Laws of Indices (Cambridge (CIE) IGCSE Maths)

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Written by: Jamie Wood

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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	$6^{- 1} = \frac{1}{6}$ $11^{- 3} = \frac{1}{11^{3}}$
${(\frac{a}{b})}^{- n} = {(\frac{b}{a})}^{n} = \frac{b^{n}}{a^{n}}$	A fraction to a negative power, is the reciprocal of the fraction, to the positive power	${(\frac{2}{5})}^{- 3} = {(\frac{5}{2})}^{3} = \frac{5^{3}}{2^{3}} = \frac{125}{8}$
$a^{\frac{1}{n}} = \sqrt[n]{a}$	The fractional power $\frac{1}{n}$ is the n^th root ( $\sqrt[n]{}$ )	$25^{\frac{1}{2}} = \sqrt[2]{25} = 5$ $27^{\frac{1}{3}} = \sqrt[3]{27} = 3$
$a^{- \frac{1}{n}} = {(a^{\frac{1}{n}})}^{- 1} = {(\sqrt[n]{a})}^{- 1} = \frac{1}{\sqrt[n]{a}}$	A negative, fractional power is one over a root	$64^{- \frac{1}{2}} = \frac{1}{\sqrt[2]{64}} = \frac{1}{8}$ $125^{- \frac{1}{3}} = \frac{1}{\sqrt[3]{125}} = \frac{1}{5}$
$a^{\frac{m}{n}} = a^{\frac{1}{n} \times m} = {(a^{\frac{1}{n}})}^{m} = {(a^{m})}^{\frac{1}{n}}$	The fractional power $\frac{m}{n}$ is the n^th root all to the power m, ${(\sqrt[n]{})}^{m}$ , or the n^th root of the power m, $\sqrt[n]{{()}^{m}}$ (both are the same)	$8^{\frac{2}{3}} = {(8^{\frac{1}{3}})}^{2} = {(\sqrt[3]{8})}^{2} = 2^{2} = 4$ $8^{\frac{2}{3}} = {(8^{2})}^{\frac{1}{3}} = \sqrt[3]{64} = 4$

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

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

$\frac{1}{16}$

(d) Without using a calculator, find the value of $8^{- \frac{1}{3}}$

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

$8^{- \frac{1}{3}} = \frac{1}{8^{\frac{1}{3}}}$

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

$\frac{1}{8^{\frac{1}{3}}} = \frac{1}{\sqrt[3]{8}}$

The cube root of 8 is 2

$\frac{1}{2}$

(e) Without using a calculator, find the value of $81^{\frac{3}{4}}$ .

Use the law of indices ${(a^{m})}^{n} = a^{m n}$ we can rewrite the expression in two ways

$81^{\frac{3}{4}} = {(81^{3})}^{\frac{1}{4}}$ or ${(81^{\frac{1}{4}})}^{3}$

Both forms are equivalent, but ${(81^{3})}^{\frac{1}{4}}$ would require calculating 81 cubed, so use the second form instead

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

${(81^{\frac{1}{4}})}^{3} = {(\sqrt[4]{81})}^{3}$

The 4th root of 81 is 3 as 3×3×3×3=3⁴=81

${(3)}^{3}$

Lastly, calculate or recall 3 cubed

Last updated: 17 October 2024

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