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Non-calculator 1 mark
Write down the value of \(12^{0}\)
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\(1\)Any number except 0, raised to the power 0, equals 1.
\(12^{0} = 1\)
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Non-calculator 2 marks(a)
Write down the value of \(7^{-1 }\)
(b)Find the value of \(4^{-2 }\)
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(a) \(\frac{1}{ 7 }\)
(b) \(\frac{1}{ 16 }\)A negative index means one over the positive power.
\(7^{-1 } = \frac{1}{ 7^{ 1 } } = \frac{1}{ 7 }\)
A negative index means one over the positive power.
\(4^{-2 } = \frac{1}{ 4^{ 2 } } = \frac{1}{ 16 }\)
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Non-calculator 2 marks(a)
\(a^{ 6 } \times a^n = a^{ 15 }\)
Work out the value of \(n\).
(b)\(a^{ 8 } \div a^n = a^{ 3 }\)
Work out the value of \(n\).
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(a) \(9\)
(b) \(5\)Multiplying powers of the same letter adds the indices.
\(6 + n = 15\), so \(n = 15 - 6 = 9\)
Dividing powers of the same letter subtracts the indices.
\(8 - n = 3\), so \(n = 8 - 3 = 5\)
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Non-calculator 2 marks
Express \(\frac{ 7^{ 5 } \times 7^{ 7 } }{ 7^{ 14 } }\) as a single power of 7, in the form \(7^n\) where \(n\) is an integer.
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\(7^{ -2 }\)The bases are all 7, so use the index laws. Do not work out the numbers.
Multiplying adds the indices:
\(7^{ 5 } \times 7^{ 7 } = 7^{ 5 + 7 } = 7^{ 12 }\)
Dividing subtracts the index of the divisor:
\(7^{ 12 } \div 7^{ 14 } = 7^{ 12 - 14 } = 7^{ -2 }\)
- M1adds the indices of the product, \(5 + 7 = 12\), or subtracts the divisor's index 14
- A1\(7^{ -2 }\)
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Non-calculator 4 marks(a)
\(n^{\frac{1}{2}} = 8\)
Find the value of \(n\).
(b)\(m^{\frac{1}{3}} = 6\)
Find the value of \(m\).
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(a) \(64\)
(b) \(216\)A power of \(\frac{1}{2}\) means the square root, so the square root of \(n\) is 8.
Squaring both sides undoes it: \(n = 8^2 = 64\)
A power of \(\frac{1}{3}\) means the cube root, so the cube root of \(m\) is 6.
Cubing both sides undoes it: \(m = 6^3 = 216\)
- (a) M1both sides squared: \(8^2\)
- (a) A1\(n = 64\)
- (b) M1both sides cubed: \(6^3\)
- (b) A1\(m = 216\)
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Non-calculator 1 mark
Hafsa says that \(\frac{1}{ 2 }\) of 36 is 18, so \(36^{\frac{1}{ 2 }} = 18\).
Hafsa is wrong. Work out the correct value of \(36^{\frac{1}{ 2 }}\).
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\(6\)An index of \(\frac{1}{ 2 }\) means "take the square root", not "divide by 2".
Hafsa worked out \(\frac{ 36 }{ 2 } = 18\), which is a different calculation from raising 36 to the power \(\frac{1}{ 2 }\).
The square root of 36 is the number that gives 36 when raised to the power 2: \(6^{ 2 } = 36\).
So \(36^{\frac{1}{ 2 }} = 6\), not 18.
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Non-calculator 5 marks(a)
Find the value of \(16^{\frac{1}{ 4 } }\)
(b)Find the value of \(256^{\frac{ 3 }{ 4 } }\)
(c)Find the value of \(25^{\frac{3}{2} }\)
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(a) \(2\)
(b) \(64\)
(c) \(125\)The bottom of the index is the root, so this is the fourth root of 16.
\(2^{ 4 } = 16\), so \(16^{\frac{1}{ 4 } } = 2\)
The bottom of the index is the root and the top is the power. Take the root first — the numbers stay smaller that way.
\(4^{ 4 } = 256\), so the fourth root of 256 is 4.
