Exponents and surds

• $a^m\times a^n=a^{m+n}$; $a^m÷a^n=a^{m-n}$; $(a^m{)}^n=a^{mn}$
• $a^0=1$; $a^{-n}=\frac{1}{a^n}$
• $a^{\frac{m}{n}}=\sqrt[n]{a^m}={\left(\sqrt[n]{a}\right)}^m$

Work with prime bases: $16=2^4$, $27=3^3$.

Example. $16^{\frac{3}{4}}÷27^{\frac{2}{3}}=2^3÷3^2=\frac{8}{9}$

With variables in the exponent.

$\frac{9^{n+1}\cdot 3^{n-2}}{27^n}=\frac{3^{2n+2}\cdot 3^{n-2}}{3^{3n}}=3^0=1$

A surd is a root that does not simplify to a rational number, such as $\sqrt{2}$ or $\sqrt[3]{5}$.

Take out the largest perfect square factor:

$\sqrt{75}=\sqrt{25\times 3}=5\sqrt{3}$

$\sqrt{72}=\sqrt{36\times 2}=6\sqrt{2}$

Remember: $\sqrt{ab}=\sqrt{a}\sqrt{b}$, but $\sqrt{a+b}\ne \sqrt{a}+\sqrt{b}$.

Add and subtract like surds only. Simplify first to make them alike:

$\sqrt{75}+\sqrt{27}-\sqrt{12}=5\sqrt{3}+3\sqrt{3}-2\sqrt{3}=6\sqrt{3}$

Multiply term by term, like algebraic brackets:

$(3\sqrt{2}-1)(3\sqrt{2}+1)=18-1=17$

This is a difference of squares, and it removes the surd. That is the idea behind rationalising.

Divide with $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$: $\frac{\sqrt{48}}{\sqrt{3}}=\sqrt{16}=4$.

To remove a surd from the denominator:

• One surd: multiply top and bottom by that surd.

$\frac{12}{\sqrt{3}}=\frac{12\sqrt{3}}{3}=4\sqrt{3}$

• Two terms: multiply top and bottom by the conjugate (change the sign between the terms).

$\frac{6}{\sqrt{5}-\sqrt{2}}\times \frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{6(\sqrt{5}+\sqrt{2})}{5-2}=2(\sqrt{5}+\sqrt{2})$