If you work through this section you should be able to:

- Understand what is meant by a power or index.
- Evaluate expressions involving powers.
- Evaluate expressions using the rules of indices.

A power or an exponent of a number states how many times to multiply the number by itself. It is written as a small superscripted number on the top right of a number, for example 2^{5} = 2×2×2×2×2. The superscript 5 is called a power or index and the number 2 is called the base. It is pronounced “two raised to the power of five”.

Raising to the power of 2 is called squaring: “4 squared” is 4^{2} = 4 × 4 = 16;

and to the power of 3 is called cubing: “5 cubed” is 5^{3} = 5 × 5 × 5 = 125.

Note: Any number raised to the power of 1 is itself, e.g. 6^{1} = 6.

Any number raised to the power of 0 has the value to 1, e.g. 9^{0} = 1.

According to the BEDMAS rule, exponentiation is done after brackets and before multiplication.

So: 3 × 2^{3} = 3 × 8 = 24 — but (3 × 2)^{3} = 6^{3} = 216.

- −4
^{2}= −4 × 4 = −16 - (−4)
^{2}= (−4) × (−4) = 16 - 2(3 × 5
^{2}− 8) = 2(3 × 25 − 8)

= 2 (75 − 8)

= 134 - 3(6 + 2 × 3
^{2})^{2}= 3(6 + 2 × 9)^{2}

= 3(6 + 18)^{2}

= 3(24)^{2}

= 3 × 576

= 1728

Negative powers denote the reciprocal. The positive powers are calculated first then the reciprocal is taken.

So: ${3}^{-2}=\frac{1}{{3}^{2}}=\frac{1}{9}$

If ${x}^{a}=b$ then $x$ is the ${a}^{\mathrm{th}}$ root of $b$. For example, if 6^{2} = 36, then 6 is the second root of 36. The second root is usually called the square root. It is written as 6 = $\sqrt{36}$.

The third root is usually called the cube root. For example, 3 = $\sqrt[3]{27}$.

Note also that when −6 is squared we again obtain 36, that is (−6)^{2} = 36. This means that 36 has another square root, −6. Therefore, $\sqrt{36}$ = ±6

Fractional powers can be written as ${a}^{\frac{m}{n}}$. A fractional power can be broken into two parts: ${m}^{\mathrm{th}}$ power and ${n}^{\mathrm{th}}$ root. You can either do the power first then take the root, or alternatively take the root first and then do the power. Therefore:

${a}^{\frac{m}{n}}$ = $\sqrt[n]{{a}^{m}}$ = ${\sqrt[n]{a}}^{m}$

So: ${81}^{\frac{3}{4}}$ = ${\left(\sqrt[4]{81}\right)}^{3}$ = 3^{3} = 27

- $\sqrt[3]{64}=\sqrt[3]{{4}^{3}}=4$
- ${25}^{-\frac{1}{2}}=\frac{1}{\sqrt{25}}=\frac{1}{5}=0.2$
- ${8}^{\frac{2}{3}}={\left(\sqrt[3]{8}\right)}^{2}={2}^{2}=4$
- ${125}^{-\frac{2}{3}}=\frac{1}{{\left(\sqrt[3]{125}\right)}^{2}}=\frac{1}{{5}^{2}}=\frac{1}{25}=0.04$

The rules of indices can be used to manipulate powers related expressions. The rules of indices are given as below:

${a}^{n}$ x ${a}^{m}={a}^{n+m}$

${a}^{n}\xf7{a}^{m}={a}^{n-m}$

${\left({a}^{n}\right)}^{m}={a}^{nm}$

${a}^{-n}=\frac{1}{{a}^{n}}$

${a}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}=\sqrt[n]{a}$

${a}^{\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}={\left({a}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}\right)}^{n}={\left(\sqrt[m]{a}\right)}^{n}$

- 4
^{3}x 4^{5}= 4^{3+5}= 4^{8} - X
^{2}x X^{7}÷ X^{4}= X^{2+7-4}

= X^{5} - (3
^{5})^{2}= 3^{10} - s
^{3}÷ t^{−4}× (s^{−3}t^{−2})^{3}= s^{3}÷ t^{−4}× s^{−9}× t^{−6}

= s^{3−9}× t^{4−6}

= s^{−6}t^{−2} - ${8}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\xf78$ = ${8}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.-1}$

${8}^{-\frac{1}{3}}$

= $\frac{1}{\sqrt[3]{8}}$

= $\frac{1}{2}$

= 0.5

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