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Non-calculator 1 mark
Write down the value of \(12^{0}\)
Show answer
\(1\)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 \(4^{-2 }\)
Show answer
(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 \(n\).
Show answer
(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.
Show answer
\(7^{ -2 }\)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:
\(7^{ 12 } \div 7^{ 14 } = 7^{ 12 - 14 } = 7^{ -2 }\)
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Non-calculator 3 marks
Write \(64 \div (4^{ m })^{3}\) as a power of 2.
Give your answer in terms of \(m\).
Show answer
\(2^{ 6 - 6m }\)Neither base is 2, so write each number as a power of 2 first.
\(64 = 2^{ 6 }\) and \(4 = 2^{ 2 }\)
The letter term becomes
\((4^{ m })^{3} = ((2^{ 2 })^{ m })^{3}\)
A power of a power multiplies the indices, so this is \(2^{ 6m }\).
Dividing powers of the same base subtracts the indices:
\(2^{ 6 } \div 2^{ 6m } = 2^{ 6 - 6m }\)
- M1\(64 = 2^{ 6 }\)
- M1\((4^{ m })^{3}\) as a power of 2, index \(6m\)
- A1combined index \(6 - 6m\)
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Non-calculator 1 mark
Write down the value of \(40^{0}\)
Show answer
\(1\)Any number except 0, raised to the power 0, equals 1.
\(40^{0} = 1\)
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Non-calculator 2 marks(a)
Write down the value of \(7^{-1 }\)
(b)Find the value of \(2^{-2 }\)
Show answer
(a) \(\frac{1}{ 7 }\)
(b) \(\frac{1}{ 4 }\)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.
\(2^{-2 } = \frac{1}{ 2^{ 2 } } = \frac{1}{ 4 }\)
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Non-calculator 2 marks(a)
\(p^{ 9 } \times p^n = p^{ 17 }\)
Find the value of \(n\).
(b)\(p^{ 14 } \div p^n = p^{ 12 }\)
Find the value of \(n\).
Show answer
(a) \(8\)
(b) \(2\)Multiplying powers of the same letter adds the indices.
\(9 + n = 17\), so \(n = 17 - 9 = 8\)
Dividing powers of the same letter subtracts the indices.
\(14 - n = 12\), so \(n = 14 - 12 = 2\)
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Non-calculator 2 marks
Express \(\frac{ 2^{ 7 } \times 2^{ 8 } }{ 2^{ 11 } }\) as a single power of 2, in the form \(2^n\) where \(n\) is an integer.
Show answer
\(2^{ 4 }\)The bases are all 2, so use the index laws. Do not work out the numbers.
Multiplying adds the indices:
\(2^{ 7 } \times 2^{ 8 } = 2^{ 7 + 8 } = 2^{ 15 }\)
Dividing subtracts the index of the divisor:
\(2^{ 15 } \div 2^{ 11 } = 2^{ 15 - 11 } = 2^{ 4 }\)
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Non-calculator 3 marks
Write \(8 \times (4^{ m })^{2}\) as a power of 2.
Give your answer in terms of \(m\).
Show answer
\(2^{ 3 + 4m }\)Neither base is 2, so write each number as a power of 2 first.
\(8 = 2^{ 3 }\) and \(4 = 2^{ 2 }\)
The letter term becomes
\((4^{ m })^{2} = ((2^{ 2 })^{ m })^{2}\)
A power of a power multiplies the indices, so this is \(2^{ 4m }\).
Multiplying powers of the same base adds the indices:
\(2^{ 3 } \times 2^{ 4m } = 2^{ 3 + 4m }\)
- M1\(8 = 2^{ 3 }\)
- M1\((4^{ m })^{2}\) as a power of 2, index \(4m\)
- A1combined index \(3 + 4m\)