GCSEFurther MathsAQAFlashcards2. Algebra & FunctionsIndices

Indices (AQA GCSE Further Maths): Flashcards

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Complete the three index laws:
$a^{m} \times a^{n} = a^{______}$
$a^{m} \div a^{n} = a^{______}$
${(a^{m})}^{n} = a^{______}$

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Cards in this collection (11)

Complete the three index laws:
$a^{m} \times a^{n} = a^{______}$
$a^{m} \div a^{n} = a^{______}$
${(a^{m})}^{n} = a^{______}$
The completed laws are:
$a^{m} \times a^{n} = a^{m + n}$
$a^{m} \div a^{n} = a^{m - n}$
${(a^{m})}^{n} = a^{m n}$
Multiplying adds the powers, dividing subtracts them, and a power of a power multiplies them.
Why is $a^{0} = 1$ for any non-zero $a$ ?
Because $a^{0} = a^{2 - 2} = a^{2} \div a^{2}$ , and anything non-zero divided by itself is 1.
The non-zero condition matters, because $0^{0}$ is not defined.
What does a negative index mean?
A negative index means one over the positive power, so $a^{- n} = \frac{1}{a^{n}}$ .
Applied to a fraction it flips it instead: ${(\frac{a}{b})}^{- n} = {(\frac{b}{a})}^{n}$ .
Complete the meaning of a fractional index, filling in the missing power and root:
$a^{\frac{m}{n}} = {(\sqrt[n]{a})}^{______} = \sqrt[______]{a^{m}}$
The completed rule is:
$a^{\frac{m}{n}} = {(\sqrt[n]{a})}^{m} = \sqrt[n]{a^{m}}$
The denominator is the root and the numerator is the power. Either order works, so take the root first when that keeps the numbers smaller.
True or False?
$2^{3} \times 5^{2}$ can be simplified using the index laws.
False.
The index laws only apply to terms with the same base, and here the bases are 2 and 5.
There is nothing to do but work each part out separately, giving $8 \times 25 = 200$ .
How can you simplify $9^{4} \div 3^{7}$ ?
Change to a common base, using $9^{4} = {(3^{2})}^{4} = 3^{8}$ .
Then $3^{8} \div 3^{7} = 3^{1} = 3$ .
Given $y = \frac{1}{27} x^{3}$ , express $\frac{1}{9} y^{- 2}$ in the form $k x^{n}$ .
Substituting gives $\frac{1}{9} {(\frac{1}{27} x^{3})}^{- 2} = \frac{1}{9} {(\frac{1}{27})}^{- 2} {(x^{3})}^{- 2}$ .
The negative power flips the fraction, so ${(\frac{1}{27})}^{- 2} = 27^{2} = 729$ , giving $\frac{729}{9} x^{- 6} = 81 x^{- 6}$ .
If $a^{x} = a^{y}$ , what can you say about $x$ and $y$ ?
The powers must be equal, so $x = y$ , provided the base is positive and not 1.
This only works with nothing else in the way: it fails for $3 a^{x} = a^{y}$ or for $a^{x} = a^{y} - 2$ .
How do you solve $4^{x} = 8^{y}$ for a relationship between $x$ and $y$ ?
Write both sides over base 2, giving ${(2^{2})}^{x} = {(2^{3})}^{y}$ and so $2^{2 x} = 2^{3 y}$ .
Equating the powers then gives $2 x = 3 y$ .
Solve $x^{\frac{2}{3}} = 4$ .
Rewrite the left side as ${(\sqrt[3]{x})}^{2} = 4$ , so $\sqrt[3]{x} = \pm 2$ .
Cubing both sides then gives $x = \pm 8$ .
Solve $\sqrt[3]{x^{5} + 40} = 2$ .
Cubing both sides gives $x^{5} + 40 = 8$ , so $x^{5} = - 32$ .
Taking the fifth root gives $x = - 2$ , which is allowed because the root is odd.

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