Rational Exponents and Radicals

Start with the simplest case, an exponent of $\frac{1}{n}$. Whatever $5^{1/3}$ turns out to be, the power-of-a-power law should still hold, so ${\left(5^{1/3}\right)}^3=5^{\frac{1}{3}\cdot 3}=5^1=5$. A number whose cube is 5 is, by definition, the cube root of 5. The same argument works for every $n$: ${\left(a^{1/n}\right)}^n=a$, so $a^{1/n}=\sqrt[n]{a}$. Nothing new was invented here; the meaning was forced on us by the laws we already trust. When $n$ is even there are two real numbers whose $n$th power is $a$ (both $4$ and $-4$ square to 16), and $a^{1/n}$ always means the non-negative one, called the principal root. So $16^{1/2}=4$, never $-4$. A few to know on sight: $49^{1/2}=7$, $125^{1/3}=5$, $81^{1/4}=3$, $32^{1/5}=2$ and $1000^{1/3}=10$.

Now allow any numerator. Because $\frac{m}{n}=\frac{1}{n}\cdot m=m\cdot \frac{1}{n}$, the power-of-a-power law splits $a^{m/n}$ two ways: take the $n$th root and then raise to the power $m$, or raise to the power $m$ and then take the $n$th root. Both give the same value (for $a\ge 0$), so you can choose whichever is easier, and by hand the root first is almost always easier, as the figure shows with $64^{2/3}$. The same idea converts between radical form and exponent form in either direction: the index of the root becomes the denominator and the power becomes the numerator. A root that sits in a denominator turns into a negative exponent.

1. Negative exponent? Take the reciprocal first so the exponent becomes positive: $16^{-3/4}=\frac{1}{16^{3/4}}$ and ${\left(\frac{2}{3}\right)}^{-1/2}={\left(\frac{3}{2}\right)}^{1/2}$. The answer does not become negative.
2. Decimals? Rewrite decimal bases and decimal exponents as fractions: $0.008=\frac{8}{1000}=\frac{1}{125}$, $0.0016=\frac{16}{10000}$, and $32^{0.4}=32^{2/5}$.
3. Fraction base? The exponent applies to the numerator and the denominator separately: ${\left(\frac{27}{64}\right)}^{1/3}=\frac{\sqrt[3]{27}}{\sqrt[3]{64}}=\frac{3}{4}\text{.}$
4. Root first (the denominator of the exponent), so the numbers shrink: $\sqrt[4]{16}=2$. Knowing the perfect squares to $15^2$, the cubes to $10^3$ and the powers of 2 to $2^{10}$ makes this step instant.
5. Then the power (the numerator): $2^3=8$, so $16^{-3/4}=\frac{1}{8}$. Finish with a reasonableness check: a base above 1 with a negative exponent must give an answer between 0 and 1.

Rational exponents obey every law from Exponent Laws, so the only new skill is fraction arithmetic in the exponents. Multiplying powers of the same base adds the fractions (find a common denominator), dividing subtracts them, and a power of a power multiplies them. For example, ${\left(27x^{3/2}\right)}^{2/3}=27^{2/3}{x}^{\frac{3}{2}\cdot \frac{2}{3}}=9x\text{:}$ the coefficient gets the exponent too, and $\frac{3}{2}\cdot \frac{2}{3}=1$. To simplify a product of different radicals, convert every radical to exponent form, combine, then convert back if the question asks for radical form: $\sqrt{x}\cdot \sqrt[4]{x^3}=x^{2/4}\cdot x^{3/4}=x^{5/4}\text{,}$ and since $\frac{5}{4}=1+\frac{1}{4}$, that is $x\sqrt[4]{x}$. Radical form cannot do this directly, because  ​ and $\sqrt[4]{}$ are different roots, but exponents with a common denominator combine easily.