Algebraic Roots & Indices

Index laws are rules you can use when doing operations with powers
- They work with both numbers and algebra
The index laws you need to know are:
$\begin{array}{rcl} a^{m} \times a^{n} & = & a^{m + n} \\ a^{m} \div a^{n} & = & \frac{a^{m}}{a^{n}} = a^{m - n} \\ {(a^{m})}^{n} & = & a^{m n} \\ {(a b)}^{n} & = & a^{n} b^{n} \\ a^{1} & = & a \\ a^{0} & = & 1 \\ a^{- n} & = & \frac{1}{a^{n}} \end{array}$
For example
$\begin{array}{rcl} p^{2} \times p^{5} & = & p^{7} \\ \frac{x^{6}}{x^{2}} & = & x^{4} \\ {(w^{10})}^{3} & = & w^{30} \\ {(c d)}^{8} & = & c^{8} d^{8} \\ q^{- 3} & = & \frac{1}{q^{3}} \end{array}$
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 (Edexcel GCSE Maths: Foundation)