Indices | AQA GCSE Further Maths Revision Notes 2020

Laws of Indices

What are the laws of indices?

There are lots of very important laws (or rules)
It is important that you know and can apply these
Understanding the explanations will help you remember them

Law	Description	Why
$a^{1} = a$	anything to the power 1 is itself	$6^{1} = 6$
$a^{m} \times a^{n} = a^{m + n}$	to multiply indices with the same base, add their powers	$4^{3} \times 4^{2} = (4 \times 4 \times 4) \times (4 \times 4) = 4^{5}$
$a^{m} \div a^{n} = \frac{a^{m}}{a^{n}} = a^{m - n}$	to divide indices with the same base, subtract their powers	$7^{5} \div 7^{3} = \frac{7 \times 7 \times 7 \times 7 \times 7}{7 \times 7} = 7^{3}$
${(a^{m})}^{n} = a^{m n}$	to raise indices to a new power, multiply their powers	${(14^{3})}^{2} = (14 \times 14 \times 14) \times (14 \times 14 \times 14) = 14^{6}$
$a^{0} = 1$	anything to the power 0 is 1	$8^{0} = 8^{2 - 2} = 8^{2} \div 8^{2} = \frac{8^{2}}{8^{2}} = 1$
$a^{- n} = \frac{1}{a^{n}}$	a negative power is "1 over" the positive power	$11^{- 3} = 11^{0 - 3} = 11^{0} \div 11^{3} = \frac{11^{0}}{11^{3}} = \frac{1}{11^{3}}$
$a^{\frac{1}{n}} = \sqrt[n]{a}$	a power of an nth is an nth root	${(5^{\frac{1}{2}})}^{2} = 5^{\frac{1}{2} \times 2} = 5^{1} = 5 so 5^{\frac{1}{2}} = \sqrt{5}$
$a^{\frac{m}{n}} = {(\sqrt[n]{a})}^{m} = \sqrt[n]{a^{m}}$	a fractional power of m over n means either - do the the nth root first, then raise it to the power m or - raise it to the power m, then take the nth root (depending on what's easier)	$9^{\frac{3}{2}} = 9^{\frac{1}{2} \times 3} = {(9^{\frac{1}{2}})}^{3} = {(\sqrt{9})}^{3} or 9^{\frac{3}{2}} = 9^{3 \times \frac{1}{2}} = {(9^{3})}^{\frac{1}{2}} = \sqrt{9^{3}}$
${(\frac{a}{b})}^{n} = \frac{a^{n}}{b^{n}}$	a power outside a fraction applies to both the numerator and the denominator	${(\frac{5}{6})}^{3} = \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^{3}}{6^{3}}$
${(\frac{a}{b})}^{- n} = {(\frac{b}{a})}^{n} = \frac{b^{n}}{a^{n}}$	flipping the fraction inside changes a negative power into a positive power	${(\frac{5}{6})}^{- 2} = \frac{1}{{(\frac{5}{6})}^{2}} = 1 \div {(\frac{5}{6})}^{2} = 1 \div \frac{5^{2}}{6^{2}} = 1 \times \frac{6^{2}}{5^{2}} = \frac{6^{2}}{5^{2}} = {(\frac{6}{5})}^{2}$

Changing the base of a term

Sometimes expressions involve different base values
You can use index laws to change the base of a term to simplify an expression involving terms with different bases
- For example $9^{4} = {(3^{2})}^{4} = 3^{2 \times 4} = 3^{8}$
- Using the above can then help with problems like $9^{4} \div 3^{7} = 3^{8} \div 3^{7} = 3^{8 - 7} = 3^{1} = 3$

Exam Tip

Index laws only work with terms that have the same base, so something like 2³ × 5² cannot be simplified using index laws

Worked example

Index laws example, A Level & AS Level Pure Maths Revision Notes

Solving Equations With Indices

How do I solve equations with indices using a change of base?

