Laws of Indices (Cambridge (CIE) IGCSE International Maths)

Revision Note

Author

Jamie Wood

Expertise

Maths

Laws of Indices

What are the laws of indices?

Index laws are rules you can use when doing operations with powers
- They work with both numbers and algebra

Law	Description	How it works
$a^{1} = a$	Anything to the power of 1 is itself	$6^{1} = 6$
$a^{0} = 1$	Anything to the power of 0 is 1	$8^{0} = 1$
$a^{m} \times a^{n} = a^{m + n}$	To multiply indices with the same base, add their powers	$4^{3} \times 4^{2} = (4 \times 4 \times 4) \times (4 \times 4) = 4^{5}$
$a^{m} \div a^{n} = \frac{a^{m}}{a^{n}} = a^{m - n}$	To divide indices with the same base, subtract their powers	$7^{5} \div 7^{2} = \frac{7 \times 7 \times 7 \times 7 \times 7}{7 \times 7} = 7^{3}$
${(a^{m})}^{n} = a^{m n}$	To raise indices to a new power, multiply their powers	${(14^{3})}^{2} = (14 \times 14 \times 14) \times (14 \times 14 \times 14) = 14^{6}$
${(a b)}^{n} = a^{n} b^{n}$	To raise a product to a power, apply the power to both numbers, and multiply	${(3 \times 4)}^{2} = 3^{2} \times 4^{2}$
${(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}$	To raise a fraction to a power, apply the power to both the numerator and denominator	${(\frac{3}{4})}^{2} = \frac{3^{2}}{4^{2}} = \frac{9}{16}$
$a^{- 1} = \frac{1}{a}$ $a^{- n} = \frac{1}{a^{n}}$	A negative power is the reciprocal	$6^{- 1} = \frac{1}{6}$ $11^{- 3} = \frac{1}{11^{3}}$

How do I deal with different bases?

Index laws only work with terms that have the same base
- $2^{3} \times 5^{2}$ cannot be simplified using index laws
Sometimes expressions involve different base values, but one is related to the other by a power
- e.g. $2^{5} \times 4^{3}$
You can use powers to rewrite one of the bases
- $2^{5} \times 4^{3} = 2^{5} \times {(2^{2})}^{3}$
- This can then be simplified more easily, as the two bases are now the same
- $2^{5} \times {(2^{2})}^{3} = 2^{5} \times 2^{6} = 2^{11}$

Worked Example

(a) Find the value of $x$ when $6^{10} \times 6^{x} = 6^{2}$

Using the law of indices $a^{m} \times a^{n} = a^{m + n}$ we can rewrite the left hand side

$6^{10} \times 6^{x} = 6^{10 + x}$

So the equation is now

$6^{10 + x} = 6^{2}$

Comparing both sides, the bases are the same, so we can say that

$10 + x = 2$

Subtract 10 from both sides

$x = - 8$

(b) Find the value of $n$ when $5^{n} \div 5^{4} = 5^{6}$

Using the law of indices $a^{m} \div a^{n} = a^{m - n}$ we can rewrite the left hand side

$5^{n} \div 5^{4} = 5^{n - 4}$

So the equation is now

$5^{n - 4} = 5^{6}$

Comparing both sides, the bases are the same, so we can say that

$n - 4 = 6$

Add 4 to both sides

$n = 10$

Using the law of indices $a^{- n} = \frac{1}{a^{n}}$ we can rewrite the expression

$2^{- 4} = \frac{1}{2^{4}}$

$2^{4} = 2 \times 2 \times 2 \times 2 = 16$ so we can rewrite the expression

$\frac{1}{2^{4}} = \frac{1}{16}$

$\frac{1}{16}$

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