Laws of indices and surdsEdexcel International A Level Maths: Revision notes

An index (power) tells you how many times the base is multiplied by itself. For any base $a$ and rational $m,n$: $a^m\times a^n=a^{m+n},\qquad \frac{a^m}{a^n}=a^{m-n},\qquad (a^m)^n=a^{mn}.$ Also $(ab)^n=a^nb^n$, $a^0=1$ and $a^{-n}=\frac{1}{a^n}$. The laws only apply when the bases are the same, or when you factorise a product raised to a power. Example: $\dfrac{(3x^2)^3}{9x^4}=\dfrac{27x^6}{9x^4}=3x^2$. Raise the number and each power to the outside index, then divide the numbers and subtract the indices.

$a^0=1,\qquad a^{-n}=\frac1{a^n},\qquad \left(\frac ab\right)^{-n}=\left(\frac ba\right)^n.$ A negative index means a reciprocal, not a negative number. So $2^{-3}=\frac18$ and $\left(\frac23\right)^{-2}=\frac94$. Negative indices let you write fractions as powers: $\frac{5}{x^3}=5x^{-3}$, but note $\frac{1}{5x^3}=\frac15x^{-3}$, because the index belongs to $x$ only.

Fractional indices are roots: $a^{\frac1n}=\sqrt[n]{a},\qquad a^{\frac mn}=\left(\sqrt[n]{a}\right)^m=\sqrt[n]{a^m},\qquad a^{-\frac mn}=\frac{1}{a^{\frac mn}}.$ Take the root first (it keeps the numbers small), then the power. Examples:

• $16^{\frac34}=(16^{\frac14})^3=2^3=8$
• $27^{-\frac23}=\frac{1}{(27^{\frac13})^2}=\frac19$
• $\left(\frac{9}{4}\right)^{-\frac12}=\left(\frac49\right)^{\frac12}=\frac23$ The same laws of indices hold for every rational index.

Write every root as a fractional power, then use the laws. For $x>0$: $\frac{(2x^3)^2\sqrt x}{8x^5}=\frac{4x^6\cdot x^{\frac12}}{8x^5}=\frac12x^{6+\frac12-5}=\frac12x^{\frac32}.$ To solve an equation, express both sides with the same base and equate the indices. For $9^x=27$: $3^{2x}=3^3$, so $2x=3$ and $x=\frac32$. To solve $x^{\frac32}=216$ raise both sides to the reciprocal power: $x=216^{\frac23}=6^2=36$.

A surd is an irrational root such as $\sqrt2$ or $\sqrt{75}$ which cannot be written as a fraction of integers. Rules for positive $a,b$: $\sqrt{ab}=\sqrt a\sqrt b,\qquad \sqrt{\frac ab}=\frac{\sqrt a}{\sqrt b},\qquad (\sqrt a)^2=a.$ Simplify by taking out the largest square factor: $\sqrt{75}=\sqrt{25\times3}=5\sqrt3$ and $\sqrt{72}=6\sqrt2$. Only like surds can be added or subtracted: $\sqrt{72}-\sqrt{18}=6\sqrt2-3\sqrt2=3\sqrt2$. Expand brackets as normal: $(3+\sqrt5)^2=9+6\sqrt5+5=14+6\sqrt5$. The answer must be exact, not a decimal.

To rationalise means to remove surds from the denominator.

• Single surd: multiply top and bottom by that surd, $\frac{6}{\sqrt3}=\frac{6\sqrt3}{3}=2\sqrt3$.
• Two terms: multiply top and bottom by the conjugate (change the sign between the terms). Then $(a+\sqrt b)(a-\sqrt b)=a^2-b$, which is rational. Example: $\frac{10}{\sqrt7-2}=\frac{10(\sqrt7+2)}{(\sqrt7-2)(\sqrt7+2)}=\frac{10\sqrt7+20}{7-4}=\frac{10\sqrt7+20}{3}.$ Always multiply the numerator by the same expression, expand carefully, and simplify the final fraction.