If two terms over the same base are equal, their their powers are equal
- If $a^{x} = a^{y}$ then $x = y$

- This is not true if there are other coefficients or terms in the way
  - e.g. $3 a^{x} = a^{y}$ or $a^{x} = a^{y} - 2$
- This is not true if the bases are different
  - e.g. $a^{x} = b^{y}$ ...
  - ...unless they can be changed into a common base
$4^{x} = 8^{y}$ have bases 4 and 8 which can be written as 2² and 2³ (over a common base of 2)

$\begin{array}{rcl} 4^{x} & = & 8^{y} \\ {(2^{2})}^{x} & = & {(2^{3})}^{y} \\ 2^{2 x} & = & 2^{3 y} so 2 x = 3 y \end{array}$

- Put brackets around your change-of-base to be able to use the index law ${(a^{n})}^{m} = a^{n m}$

How do I solve equations with rational powers of x?

Use $a^{\frac{m}{n}} = {(\sqrt[n]{a})}^{m} = \sqrt[n]{a^{m}}$ to convert a rational power into a root
- If $\sqrt[n]{x} = y$ then $x = y^{n}$
  - raise both sides to the power n
$x^{\frac{2}{5}} = 4$ becomes $\sqrt[5]{x^{2}} = 4$ which raises to the power 5 to give $x^{2} = 4^{5} = 1024$
- solutions are $x = \pm \sqrt{1024} = \pm 32$
- don't forget plus-or-minus rooting, $\pm \sqrt[n]{}$ , when n is even
Sometimes there is more than just x in a root
- treat roots like brackets
- $\sqrt[3]{x^{5} + 40} = 2$ cubes to $x^{5} + 40 = 2^{3} = 8$ , which rearranges to $x^{5} = - 32$
  - solution is $x = \sqrt[5]{- 32} = - 2$
  - you can root a negative if its an odd root (not even)
  - $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$ so $\sqrt[3]{x^{5} + 40} \neq \sqrt[3]{x^{5}} + \sqrt[3]{40}$

Exam Tip

In a non-calculator paper, choose which way to use $a^{\frac{m}{n}} = {(\sqrt[n]{a})}^{m} = \sqrt[n]{a^{m}}$ to make calculations easier
- $x^{\frac{3}{4}} = 8$ can be written $\sqrt[4]{x^{3}} = 8$ or ${(\sqrt[4]{x})}^{3} = 8$ but the second way is easier to solve (the first way requires you to work out 8⁴then cube-root it!)

Worked example

Solve the following equations

(a)

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

\begin{array}{rcl} 8^{x} & = & \frac{1}{4^{x}} \times \sqrt{2} \end{array}

Rewrite the right-hand-side of the equation using indices.

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

Change the bases so that all of the bases are the same using $2^{2} = 4$ and $2^{3} = 8$ .

${(2^{3})}^{x} = {(2^{2})}^{- x} \times 2^{\frac{1}{2}}$

Rewrite using the brackets law.

${2^{3}}^{x} = {2^{- 2}}^{x} \times 2^{\frac{1}{2}}$

Rewrite the right-hand side using the law of indices: $a^{x} \times a^{y} = a^{x + y}$

. ${2^{3}}^{x} = 2^{(- 2 x + \frac{1}{2})}$

Now that the bases are the same with no extra components the powers can be equated and solved.

$\begin{array}{rcl} 3 x & = & - 2 x + \frac{1}{2} \\ 5 x & = & \frac{1}{2} \\ 10 x & = & 1 \end{array}$

$x = \frac{1}{10}$

(b)

\begin{matrix}  \end{matrix}

3 x^{\frac{2}{3}} - 12 = 0

Isolate the $x$ term by adding 12 and dividing by 3.

$\begin{array}{rcl} 3 x^{\frac{2}{3}} & = & 12 \\ x^{\frac{2}{3}} & = & 4 \end{array}$

Rewrite the left-hand side by converting the rational power into a root.

${(\sqrt[3]{x})}^{2} = 4$

Square root both sides, don't forget there could be a positive or a negative answer.

$\sqrt[3]{x} = \pm \sqrt{4} = \pm 2$

Cube both sides, remember to cube both the positive or a negative solutions.

$x = {(\pm 2)}^{3} = \pm 8$

$x = \pm 8$

Indices (AQA GCSE Further Maths)

Revision Note

Laws of Indices

What are the laws of indices?

Changing the base of a term

Exam Tip

Worked example

Solving Equations With Indices

How do I solve equations with indices using a change of base?

How do I solve equations with rational powers of x?

Exam Tip

Worked example

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Author: Mark