d^{ 6 }\)
- (b) M1two of: numbers \(6\), \(c^{ 4 }\), \(d^{ 4 }\)
- (b) A1\(6c^{ 4 }d^{ 4 }\)