Algebraic Roots & 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	$a^{1} = a$
$a^{0} = 1$	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	${(f \times g)}^{2} = f^{2} \times g^{2}$
${(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}$	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)

\begin{matrix}  \end{matrix}

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

x

must equal the power on the right-hand side

x = - 3

Algebraic Roots & Indices (CIE IGCSE Maths: Core)