Algebraic Roots & Indices (Cambridge (CIE) IGCSE Maths)

Revision Note

Reviewed by: Dan Finlay

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	$x^{1} = x$
$a^{0} = 1$	Anything to the power of 0 is 1	$b^{0} = 1$
$a^{m} \times a^{n} = a^{m + n}$	To multiply indices with the same base, add their powers	$c^{3} \times c^{2} = (c \times c \times c) \times (c \times c) = c^{5}$
$a^{m} \div a^{n} = \frac{a^{m}}{a^{n}} = a^{m - n}$	To divide indices with the same base, subtract their powers	$d^{5} \div d^{2} = \frac{d \times d \times d \times d \times d}{d \times d} = d^{3}$
${(a^{m})}^{n} = a^{m n}$	To raise indices to a new power, multiply their powers	${(e^{3})}^{2} = (e \times e \times e) \times (e \times e \times e) = e^{6}$
${(a b)}^{n} = a^{n} b^{n}$	To raise a product to a power, apply the power to both numbers, and multiply	${(f \times g)}^{2} = f^{2} \times g^{2} = f^{2} g^{2}$
${(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}$	To raise a fraction to a power, apply the power to both the numerator and denominator	${(\frac{h}{i})}^{2} = \frac{h^{2}}{i^{2}}$
$a^{- 1} = \frac{1}{a}$ $a^{- n} = \frac{1}{a^{n}}$	A negative power is the reciprocal	$j^{- 1} = \frac{1}{j}$ $k^{- 3} = \frac{1}{k^{3}}$

These can be used to simplify expressions
- Work out the number and algebra parts separately
  - $(3 x^{7}) \times (6 x^{4}) = (3 \times 6) \times (x^{7} \times x^{4}) = 18 x^{7 + 4} = 18 x^{11}$
  - $\frac{6 x^{7}}{3 x^{4}} = \frac{6}{3} \times \frac{x^{7}}{x^{4}} = 2 x^{7 - 4} = 2 x^{3}$
  - ${(3 x^{7})}^{2} = {(3)}^{2} \times {(x^{7})}^{2} = 9 x^{14}$

Write both sides of the equation over the same base number
- Then work out what x should be
  $\begin{array}{rcl} 5^{x} & = & 125 \\ 5^{x} & = & 5^{3} \\ x & = & 3 \end{array}$

You might have to use negative indices
$\begin{array}{rcl} 2^{x} & = & \frac{1}{8} \\ 2^{x} & = & \frac{1}{2^{3}} \\ 2^{x} & = & 2^{- 3} \\ x & = & - 3 \end{array}$

(a) Simplify ${(u^{5})}^{5}$

Use ${(a^{m})}^{n} = a^{m n}$

${(u^{5})}^{5} = u^{5 \times 5}$

$u^{25}$

(b) If $q^{x} = \frac{q^{2} \times q^{5}}{q^{10}}$ find $x$ .

Use $a^{m} \times a^{n} = a^{m + n}$ to simplify the numerator

$q^{2} \times q^{5} = q^{2 + 5} = q^{7}$

Use $\frac{a^{m}}{a^{n}} = a^{m - n}$ to simplify the fraction

$\frac{q^{7}}{q^{10}} = q^{7 - 10} = q^{- 3}$

Write out both sides of the equation

$q^{x} = q^{- 3}$

Both sides are now over the same base of $q$

So $x$ must equal the power on the right-hand side

$x = - 3$

Last updated: 4 June 2024

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