Year 9AC9M9A01

Index Laws

Apply the exponent laws to numerical expressions with integer exponents and extend to variables.

There is no getting away from learning and applying these rules to the last detail.

The rules

Law 1 — Multiplying powers

a^{m} \times a^{n} = a^{m+n}

Same base? Keep the base and add the exponents.

Law 2 — Dividing powers

\frac{a^{m}}{a^{n}} = a^{m-n}\ \ (m>n) \qquad \frac{a^{m}}{a^{n}} = \frac{1}{a^{n-m}}\ \ (m<n)

Same base? Keep the base and subtract the exponents. Bigger exponent minus smaller.

Law 3 — Power of a power

\left(a^{m}b^{m}\right)^{n} = a^{mn}b^{mn} \qquad \left(\frac{a^{m}}{b^{m}}\right)^{n} = \frac{a^{mn}}{b^{mn}}

A power raised to another exponent? Multiply the exponents.

Law 4 — Zero exponent

a^{0} = 1

Any non-zero number raised to the power of 0 is 1.

Law 5 — Negative exponents

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

A negative exponent flips the term to the other side of the fraction bar.

Useful shortcut

\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n}

Flip the fraction, drop the minus.

e.g.

\frac{5^{2x} \times 10^{4x}}{2^{2} \cdot 25^{x+1}}

Change all bases into prime numbers $\frac{5^{2x} \times (2 \times 5)^{4x}}{(2)^{2}\left(5^{2}\right)^{x+1}}$
Multiply out to lose the brackets $= \frac{5^{2x} \times 2^{4x} \times 5^{4x}}{2^{2} \cdot 5^{2x+2}}$
Simplify top and bottom $= \frac{2^{4x} \times 5^{6x}}{2^{2} \times 5^{2x+2}}$
Cancel like terms $= 2^{4x-2} \times 5^{6x-(2x+2)}$
Simplify if possible $= 2^{4x-2} \times 5^{4x-2}$

\frac{x^{-2}y^{4}}{3x^{4}y^{-4}}

Answer

\frac{y^{8}}{3x^{6}}

\frac{xy^{3}}{2yx^{-1}}

Answer

\frac{x^{2}y^{2}}{2}

\frac{x^{0}y}{3x^{2}y}

Answer

\frac{1}{3x^{2}}

\frac{2x^{-2}}{4xy^{0}}

Answer

\frac{1}{2x^{3}}

\frac{2xy^{3}}{y^{2} \times 2x^{-1}y}

Answer

x^{2}

\left(\frac{x^{6}}{3x^{2}y^{3} \times 2y^{4}}\right)^{-2}

Answer

\frac{36y^{14}}{x^{8}}

\frac{x^{2}y^{2} \times \left(x^{2}y^{2}\right)^{2}}{2xy^{5}}

Answer

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

\frac{2x^{4}y^{-2} \times \left(y^{-3}\right)}{\left(y^{4}\right)^{0}}

Answer

\frac{2x^{4}}{y^{5}}

\left(\frac{xy^{2} \times 2y^{-1}x^{4}}{y^{-3}x^{4}}\right)^{-4}

Answer

\frac{1}{16x^{4}y^{16}}

\left(\frac{2x^{4}y^{3} \times 2x^{-1}y^{3}}{2x}\right)^{0}

Answer

1