What could \(x^{1/2}\) mean? We'd like the laws of exponents to keep working when the exponent is a fraction. If they do, then
\[\left(16^{\frac{1}{2}}\right)^2 = 16^{\frac{1}{2} \cdot 2} = 16^1 = 16\]
So \(16^{\frac{1}{2}}\) is a number whose square is \(16\) :\(\sqrt{16} = 4\) .\(\frac{1}{2}\) means a square root, \(\frac{1}{3}\) means a cube root, and so on.
DefinitionRational exponents
For \(a > 0\) ,\(m\) and a whole number \(n \ge 2\) :
\[a^{\frac{1}{n}} = \sqrt[n]{a} \qquad\qquad a^{\frac{m}{n}} = \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m\]
The denominator is the index of the root; the numerator is the power.
Taking the root first usually keeps the numbers small.
ExampleEvaluate four powers
\[27^{\frac{2}{3}} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9\]
\[81^{\frac{3}{4}} = \left(\sqrt[4]{81}\right)^3 = 3^3 = 27\]
\[8^{-\frac{2}{3}} = \frac{1}{8^{\frac{2}{3}}} = \frac{1}{\left(\sqrt[3]{8}\right)^2} = \frac{1}{4}\]
A negative exponent still means "take the reciprocal", exactly as with integer exponents.
Why the base has to be positive With a negative base, equal fractions could give different answers:
\[(-1)^{\frac{1}{3}} = \sqrt[3]{-1} = -1 \qquad \text{but} \qquad (-1)^{\frac{2}{6}} = \sqrt[6]{(-1)^2} = \sqrt[6]{1} = 1\]
Since \(\frac{1}{3} = \frac{2}{6}\) ,positive bases .
The laws of exponents still work RuleLaws of exponents (for \(a, b > 0\) and rational \(r, s\) )
\[a^r \cdot a^s = a^{r + s} \qquad\qquad \frac{a^r}{a^s} = a^{r - s} \qquad\qquad \left(a^r\right)^s = a^{rs}\]
\[(ab)^r = a^r b^r \qquad\qquad a^{-r} = \frac{1}{a^r}\]
This is the payoff: rewriting radicals as powers turns radical problems into fraction arithmetic .
ExampleA quotient of radicals with different indices
\[\frac{\sqrt{x}}{\sqrt[3]{x}} = \frac{x^{\frac{1}{2}}}{x^{\frac{1}{3}}} = x^{\frac{1}{2} - \frac{1}{3}} = x^{\frac{1}{6}} = \sqrt[6]{x}\]
ExampleSimplify \(x^{\frac{1}{2}} \cdot x^{\frac{1}{4}}\) and \(\left(x^{\frac{3}{2}}\right)^{\frac{2}{3}}\)
\[x^{\frac{1}{2}} \cdot x^{\frac{1}{4}} = x^{\frac{2}{4} + \frac{1}{4}} = x^{\frac{3}{4}} \qquad\qquad \left(x^{\frac{3}{2}}\right)^{\frac{2}{3}} = x^{1} = x\]
Watch outAdd or multiply?
When you multiply powers with the same base, add the exponents: \(x^{\frac{1}{2}} \cdot x^{\frac{1}{4}} = x^{\frac{3}{4}}\) , not \(x^{\frac{1}{8}}\) .power of a power .
Practice Practice
Simplify \(\frac{x^{\frac{2}{3}} \cdot x^{\frac{5}{6}}}{x^{\frac{1}{2}}}\) , with \(x > 0\) .
Show Hide the solutionUse sixths:
\[x^{\frac{4}{6} + \frac{5}{6} - \frac{3}{6}} = x^{\frac{6}{6}} = x\]
Practice
Write \(\sqrt[3]{x^2\sqrt{x}}\) as a single power of \(x\) ,\(x > 0\) .
Show Hide the solution\[\left(x^2 \cdot x^{\frac{1}{2}}\right)^{\frac{1}{3}} = \left(x^{\frac{5}{2}}\right)^{\frac{1}{3}} = x^{\frac{5}{6}}\]
Quick check
What is \(125^{\frac{1}{3}}\) ?
A \(41.\overline{6}\) B \(5\) C \(25\) D \(\frac{1}{125}\)
An exponent of \(\frac{1}{3}\) means a cube root, and \(5^3 = 125\) .
Quick check
What is \(16^{-\frac{3}{4}}\) ?
A \(-8\) B \(8\) C \(\frac{1}{8}\) D \(-\frac{1}{8}\)
\(16^{\frac{3}{4}} = \left(\sqrt[4]{16}\right)^3 = 2^3 = 8\) , and the negative exponent gives the reciprocal.