2.2.1 Algebraic Roots & Indices | CIE IGCSE Maths: Extended Revision Notes 2023

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:

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

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

In more complicated questions you might have to use negative and fractional 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)

\begin{matrix} ^{} \end{matrix}

Simplify

\frac{(3 x^{2}) (2 x^{3} y^{2})}{(6 x^{2} y)}

Multiply out the brackets in the numerator.
Rearrange the numerator so that you are multiplying the numbers together, the $x$ terms together and the $y$ terms together.

$\frac{3 \times 2 \times x^{2} \times x^{3} \times y^{2}}{6 x^{2} y}$

Simplify the numerator.
Multiply the constants together and add the powers of the $x$ terms together.

$\frac{6 x^{5} y^{2}}{6 x^{2} y}$

Divide the constants.
Subtract the power of the $x$ term in the denominator from the $x$ term in the numerator: $x^{5 - 2} = x^{3}$ .
Subtract the power of the $y$ term in the denominator from the $y$ term in the numerator: $y^{2 - 1} = y^{1}$ .

$x^{3} y$

(b)

\begin{matrix}  \end{matrix}

Simplify

{(\frac{54 x^{7}}{2 x^{4}})}^{- \frac{1}{3}}

Simplify the expression inside the brackets.
Cancel down the constants.
Subtract the power of the $x$ term in the denominator from the $x$ term in the numerator: $x^{7 - 4} = x^{3}$ .

${(27 x^{3})}^{- \frac{1}{3}}$

Apply the negative index outside the brackets by 'flipping' the fraction inside the brackets.

${(\frac{1}{27 x^{3}})}^{\frac{1}{3}}$

Apply the fractional index outside the brackets to everything inside the brackets.

$\frac{1^{\frac{1}{3}}}{27^{\frac{1}{3}} x^{3 \times \frac{1}{3}}}$

Simplify.

$\frac{1}{3 x}$

Algebraic Roots & Indices (CIE IGCSE Maths: Extended)

Revision Note

Algebraic Roots & Indices

Can I use the laws of indices with algebra?

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

Worked example

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