Now raise that to the power 3: \(4^{ 3 } = 64\)
\(256^{\frac{ 3 }{ 4 } } = 64\)
An index of \(\frac{3}{2}\) means square root, then cube.
\(\sqrt{ 25 } = 5\), and \(5^{3} = 125\)
\(25^{\frac{3}{2} } = 125\)
- (b) M1take the fourth root of 256, which is 4
- (b) A1\(4^{ 3 } = 64\)
- (c) M1\(\sqrt{ 25 } = 5\)
- (c) A1\(5^{3} = 125\)
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Non-calculator 5 marks(a)
Write \(\sqrt{ p }\) as a power of \(p\)
(b)Write \(\frac{ 8 }{ p^{ 5 } }\) in the form \(ap^n\)
(c)Write \(p\sqrt{ p }\) as a power of \(p\)
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(a) \(p^{\frac{1}{2} }\)
(b) \(8p^{-5 }\)
(c) \(p^{\frac{3}{2} }\)A square root is an index of \(\frac{1}{2}\).
\(\sqrt{ p } = p^{\frac{1}{2} }\)
Only the power moves. Taking \(p^{ 5 }\) from the bottom of the fraction to the top changes the sign of its index, and the 8 stays in front.
\(\frac{ 8 }{ p^{ 5 } } = 8 \times \frac{1}{ p^{ 5 } } = 8p^{-5 }\)
\(p\) is \(p^{1}\), and \(\sqrt{ p }\) is \(p^{\frac{1}{2} }\).
Multiplying powers of the same letter adds the indices:
\(p^{1} \times p^{\frac{1}{2} } = p^{1 + \frac{1}{2} } = p^{\frac{3}{2} }\)
- (b) M1power moved from the bottom to the top with the sign of its index changed, or \(8 \times \frac{1}{ p^{ 5 } }\)
- (b) A1\(8p^{-5 }\)
- (c) M1\(\sqrt{ p }\) written as \(p^{\frac{1}{2} }\), with \(p\) as \(p^{1}\)
- (c) A1indices added to give \(p^{\frac{3}{2} }\)
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Non-calculator 3 marks
Write \(4 \div (32^{ n })^{3}\) as a power of 2.
Give your answer in terms of \(n\).
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\(2^{ 2 - 15n }\)Neither base is 2, so write each number as a power of 2 first.
\(4 = 2^{ 2 }\) and \(32 = 2^{ 5 }\)
The letter term becomes
\((32^{ n })^{3} = ((2^{ 5 })^{ n })^{3}\)
A power of a power multiplies the indices, so this is \(2^{ 15n }\).
Dividing powers of the same base subtracts the indices:
\(2^{ 2 } \div 2^{ 15n } = 2^{ 2 - 15n }\)
- M1\(4 = 2^{ 2 }\)
- M1\((32^{ n })^{3}\) as a power of 2, index \(15n\)
- A1combined index \(2 - 15n\)
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Non-calculator 4 marks(a)
Work out \(9^{-\frac{1}{2} }\)
Give your answer as a fraction.
(b)Work out \(27^{-\frac{ 2 }{ 3 } }\)
Give your answer as a fraction.
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(a) \(\frac{1}{ 3 }\)
(b) \(\frac{1}{ 9 }\)A negative index means one over the matching positive power.
\(9^{-\frac{1}{2} } = \frac{1}{ 9^{\frac{1}{2} } }\)
\(9^{\frac{1}{2} }\) is the square root of 9, which is 3.
\(9^{-\frac{1}{2} } = \frac{1}{ 3 }\)
Deal with the minus sign first: a negative index means one over the matching positive power.
\(27^{-\frac{ 2 }{ 3 } } = \frac{1}{ 27^{\frac{ 2 }{ 3 } } }\)
The cube root of 27 is 3, and \(3^{ 2 } = 9\).
\(27^{-\frac{ 2 }{ 3 } } = \frac{1}{ 9 }\)
- (a) M1negative index as the reciprocal \(\frac{1}{ 9^{\frac{1}{2} } }\)
- (a) A1square root 3, so \(\frac{1}{ 3 }\)
- (b) M1negative index as the reciprocal \(\frac{1}{ 27^{\frac{ 2 }{ 3 } } }\)
- (b) A1\(\frac{1}{ 9 }\)