Algebraic Roots & Indices

Can I use the laws of indices with algebra?

Laws of indices work with numerical and algebraic terms
These can be used to simplify expressions where terms are multiplied or divided
- Deal with the number and algebraic parts separately
  - $(3 x^{7}) \times (6 x^{4}) = (3 \times 6) \times (x^{7} \times x^{4}) = 18 x^{11}$
  - $\frac{3 x^{7}}{6 x^{4}} = \frac{3}{6} \times \frac{x^{7}}{x^{4}} = \frac{1}{2} x^{3}$
  - ${(3 x^{7})}^{2} = {(3)}^{2} \times {(x^{7})}^{2} = 9 x^{14}$
The index laws you need to know and use are summarised here:
$\begin{array}{rcl} a^{m} \times a^{n} & = & a^{m + n} \\ a^{m} \div a^{n} & = & \frac{a^{m}}{a^{n}} = a^{m - n} \\ {(a^{m})}^{n} & = & a^{m n} \\ {(a b)}^{n} & = & a^{n} b^{n} \\ a^{1} & = & a \\ a^{0} & = & 1 \\ a^{- n} & = & \frac{1}{a^{n}} \end{array}$

How can I solve equations when the unknown is in the index?

If two powers (bigger than 1) are equal and the base numbers are the same then the indices must be the same
- If $a^{x} = a^{y}$ then $x = y$
If the unknown is part of the index then write both sides with the same base number
- Then you can ignore the base number and make the indices equal and solve that equation

$\begin{array}{rcl} 5^{x + 1} & = & 125 \\ 5^{x + 1} & = & 5^{3} \\ x + 1 & = & 3 \\ x & = & 2 \end{array}$

In more complicated questions you might have to use negative indices
- You may also have to rewrite both sides with the same base number

$\begin{array}{rcl} 8^{x} & = & \frac{1}{4} \\ {(2^{3})}^{x} & = & \frac{1}{2^{2}} \\ 2^{3 x} & = & 2^{- 2} \\ 3 x & = & - 2 \\ x & = & - \frac{2}{3} \end{array}$

Worked example

(a)

Simplify

{(u^{5})}^{5}

Using the law of indices

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

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

u^{25}

(b)

\frac{}{}

Show clearly how

{(\frac{x}{5})}^{- 2}

can be rewritten as

\frac{25}{x^{2}}

Using the law of indices

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

we can rewrite the expression

{(\frac{x}{5})}^{- 2} = {({(\frac{x}{5})}^{- 1})}^{2}

A power of -1, means the reciprocal (1 divided by the number)

{({(\frac{x}{5})}^{- 1})}^{2} = {(\frac{5}{x})}^{2}

The power of 2 is applied to both the top and bottom of the fraction

{(\frac{5}{x})}^{2} = \frac{5^{2}}{x^{2}}

5 squared is 25

\frac{25}{x^{2}}

Algebraic Roots & Indices (CIE IGCSE Maths: Core)

Revision Note