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

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Algebraic Roots & Indices

What are the laws of 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	$x^{1} = x$
$a^{0} = 1$	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	$j^{- 1} = \frac{1}{j}$ $k^{- 3} = \frac{1}{k^{3}}$
${(\frac{a}{b})}^{- n} = {(\frac{b}{a})}^{n} = \frac{b^{n}}{a^{n}}$	A fraction to a negative power, is the reciprocal of the fraction, to the positive power	${(\frac{l}{m})}^{- 3} = {(\frac{m}{l})}^{3} = \frac{m^{3}}{l^{3}}$
$a^{\frac{1}{n}} = \sqrt[n]{a}$	The fractional power $\frac{1}{n}$ is the n^th root ( $\sqrt[n]{}$ )	$n^{\frac{1}{2}} = \sqrt[2]{n}$ $p^{\frac{1}{3}} = \sqrt[3]{p}$
$a^{- \frac{1}{n}} = {(a^{\frac{1}{n}})}^{- 1} = {(\sqrt[n]{a})}^{- 1} = \frac{1}{\sqrt[n]{a}}$	A negative, fractional power is one over a root	$q^{- \frac{1}{2}} = \frac{1}{\sqrt[2]{q}}$ $r^{- \frac{1}{3}} = \frac{1}{\sqrt[3]{r}}$
$a^{\frac{m}{n}} = a^{\frac{1}{n} \times m} = {(a^{\frac{1}{n}})}^{m} = {(\sqrt[n]{a})}^{m} = {(a^{m})}^{\frac{1}{n}} = \sqrt[n]{a^{m}}$	The fractional power $\frac{m}{n}$ is the n^th root all to the power m, ${(\sqrt[n]{})}^{m}$ , or the n^th root of the power m, $\sqrt[n]{{()}^{m}}$ (both are the same)	$s^{\frac{2}{3}} = {(s^{\frac{1}{3}})}^{2} = {(\sqrt[3]{s})}^{2}$ $s^{\frac{2}{3}} = {(s^{2})}^{\frac{1}{3}} = \sqrt[3]{s^{2}}$

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

How do I find an unknown inside a power?

A term may have a power involving an unknown
- E.g. $7^{4 x}$
If both sides of an equation have the same base number, then the powers must be equal
- E.g. If $4^{3 x} = 4^{9}$ then $3 x = 9$
- And $x = 3$
You may have to do some simplifying first to reach this point
- E.g. $3^{2 x} \times 3^{4} = 3^{18}$ simplifies to $3^{2 x + 4} = 3^{18}$
- Therefore $2 x + 4 = 18$
- And $x = 7$

Worked Example

(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

$x = - 3$

Worked Example

(a) Rewrite $\frac{1}{\sqrt[3]{x^{4}}}$ in the form $x^{n}$ where $n$ is a negative fraction.

Use $a^{\frac{1}{n}} = \sqrt[n]{a}$ to rewrite the cube-root as a power of $\frac{1}{3}$

$\frac{1}{{(x^{4})}^{\frac{1}{3}}}$

Use ${(a^{m})}^{n} = a^{m n}$ to simplify the denominator

$\frac{1}{x^{\frac{4}{3}}}$

Use $a^{- n} = \frac{1}{a^{n}}$ to rewrite as a term with a negative fraction as the power

$x^{- \frac{4}{3}}$

(b) Find the value of the constants $m$ and $a$ given that ${(a x^{6})}^{\frac{1}{m}} = 8 x^{3}$ .

Use ${(a b)}^{n} = a^{n} b^{n}$ to rewrite the left hand side
Remember to apply the power to both $a$ and $x^{6}$

$a^{\frac{1}{m}} \times x^{\frac{6}{m}} = 8 x^{3}$

Both sides of the equation have a constant part, $a^{\frac{1}{m}}$ and $8$
And both sides of the equation have a part in terms of $x$

The two sides of the equation are equal, so set the respective parts equal to one another

First,

$x^{\frac{6}{m}} = x^{3}$

The bases are the same, therefore the powers are equal

$\frac{6}{m} = 3$

Solve to find $m$

$m = 2$

Then set the constant parts of both sides equal to one another

$a^{\frac{1}{m}} = 8$

We now know that $m = 2$ , so substitute this in

$a^{\frac{1}{2}} = 8$

Use $a^{\frac{1}{n}} = \sqrt[n]{a}$ to rewrite as a square root

$\sqrt[2]{a} = 8$

Find $a$ by squaring both sides

$a = 64$

Last updated: 17 October 2024

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

Revision Note

Algebraic Roots & Indices

What are the laws of indices?

How do I find an unknown inside a power?

You've read 0 of your 5 free revision notes this week

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