Big Fact: Radicals are really just exponents! Whoahhhhh!
Proof: Consider two equations $(\sqrt{x})^2=x$.
Using rules of exponents (gulp!), we now simplify $(x^{\frac{1}{2}})^2$
Conclusion?
Conclusion? $$x^{\frac{1}{2}}=\sqrt{x}$$
Duhn! Duhn! Duuuuhhhun!
Definition: Rational Exponents.
$$x^{\frac{m}{n}}=\sqrt[n]{x^m}$$
Equivalently,
$$x^{\frac{m}{n}}=\left(\sqrt[n]{x}\right)^m$$
Examples
$16^{\frac{3}{4}}$
$36^{-\frac{3}{2}}$
$(0.49)^{-\frac{3}{2}}$
$\left(\frac{125}{216}\right)^{\frac{2}{3}}$
Recall Your Rules Exponents: All the familiar (I hope) rules of exponents hold for rational exponents.
The Product Rule: $x^m \cdot x^n=x^{m+n}$
The Quotient Rule: $\frac{x^m}{x^n}=x^{m-n}$
The Power Rule: $(x^m)^n=x^{m \cdot n}$
Power of a Product: $(xy)^n=x^n y^n$
Power of a Quotient: $\left(\frac{x}{y}\right)^n=\frac{x^n}{y^n}$
Upstairs/Downstairs 1: $\frac{1}{x^{-n}}=x^{n}$
Upstairs/Downstairs 2: $x^{-n}=\frac{1}{x^n}$
Fraction Flip: $\left(\frac{x}{y}\right)^{-n}=\left(\frac{y}{x}\right)^{n}$
A Veritable, Venerable Variety of Examples
$\lambda ^{\frac{5}{2}}\lambda ^{\frac{3}{4}}$
$(2^{\frac{1}{2}} - 11^{\frac{1}{2}})^2$
$\left(\frac{16 \zeta ^{\frac{5}{2}}}{625 \zeta ^{\frac{3}{5}}}\right)^{\frac{3}{4}}$
$\frac{\left(125 \eta ^{\frac{3}{2}} j ^{\frac{1}{4}}\right) ^{\frac{1}{2}}\left(5 \eta ^{\frac{3}{5}} j ^{\frac{1}{5}}\right) ^{\frac{1}{2}}}{\left(\eta ^{-\frac{3}{5}} j ^{-\frac{1}{4}}\right) ^{-\frac{5}{3}}}$